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Chapter8 In everyday life, we see some objects at rest and others in motion. Birds fly, fish swim, blood flows through veins and arteries and cars move. Atoms, molecules, planets, stars and galaxies are all in motion. We often perceive an object to be in motion when its position changes with time. However, there are situations where the motion is inferred through indirect evidences. For example, we infer the motion of air by observing the movement of dust and the movement of leaves and branches of trees. What causes the phenomena of sunrise, sunset and changing of seasons? Is it due to the motion of the earth? If it is true, why don't we directly perceive the motion of the earth? An object may appear to be moving for one person and stationary for some other. For the passengers in a moving bus, the roadside trees appear to be moving backwards. A person standing on the road–side perceives the bus alongwith the passengers as moving. However, a passenger inside the bus sees his fellow passengers to be at rest. What do these observations indicate? Most motions are complex. Some objects may move in a straight line, others may take a circular path. Some may rotate and a few others may vibrate. There may be situations involving a combination of these. In this chapter, we shall first learn to describe the motion of objects along a straight line. We shall also learn to express such motions through simple equations and graphs. Later, we shall discuss ways of describing circular motion. MOTION Activity ______________ 8.1 • Discuss whether the walls of your classroom are at rest or in motion. Activity ______________ 8.2 • Have you ever experienced that the train in which you are sitting appears to move while it is at rest? • Discuss and share your experience. Think and Act We sometimes are endangered by the motion of objects around us, especially if that motion is erratic and uncontrolled as observed in a flooded river, a hurricane or a tsunami. On the other hand, controlled motion can be a service to human beings such as in the generation of hydro-electric power. Do you feel the necessity to study the erratic motion of some objects and learn to control them? 8.1 Describing Motion We describe the location of an object by specifying a reference point. Let us understand this by an example. Let us assume that a school in a village is 2 km north of the railway station. We have specified the position of the school with respect to the railway station. In this example, the railway station is the reference point. We could have also chosen other reference points according to our convenience. Therefore, to describe the position of an object we need to specify a reference point called the origin. 8.1.1 MOTION ALONG A STRAIGHT LINE The simplest type of motion is the motion along a straight line. We shall first learn to describe this by an example. Consider the motion of an object moving along a straight path. The object starts its journey from O which is treated as its reference point (Fig. 8.1). Let A, B and C represent the position of the object at different instants. At first, the object moves through C and B and reaches A. Then it moves back along the same path and reaches C through B. while the magnitude of displacement = 35 km. Thus, the magnitude of displacement (35 km) is not equal to the path length (85 km). Further, we will notice that the magnitude of the displacement for a course of motion may be zero but the corresponding distance covered is not zero. If we consider the object to travel back to O, the final position concides with the initial position, and therefore, the displacement is zero. However, the distance covered in this journey is OA + AO = 60 km + 60 km = 120 km. Thus, two different physical quantities — the distance and the Fig. 8.1: Positions of an object on a straight line path The total path length covered by the object is OA + AC, that is 60 km + 35 km = 95 km. This is the distance covered by the object. To describe distance we need to specify only the numerical value and not the direction of motion. There are certain quantities which are described by specifying only their numerical values. The numerical value of a physical quantity is its magnitude. From this example, can you find out the distance of the final position C of the object from the initial position O? This difference will give you the numerical value of the displacement of the object from O to C through A. The shortest distance measured from the initial to the final position of an object is known as the displacement. Can the magnitude of the displacement be equal to the distance travelled by an object? Consider the example given in (Fig. 8.1). For motion of the object from O to A, the distance covered is 60 km and the magnitude of displacement is also 60 km. During its motion from O to A and back to B, the distance covered = 60 km + 25 km = 85 km displacement, are used to describe the overall motion of an object and to locate its final position with reference to its initial position at a given time. Activity ______________ 8.3 • Take a metre scale and a long rope. • Walk from one corner of a basket-ball court to its oppposite corner along its sides. • Measure the distance covered by you and magnitude of the displacement. • What difference would you notice between the two in this case? Activity ______________ 8.4 • Automobiles are fitted with a device that shows the distance travelled. Such a device is known as an odometer. A car is driven from Bhubaneshwar to New Delhi. The difference between the final reading and the initial reading of the odometer is 1850 km. • Find the magnitude of the displacement between Bhubaneshwar and New Delhi by using the Road Map of India. Questions 1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example. 2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position? 3. Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object. 8.1.2 UNIFORM MOTION AND NONUNIFORM MOTION Consider an object moving along a straight line. Let it travel 5 m in the first second, 5 m more in the next second, 5 m in the third second and 5 m in the fourth second. In this case, the object covers 5 m in each second. As the object covers equal distances in equal intervals of time, it is said to be in uniform motion. The time interval in this motion should be small. In our day-to-day life, we come across motions where objects cover unequal distances in equal intervals of time, for example, when a car is moving on a crowded street or a person is jogging in a park. These are some instances of non-uniform motion. Activity ______________8.5 • The data regarding the motion of two different objects A and B are given in Table 8.1. • Examine them carefully and state whether the motion of the objects is uniform or non-uniform. 8.2 Measuring the Rate of Motion Look at the situations given in Fig. 8.2. If the bowling speed is 143 km h–1 in Fig. 8.2(a) what does it mean? What do you understand from the signboard in Fig. 8.2(b)? Different objects may take different amounts of time to cover a given distance. Some of them move fast and some move slowly. The rate at which objects move can be different. Also, different objects can move at the same rate. One of the ways of measuring the rate of motion of an object is to find out the distance travelled by the object in unit time. This quantity is referred to as speed. The SI unit of speed is metre per second. This is represented by the symbol m s–1 or m/s.The other units of speed include centimetre per second (cm s–1) and kilometre per hour (km h–1). To specify the speed of an object, we require only its magnitude. The speed of an object need not be constant. In most cases, objects will be in non-uniform motion. Therefore, we describe the rate of motion of such objects in terms of their average speed. The average speed of an object is obtained by dividing the total distance travelled by the total time taken. That is, Total distance travelled average speed = Total time taken If an object travels a distance s in time t then its speed v is, s v = (8.1)t Let us understand this by an example. A car travels a distance of 100 km in 2 h. Its average speed is 50 km h–1. The car might not have travelled at 50 km h–1 all the time. Sometimes it might have travelled faster and sometimes slower than this. Example 8.1 An object travels 16 m in 4 s and then another 16 m in 2 s. What is the average speed of the object? Solution: Total distance travelled by the object = 16 m + 16 m = 32 m Total time taken = 4 s + 2 s = 6 s Therefore, the average speed of the object is 5.33 m s–1. Total distance travelled Average speed = Total time taken 32 m = 6s = 5.33 m s–1 8.2.1 SPEED WITH DIRECTION The rate of motion of an object can be more comprehensive if we specify its direction of motion along with its speed. The quantity that specifies both these aspects is called velocity. Velocity is the speed of an object moving in a definite direction. The velocity of an object can be uniform or variable. It can be changed by changing the object's speed, direction of motion or both. When an object is moving along a straight line at a variable speed, we can express the magnitude of its rate of motion in terms of average velocity. It is calculated in the same way as we calculate average speed. In case the velocity of the object is changing at a uniform rate, then average velocity is given by the arithmetic mean of initial velocity and final velocity for a given period of time. That is, initialvelocity+ final velocity average velocity = 2 u+v Mathematically, vav = (8.2)2 where vav is the average velocity, u is the initial velocity and v is the final velocity of the object. Speed and velocity have the same units, that is, m s–1 or m/s. Activity ______________ 8.6 • Measure the time it takes you to walk from your house to your bus stop or the school. If you consider that your average walking speed is 4 km h–1, estimate the distance of the bus stop or school from your house. Activity ______________8.7 • At a time when it is cloudy, there may be frequent thunder and lightning. The sound of thunder takes some time to reach you after you see the lightning. • Can you answer why this happens? • Measure this time interval using a digital wrist watch or a stop watch. • Calculate the distance of the nearest point of lightning. (Speed of sound in air = 346 m s-1.) Questions 1. Distinguish between speed and velocity. 2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed? 3. What does the odometer of an automobile measure? 4. What does the path of an object look like when it is in uniform motion? 5. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 108 –1m s. Example 8.2 The odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of the trip. If the trip took 8 h, calculate the average speed of the car in km h–1 and m s–1. Solution: Distance covered by the car, s = 2400 km – 2000 km = 400 km Time elapsed, t = 8h Average speed of the car is, s 400 km v = av t 8h = 50km h–1 km 1000m 1h 50× × h 1km 3600s = = 13.9 m s–1 The average speed of the car is 50 km h–1 or 13.9 m s–1. Example 8.3 Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Usha. Solution: Total distance covered by Usha in 1 min is 180 m. Displacement of Usha in 1 min = 0 m Total distance covered Average speed = Totaltimetaken 180m 180 m 1 min =×= 1min 1min 60s = 3 m s-1 Displacement Average velocity = Totaltimetaken 0m = 60 s = 0 m s–1 The average speed of Usha is 3 m s–1 and her average velocity is 0 m s–1. 8.3 Rate of Change of Velocity During uniform motion of an object along a straight line, the velocity remains constant with time. In this case, the change in velocity of the object for any time interval is zero. However, in non-uniform motion, velocity varies with time. It has different values at different instants and at different points of the path. Thus, the change in velocity of the object during any time interval is not zero. Can we now express the change in velocity of an object? To answer such a question, we have to introduce another physical quantity called acceleration, which is a measure of the change in the velocity of an object per unit time. That is, change in velocity acceleration = time taken If the velocity of an object changes from an initial value u to the final value v in time t, the acceleration a is, v–u a= (8.3)t This kind of motion is known as accelerated motion. The acceleration is taken to be positive if it is in the direction of velocity and negative when it is opposite to the direction of velocity. The SI unit of acceleration is m s–2 . If an object travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time, then the acceleration of the object is said to be uniform. The motion of a freely falling body is an example of uniformly accelerated motion. On the other hand, an object can travel with non-uniform acceleration if its velocity changes at a non-uniform rate. For example, if a car travelling along a straight road increases its speed by unequal amounts in equal intervals of time, then the car is said to be moving with non-uniform acceleration. Activity ______________ 8.8 • In your everyday life you come across a range of motions in which (a) acceleration is in the direction of motion, (b) acceleration is against the direction of motion, (c) acceleration is uniform, (d) acceleration is non-uniform. • Can you identify one example each of the above type of motion? Example 8.4 Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m s–1 in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m s-1 in the next 5 s. Calculate the acceleration of the bicycle in both the cases. Solution: In the first case: initial velocity, u = 0 ; final velocity, v = 6 m s–1 ; time, t = 30 s . From Eq. (8.3), we have v–u a= t Substituting the given values of u,v and t in the above equation, we get –1 –16m s –0m s a = 30 s = 0.2 m s–2 In the second case: initial velocity, u = 6 m s–1; final velocity, v = 4 m s–1; time, t = 5 s. –1 –14m s –6m s Then, a = 5s = –0.4 m s–2 . The acceleration of the bicycle in the first case is 0.2 m s–2 and in the second case, it is –0.4 m s–2. Questions 1. When will you say a body is in (i) uniform acceleration? (ii) nonuniform acceleration? 2. A bus decreases its speed from 80 km h–1 to 60 km h–1 in 5 s. Find the acceleration of the bus. 3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h–1 in 10 minutes. Find its acceleration. 8.4 Graphical Representation of Motion Graphs provide a convenient method to present basic information about a variety of events. For example, in the telecast of a one-day cricket match, vertical bar graphs show the run rate of a team in each over. As you have studied in mathematics, a straight line graph helps in solving a linear equation having two variables. To describe the motion of an object, we can use line graphs. In this case, line graphs show dependence of one physical quantity, such as distance or velocity, on another quantity, such as time. 8.4.1 DISTANCE–TIME GRAPHS The change in the position of an object with time can be represented on the distance-time graph adopting a convenient scale of choice. In this graph, time is taken along the x–axis and distance is taken along the y-axis. Distance-time graphs can be employed under various conditions where objects move with uniform speed, non-uniform speed, remain at rest etc. We know that when an object travels equal distances in equal intervals of time, it moves with uniform speed. This shows that the distance travelled by the object is directly proportional to time taken. Thus, for uniform speed, a graph of distance travelled against time is a straight line, as shown in Fig. 8.3. The portion OB of the graph shows that the distance is increasing at a uniform rate. Note that, you can also use the term uniform velocity in place of uniform speed if you take the magnitude of displacement equal to the distance travelled by the object along the y-axis. We can use the distance-time graph to determine the speed of an object. To do so, consider a small part AB of the distance-time graph shown in Fig 8.3. Draw a line parallel to the x-axis from point A and another line parallel to the y-axis from point B. These two lines meet each other at point C to form a triangle ABC. Now, on the graph, AC denotes the time interval (t2 – t1) while BC corresponds to the distance (s– s). We can see from the 2 1graph that as the object moves from the point A to B, it covers a distance (s2 – s1) in time (t– t). The speed, v of the object, therefore 2 1can be represented as 2s 1 – s v = 2t 1 – t (8.4) We can also plot the distance-time graph for accelerated motion. Table 8.2 shows the The distance-time graph for the motion of the car is shown in Fig. 8.4. Note that the shape of this graph is different from the earlier distance-time graph (Fig. 8.3) for uniform motion. The nature of this graph shows nonlinear variation of the distance travelled by the car with time. Thus, the graph shown in Fig 8.4 represents motion with non-uniform speed. 8.4.2 VELOCITY-TIME GRAPHS The variation in velocity with time for an object moving in a straight line can be represented by a velocity-time graph. In this graph, time is represented along the x-axis and the velocity is represented along the y-axis. If the object moves at uniform velocity, the height of its velocity-time graph will not change with time (Fig. 8.5). It will be a straight line parallel to the x-axis. Fig. 8.5 shows the velocity-time graph for a car moving with uniform velocity of 40 km h–1. We know that the product of velocity and time give displacement of an object moving with uniform velocity. The area enclosed by velocity-time graph and the time axis will be equal to the magnitude of the displacement. To know the distance moved by the car between time tand tusing Fig. 8.5, draw12 perpendiculars from the points corresponding to the time t and t2 on the graph. The velocity1of 40 km h–1 is represented by the height AC or BD and the time (t2 – t1) is represented by the length AB. So, the distance s moved by the car in time (t– t) can be expressed as2 1s = AC × CD = [(40 km h–1) × (t2 – t1) h] = 40 (t2– t1) km = area of the rectangle ABDC (shaded in Fig. 8.5). We can also study about uniformly accelerated motion by plotting its velocity– time graph. Consider a car being driven along a straight road for testing its engine. Suppose a person sitting next to the driver records its velocity after every 5 seconds by noting the reading of the speedometer of the car. The velocity of the car, in km h–1 as well as in m s–1, at different instants of time is shown in table 8.3. In this case, the velocity-time graph for the motion of the car is shown in Fig. 8.6. The nature of the graph shows that velocity changes by equal amounts in equal intervals of time. Thus, for all uniformly accelerated motion, the velocity-time graph is a straight line. You can also determine the distance moved by the car from its velocity-time graph. The area under the velocity-time graph gives the distance (magnitude of displacement) moved by the car in a given interval of time. If the car would have been moving with uniform velocity, the distance travelled by it would be represented by the area ABCD under the graph (Fig. 8.6). Since the magnitude of the velocity of the car is changing due to acceleration, the distance s travelled by the car will be given by the area ABCDE under the velocity-time graph (Fig. 8.6). That is, s = area ABCDE = area of the rectangle ABCD + area of the triangle ADE 1 = AB × BC + (AD × DE)2 In the case of non-uniformly accelerated motion, velocity-time graphs can have any shape. Fig. 8.7: Velocity-time graphs of an object in non-uniformly accelerated motion. Fig. 8.7(a) shows a velocity-time graph that represents the motion of an object whose velocity is decreasing with time while Fig. 8.7 (b) shows the velocity-time graph representing the non-uniform variation of velocity of the object with time. Try to interpret these graphs. Activity ______________ 8.9 • The times of arrival and departure of a train at three stations A, B and C and the distance of stations B and C from station A are given in table 8.4. •Plot and interpret the distance-time graph for the train assuming that its motion between any two stations is uniform. Activity _____________ 8.10 • Feroz and his sister Sania go to school on their bicycles. Both of them start at the same time from their home but take different times to reach the school although they follow the same route. Table 8.5 shows the distance travelled by them in different times • Plot the distance-time graph for their motions on the same scale and interpret. Questions 1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object? 2. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis? 3. What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis? 4. What is the quantity which is measured by the area occupied below the velocity-time graph? 8.5 Equations of Motion by Graphical Method When an object moves along a straight line with uniform acceleration, it is possible to relate its velocity, acceleration during motion and the distance covered by it in a certain time interval by a set of equations known as the equations of motion. Therearethree such equations. These are: v =u + at (8.5) s =ut + ½ at2 (8.6) 2 as =v2 – u2 (8.7) where u is the initial velocity of the object which moves with uniform acceleration a for time t, v is the final velocity, and s is the distance travelled by the object in time t. Eq. (8.5) describes the velocity-time relation and Eq. (8.6) represents the position-time relation. Eq. (8.7), which represents the relation between the position and the velocity, can be obtained from Eqs. (8.5) and (8.6) by eliminating t. These three equations can be derived by graphical method. 8.5.1 EQUATION FOR VELOCITY-TIME RELATION Consider the velocity-time graph of an object that moves under uniform acceleration as shown in Fig. 8.8 (similar to Fig. 8.6, but now with u ≠ 0). From this graph, you can see that initial velocity of the object is u (at point A) and then it increases to v (at point B) in time t. The velocity changes at a uniform rate a. In Fig. 8.8, the perpendicular lines BC and BE are drawn from point B on the time and the velocity axes respectively, so that the initial velocity is represented by OA, the final velocity is represented by BC and the time interval t is represented by OC. BD = BC – CD, represents the change in velocity in time interval t. Let us draw AD parallel to OC. From the graph, we observe that BC = BD + DC = BD + OA Substituting BC = v and OA = u, we get v =BD + u or BD = v – u (8.8) From the velocity-time graph (Fig. 8.8), the acceleration of the object is given by Change in velocity a = time taken BD BD = AD OC = Substituting OC = t, we get BD a = t or BD = at (8.9) Using Eqs. (8.8) and (8.9) we get v = u + at 8.5.2 EQUATION FOR POSITION-TIME RELATION Let us consider that the object has travelled a distance s in time t under uniform acceleration a. In Fig. 8.8, the distance travelled by the object is obtained by the area enclosed within OABC under the velocity-time graph AB. Thus, the distances travelled by the object is given by s = area OABC (which is a trapezium) = area of the rectangle OADC + area of the triangle ABD 1 = OA × OC + (AD × BD) (8.10)2Substituting OA =u, OC = AD =t and BD = at, we get 1 s = u × t + (t×at )2 1 or s = u t + a t 2 2 8.5.3 EQUATION FOR POSITION–VELOCITY RELATION From the velocity-time graph shown in Fig. 8.8, the distance s travelled by the object in time t, moving under uniform acceleration a is given by the area enclosed within the trapezium OABC under the graph. That is, s = area of the trapezium OABC OA +BC ×OC = 2 Substituting OA = u, BC = v and OC = t, we get u+v t s (8.11)2 From the velocity-time relation (Eq. 8.6), we get v– u t= (8.12)a Using Eqs. (8.11) and (8.12) we have v+u v-u s= 2a or 2 a s = v2 – u2 Example 8.5 A train starting from rest attains a velocity of 72 km h–1 in 5 minutes. Assuming that the acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for attaining this velocity. Solution: We have been given u = 0 ; v = 72 km h–1 = 20 m s-1 and t = 5 minutes = 300 s. (i) From Eq. (8.5) we know that v–u a= t –1 –1 20 m s – 0ms = 300 s 1 –2= ms 15 (ii) From Eq. (8.7) we have 2 a s = v2 – u2 = v2 – 0 Thus, 2 v s = 2a (20 ms –1)2 = 2×(1/15) ms –2 = 3000 m = 3 km 1 The acceleration of the train is m s– 2 15 and the distance travelled is 3 km. Example 8.6 A car accelerates uniformly –1–1from 18 km h to 36 km h in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time. Solution: We are given that u = 18 km h–1 = 5 m s–1 v = 36 km h–1 = 10 m s–1 and t =5 s . (i) From Eq. (8.5) we have v–u a= t -1 -110 ms –5 m s = 5s = 1 m s–2 (ii) From Eq. (8.6) we have 1 s = u t + a t 2 21 = 5 m s–1 × 5 s + × 1 m s–2 × (5 s)2 2 = 25 m + 12.5 m = 37.5 m –2The acceleration of the car is 1 m sand the distance covered is 37.5 m. Example 8.7 The brakes applied to a car produce an acceleration of 6 m s-2 in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time. Solution: We have been given a = – 6 m s–2 ; t = 2 s and v = 0 m s–1. From Eq. (8.5) we know that v = u + at 0 = u + (– 6 m s–2) × 2 sor u = 12 m s–1 . From Eq. (8.6) we get 1 s = u t + a t 2 21 = (12 m s–1 ) × (2 s) + (–6 m s–2 ) (2 s)2 2= 24 m – 12 m = 12 m Thus, the car will move 12 m before it stops after the application of brakes. Can you now appreciate why drivers are cautioned to maintain some distance between vehicles while travelling on the road? Q uestions 1. A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled. 2. A train is travelling at a speed of 90 km h–1. Brakes are applied so as to produce a uniform acceleration of – 0.5 m s-2. Find how far the train will go before it is brought to rest. 3. A trolley, while going down an inclined plane, has an acceleration of 2 cm s-2. What will be its velocity 3 s after the start? 4. A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 s after start? 5. A stone is thrown in a vertically upward direction with a velocity of 5 m s-1. If the acceleration of the stone during its motion is 10 m s–2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there? 8.6 Uniform Circular Motion When the velocity of an object changes, we say that the object is accelerating. The change in the velocity could be due to change in its magnitude or the direction of the motion or both. Can you think of an example when an object does not change its magnitude of velocity but only its direction of motion? (a)Rectangular track (b) Hexagonal track (c)Octagonal shaped track (d) A circular track Fig. 8.9: The motion of an athlete along closed tracks of different shapes. 110 Let us consider an example of the motion of a body along a closed path. Fig 8.9 (a) shows the path of an athlete along a rectangular track ABCD. Let us assume that the athlete runs at a uniform speed on the straight parts AB, BC, CD and DA of the track. In order to keep himself on track, he quickly changes his speed at the corners. How many times will the athlete have to change his direction of motion, while he completes one round? It is clear that to move in a rectangular track once, he has to change his direction of motion four times. Now, suppose instead of a rectangular track, the athlete is running along a hexagonal shaped path ABCDEF, as shown in Fig. 8.9(b). In this situation, the athlete will have to change his direction six times while he completes one round. What if the track was not a hexagon but a regular octagon, with eight equal sides as shown by ABCDEFGH in Fig. 8.9(c)? It is observed that as the number of sides of the track increases the athelete has to take turns more and more often. What would happen to the shape of the track as we go on increasing the number of sides indefinitely? If you do this you will notice that the shape of the track approaches the shape of a circle and the length of each of the sides will decrease to a point. If the athlete moves with a velocity of constant magnitude along the circular path, the only change in his velocity is due to the change in the direction of motion. The motion of the athlete moving along a circular path is, therefore,an example of an accelerated motion. We know that the circumference of a circle of radius r is given by 2r . If the athlete takes t seconds to go once around the circular path of radius r, the velocity v is given by 2r v= (8.13)t When an object moves in a circular path with uniform speed, its motion is called uniform circular motion. SCIENCE Activity _____________ 8.11 • Take a piece of thread and tie a small piece of stone at one of its ends. Move the stone to describe a circular path with constant speed by holding the thread at the other end, as shown in Fig. 8.10. Fig. 8.10: A stone describing a circular path with a velocity of constant magnitude. • Now, let the stone go by releasing the thread. • Can you tell the direction in which the stone moves after it is released? • By repeating the activity for a few times and releasing the stone at different positions of the circular path, check whether the direction in which the stone moves remains the same or not. What you have learnt If you carefully note, on being released the stone moves along a straight line tangential to the circular path. This is because once the stone is released, it continues to move along the direction it has been moving at that instant. This shows that the direction of motion changed at every point when the stone was moving along the circular path. When an athlete throws a hammer or a discus in a sports meet, he/she holds the hammer or the discus in his/her hand and gives it a circular motion by rotating his/her own body. Once released in the desired direction, the hammer or discus moves in the direction in which it was moving at the time it was released, just like the piece of stone in the activity described above. There are many more familiar examples of objects moving under uniform circular motion, such as the motion of the moon and the earth, a satellite in a circular orbit around the earth, a cyclist on a circular track at constant speed and so on. • Motion is a change of position; it can be described in terms of the distance moved or the displacement. • The motion of an object could be uniform or non-uniform depending on whether its velocity is constant or changing. • The speed of an object is the distance covered per unit time, and velocity is the displacement per unit time. • The acceleration of an object is the change in velocity per unit time. • Uniform and non-uniform motions of objects can be shown through graphs. • The motion of an object moving at uniform acceleration can be described with the help of three equations, namely v = u + at s = ut + ½ at2 2as = v2 – u2 where u is initial velocity of the object, which moves with uniform acceleration a for time t, v is its final velocity and s is the distance it travelled in time t. • If an object moves in a circular path with uniform speed, its motion is called uniform circular motion. Exercises 1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s? 2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph's average speeds and velocities in jogging (a) from A to B and (b) from A to C? 3. Abdul, while driving to school, computes the average speed for his trip to be 20 km h–1. On his return trip along the same route, there is less traffic and the average speed is 30 km h–1. What is the average speed for Abdul's trip? 4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s–2 for 8.0 s. How far does the boat travel during this time? 5. A driver of a car travelling at 52 km h–1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h–1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied? 6. Fig 8.11 shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions: Fig. 8.11 (a) Which of the three is travelling the fastest? (b) Are all three ever at the same point on the road? (c) How far has C travelled when B passes A? (d) How far has B travelled by the time it passes C? 7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, with what velocity will it strike the ground? After what time will it strike the ground? 8. The speed-time graph for a car is shown is Fig. 8.12. Fig. 8.12 (a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period. (b) Which part of the graph represents uniform motion of the car? 9. State which of the following situations are possible and give an example for each of these: (a) an object with a constant acceleration but with zero velocity (b) an object moving in a certain direction with an acceleration in the perpendicular direction. 10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
>CHAP_8>
In everyday life, we see some objects at rest and others in motion. Birds fly, fish swim, blood flows through veins and arteries, and cars move. Atoms, molecules, planets, stars and galaxies are all in motion. We often perceive an object to be in motion when its position changes with time. However, there are situations where the motion is inferred through indirect evidences. For example, we infer the motion of air by observing the movement of dust and the movement of leaves and branches of trees. What causes the phenomena of sunrise, sunset and changing of seasons? Is it due to the motion of the earth? If it is true, why don't we directly perceive the motion of the earth?
An object may appear to be moving for one person and stationary for some other. For the passengers in a moving bus, the roadside trees appear to be moving backwards. A person standing on the road–side perceives the bus alongwith the passengers as moving. However, a passenger inside the bus sees his fellow passengers to be at rest. What do these observations indicate?
Most motions are complex. Some objects may move in a straight line, others may take a circular path. Some may rotate and a few others may vibrate. There may be situations involving a combination of these. In this chapter, we shall first learn to describe the motion of objects along a straight line. We shall also learn to express such motions through simple equations and graphs. Later, we shall discuss ways of describing circular motion.
Activity	8.1
•	Discuss whether the walls of your classroom are at rest or in motion.
•	Have you ever experienced that the train in which you are sitting appears to move while it is at rest?
• Discuss and share your experience.
Think and Act
We sometimes are endangered by the motion of objects around us, especially if that motion is erratic and uncontrolled as observed in a flooded river, a hurricane or a tsunami. On the other hand, controlled motion can be a service to human beings such as in the generation of hydro-electric power. Do you feel the necessity to study the erratic motion of some objects and learn to control them?
8.1	Describing Motion
We describe the location of an object by specifying a reference point. Let us understand this by an example. Let us assume that a school in a village is 2 km north of the railway station. We have specified the position of the school with respect to the railway station. In this example, the railway station is the reference point. We could have also chosen other reference points according to our convenience. Therefore, to describe the position of an object we need to specify a reference point called the origin.
8.1.1	Motion along a straight line
The simplest type of motion is the motion along a straight line. We shall first learn to describe this by an example. Consider the motion of an object moving along a straight path. The object starts its journey from O which is treated as its reference point (Fig. 8.1). Let A, B and C represent the position of the object at different instants. At first, the object moves through C and B and reaches A. Then it moves back along the same path and reaches C through B.
Fig. 8.1: Positions of an object on a straight line path
The total path length covered by the object is OA + AC, that is 60 km + 35 km = 95 km. This is the distance covered by the object. To describe distance we need to specify only the numerical value and not the direction of motion. There are certain quantities which are described by specifying only their numerical values. The numerical value of a physical quantity is its magnitude. From this example, can you find out the distance of the final position C of the object from the initial position O? This difference will give you the numerical value of the displacement of the object from O to C through A. The shortest distance measured from the initial to the final position of an object is known as the displacement.
Can the magnitude of the displacement be equal to the distance travelled by an object? Consider the example given in (Fig. 8.1). For motion of the object from O to A, the distance covered is 60 km and the magnitude of displacement is also 60 km. During its motion from O to A and back to B, the distance covered = 60 km + 25 km = 85 km while the magnitude of displacement = 35 km. Thus, the magnitude of displacement (35 km) is not equal to the path length (85 km). Further, we will notice that the magnitude of the displacement for a course of motion may be zero but the corresponding distance covered is not zero. If we consider the object to travel back to O, the final position concides with the initial position, and therefore, the displacement is zero. However, the distance covered in this journey is OA + AO = 60 km + 60 km = 120 km. Thus, two different physical quantities — the distance and the displacement, are used to describe the overall motion of an object and to locate its final position with reference to its initial position at a given time.
•	Take a metre scale and a long rope.
•	Walk from one corner of a basket-ball court to its oppposite corner along its sides.
•	Measure the distance covered by you and magnitude of the displacement.
•	What difference would you notice between the two in this case?
•	Automobiles are fitted with a device that shows the distance travelled. Such a device is known as an odometer. A car is driven from Bhubaneshwar to New Delhi. The difference between the final reading and the initial reading of the odometer is 1850 km.
•	Find the magnitude of the displacement between Bhubaneshwar and New Delhi by using the Road Map of India.
1.	An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
2.	A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
3.	Which of the following is true for displacement?
(a)	It cannot be zero.
(b)	Its magnitude is greater than the distance travelled by the object.
8.1.2	Uniform motion and non-uniform motion
Consider an object moving along a straight line. Let it travel 5 m in the first second,
5 m more in the next second, 5 m in the third second and 5 m in the fourth second. In this case, the object covers 5 m in each second. As the object covers equal distances in equal intervals of time, it is said to be in uniform motion. The time interval in this motion should be small. In our day-to-day life, we come across motions where objects cover unequal distances in equal intervals of timqbe, for example, when a car is moving on a crowded street or a person is jogging in a park. These are some instances of non-uniform motion.
•	The data regarding the motion of two different objects A and B are given in Table 8.1.
•	Examine them carefully and state whether the motion of the objects is uniform or non-uniform.
8.2	Measuring the Rate of Motion
Look at the situations given in Fig. 8.2. If the bowling speed is 143 km h–1 in Fig. 8.2(a) what does it mean? What do you understand from the signboard in Fig. 8.2(b)?
Different objects may take different amounts of time to cover a given distance. Some of them move fast and some move slowly. The rate at which objects move can be different. Also, different objects can move at the same rate. One of the ways of measuring the rate of motion of an object is to find out the distance travelled by the object in unit time. This quantity is referred to as speed. The SI unit of speed is metre per second. This is represented by the symbol m s–1 or m/s. The other units of speed include centimetre per second (cm s–1) and kilometre per hour (km h–1). To specify the speed of an object, we require only its magnitude. The speed of an object need not be constant. In most cases, objects will be in non-uniform motion. Therefore, we describe the rate of motion of such objects in terms of their average speed. The average speed of an object is obtained by dividing the total distance travelled by the total time taken. That is,
average speed =
If an object travels a distance s in time t then its speed v is,
Let us understand this by an example. A car travels a distance of 100 km in 2 h. Its average speed is 50 km h–1. The car might not have travelled at 50 km h–1 all the time. Sometimes it might have travelled faster and sometimes slower than this.
Example 8.1 An object travels 16 m in 4 s and then another 16 m in 2 s. What is the average speed of the object?
Total distance travelled by the object = 16 m + 16 m = 32 m
Total time taken = 4 s + 2 s = 6 s
= 5.33 m s–1
Therefore, the average speed of the object is 5.33 m s–1.
8.2.1	Speed with direction
The rate of motion of an object can be more comprehensive if we specify its direction of motion along with its speed. The quantity that specifies both these aspects is called velocity. Velocity is the speed of an object moving in a definite direction. The velocity of an object can be uniform or variable. It can be changed by changing the object's speed, direction of motion or both. When an object is moving along a straight line at a variable speed, we can express the magnitude of its rate of motion in terms of average velocity. It is calculated in the same way as we calculate average speed.
In case the velocity of the object is changing at a uniform rate, then average velocity is given by the arithmetic mean of initial velocity and final velocity for a given period of time. That is,
where vav is the average velocity, u is the initial velocity and v is the final velocity of the object.
Speed and velocity have the same units, that is, m s–1 or m/s.
•	Measure the time it takes you to walk from your house to your bus stop or the school. If you consider that your average walking speed is 4 km h–1, estimate the distance of the bus stop or school from your house.
•	At a time when it is cloudy, there may be frequent thunder and lightning. The sound of thunder takes some time to reach you after you see the lightning.
•	Can you answer why this happens?
•	Measure this time interval using a digital wrist watch or a stop watch.
•	Calculate the distance of the nearest point of lightning. (Speed of sound in air = 346 m s-1.)
1.	Distinguish between speed and velocity.
2.	Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
3.	What does the odometer of an automobile measure?
4.	What does the path of an object look like when it is in uniform motion?
5.	During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 108 m s–1.
Example 8.2 The odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of the trip. If the trip took 8 h, calculate the average speed of the car in km h–1 and m s–1.
Example 8.2 The odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of the trip. If the trip to
ok 8 h, calculate the average speed of the car in km h –1 and m s –1 .
Distance covered by the car,
s = 2400 km – 2000 km = 400 km
Time elapsed, t = 8 h
Average speed of the car is,
The average speed of the car is 50 km h–1 or 13.9 m s–1.
Example 8.3 Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Usha.
Solution: Total distance covered by Usha in 1 min is 180 m.
Displacement of Usha in 1 min = 0 m
8.3	Rate of Change of Velocity
During uniform motion of an object along a straight line, the velocity remains constant with time. In this case, the change in velocity of the object for any time interval is zero. However, in non-uniform motion, velocity varies with time. It has different values at different instants and at different points of the path. Thus, the change in velocity of the object during any time interval is not zero. Can we now express the change in velocity of an object?
To answer such a question, we have to introduce another physical quantity called acceleration, which is a measure of the change in the velocity of an object per unit time. That is,
If the velocity of an object changes from an initial value u to the final value v in time t, the acceleration a is,
This kind of motion is known as accelerated motion. The acceleration is taken to be positive if it is in the direction of velocity and negative when it is opposite to the direction of velocity. The SI unit of acceleration is m s–2 .
If an object travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time, then the acceleration of the object is said to be uniform. The motion of a freely falling body is an example of uniformly accelerated motion. On the other hand, an object can travel with non-uniform acceleration if its velocity changes at a non-uniform rate. For example, if a car travelling along a straight road increases its speed by unequal amounts in equal intervals of time, then the car is said to be moving with non-uniform acceleration.
•	In your everyday life you come across a range of motions in which
(a)	acceleration is in the direction of motion,
(b)	acceleration is against the direction of motion,
(c)	acceleration is uniform,
(d)	acceleration is non-uniform.
•	Can you identify one example each for the above type of motion?
Example 8.4 Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m s–1 in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m s-1 in the next 5 s. Calculate the acceleration of the bicycle in both the cases.
Example 8.4 Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m s –1 in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m s -1 in the next 5 s. Calculate the acceleration of the bicycle in both the cases.
In the first case:
initial velocity, u = 0 ;
final velocity, v = 6 m s–1 ;
time, t = 30 s .
From Eq. (8.3), we have
The acceleration of the bicycle in the first case is 0.2 m s–2 and in the second case, it is –0.4 m s–2
1.	When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?
2. A bus decreases its speed from 80 km h–1 to 60 km h–1 in 5 s. Find the acceleration of the bus.
3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h–1 in 10 minutes. Find its acceleration.
8.4	Graphical Representation of Motion
Graphs provide a convenient method to present basic information about a variety of events. For example, in the telecast of a one-day cricket match, vertical bar graphs show the run rate of a team in each over. As you have studied in mathematics, a straight line graph helps in solving a linear equation having two variables.
To describe the motion of an object, we can use line graphs. In this case, line graphs show dependence of one physical quantity, such as distance or velocity, on another quantity, such as time.
8.4.1	Distance–time graphs
The change in the position of an object with time can be represented on the distance-time graph adopting a convenient scale of choice. In this graph, time is taken along the x–axis and distance is taken along the y-axis. Distance-time graphs can be employed under various conditions where objects move with uniform speed, non-uniform speed, remain at rest etc.
We know that when an object travels equal distances in equal intervals of time, it moves with uniform speed. This shows that the distance travelled by the object is directly proportional to time taken. Thus, for uniform speed, a graph of distance travelled against time is a straight line, as shown in Fig. 8.3. The portion OB of the graph shows that the distance is increasing at a uniform rate. Note that, you can also use the term uniform velocity in place of uniform speed if you take the magnitude of displacement equal to the distance travelled by the object along the y-axis.
Fig. 8.3:	Distance-time graph of an object moving with uniform speed
We can use the distance-time graph to determine the speed of an object. To do so, consider a small part AB of the distance-time graph shown in Fig 8.3. Draw a line parallel to the x-axis from point A and another line parallel to the y-axis from point B. These two lines meet each other at point C to form a triangle ABC. Now, on the graph, Ac denotes the time interval (t2 – t1) while BC corresponds to the distance (s2 – s1). We can see from the graph that as the object moves from the point A to B, it covers a distance (s2 – s1) in time (t2 – t1). The speed, v of the object, therefore can be represented as
We can also plot the distance-time graph for accelerated motion. Table 8.2 shows the distance travelled by a car in a time interval of two seconds.
Fig. 8.4: Distance-time graph for a car moving with non-uniform speed
The distance-time graph for the motion of the car is shown in Fig. 8.4. Note that the shape of this graph is different from the earlier distance-time graph (Fig. 8.3) for uniform motion. The nature of this graph shows non-linear variation of the distance travelled by the car with time. Thus, the graph shown in Fig 8.4 represents motion with non-uniform speed.
8.4.2 Velocity-time graphs
The variation in velocity with time for an object moving in a straight line can be represented by a velocity-time graph. In this graph, time is represented along the x-axis and the velocity is represented along the
y-axis. If the object moves at uniform velocity, the height of its velocity-time graph will not change with time (Fig. 8.5). It will be a straight line parallel to the x-axis. Fig. 8.5 shows the velocity-time graph for a car moving with uniform velocity of 40 km h–1.
Fig. 8.5:	Velocity-time graph for uniform motion of a car
We know that the product of velocity and time give displacement of an object moving with uniform velocity. The area enclosed by velocity-time graph and the time axis will be equal to the magnitude of the displacement.
To know the distance moved by the car between time t1 and t2 using Fig. 8.5, draw perpendiculars from the points corresponding to the time t1 and t2 on the graph. The velocity of 40 km h–1 is represented by the height AC or BD and the time (t2 – t1) is represented by the length AB.
So, the distance s moved by the car in time (t2 – t1) can be expressed as
s	=	AC × CD
=	[(40 km h–1) × (t2 – t1) h]
=	40 (t2– t1) km
=	area of the rectangle ABDC (shaded in Fig. 8.5).
We can also study about uniformly accelerated motion by plotting its velocity– time graph. Consider a car being driven along a straight road for testing its engine. Suppose a person sitting next to the driver records its velocity after every 5 seconds by noting the reading of the speedometer of the car. The velocity of the car, in km h–1 as well as in m s–1, at different instants of time is shown in table 8.3.
In this case, the velocity-time graph for the motion of the car is shown in Fig. 8.6. The nature of the graph shows that velocity changes by equal amounts in equal intervals of time. Thus, for all uniformly accelerated motion, the velocity-time graph is a straight line.
Fig. 8.6: Velocity-time graph for a car moving with uniform accelerations.
You can also determine the distance moved by the car from its velocity-time graph. The area under the velocity-time graph gives the distance (magnitude of displacement) moved by the car in a given interval of time. If the car would have been moving with uniform velocity, the distance travelled by it would be represented by the area ABCD under the graph (Fig. 8.6). Since the magnitude of the velocity of the car is changing due to acceleration, the distance s travelled by the car will be given by the area ABCDE under the velocity-time graph (Fig. 8.6).
s	=	area ABCDE
=	area of the rectangle ABCD + area of the triangle ADE
In the case of non-uniformly accelerated motion, velocity-time graphs can have any shape.
Fig. 8.7: Velocity-time graphs of an object in non-uniformly accelerated motion.
Fig. 8.7(a) shows a velocity-time graph that represents the motion of an object whose velocity is decreasing with time while Fig. 8.7 (b) shows the velocity-time graph representing the non-uniform variation of velocity of the object with time. Try to interpret these graphs.
•	The times of arrival and departure of a train at three stations A, B and C and the distance of stations B and C from station A are given in table 8.4.
•	Plot and interpret the distance-time graph for the train assuming that its motion between any two stations is uniform.
Activity	8.10
•	Feroz and his sister Sania go to school on their bicycles. Both of them start at the same time from their home but take different times to reach the school although they follow the same route. Table 8.5 shows the distance travelled by them in different times
•	Plot the distance-time graph for their motions on the same scale and interpret.
1.	What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
2.	What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
3.	What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
4.	What is the quantity which is measured by the area occupied below the velocity-time graph?
8.5	Equations of Motion by Graphical Method
When an object moves along a straight line with uniform acceleration, it is possible to relate its velocity, acceleration during motion and the distance covered by it in a certain time interval by a set of equations known as the equations of motion. For convenience, a set of three such equations are given below:
v	=	u + at (8.5)
s	=	ut + ½ at2	(8.6)
2 a	s	=	v2 – u2 (8.7)
where u is the initial velocity of the object which moves with uniform acceleration a for time t, v is the final velocity, and s is the distance travelled by the object in time t. Eq. (8.5) describes the velocity-time relation and Eq. (8.6) represents the position-time relation. Eq. (8.7), which represents the relation between the position and the velocity, can be obtained from Eqs. (8.5) and (8.6) by eliminating t. These three equations can be derived by graphical method.
8.5.1	Equation for velocity-time relation
Consider the velocity-time graph of an object that moves under uniform acceleration as shown in Fig. 8.8 (similar to Fig. 8.6, but now with u ≠ 0). From this graph, you can see that initial velocity of the object is u (at point A) and then it increases to v (at point B) in time t. The velocity changes at a uniform rate a. In Fig. 8.8, the perpendicular lines BC and BE are drawn from point B on the time and the velocity axes respectively, so that the initial velocity is represented by OA, the final velocity is represented by BC and the time interval t is represented by OC. BD = BC – CD, represents the change in velocity in time interval t.
Fig. 8.8: Velocity-time graph to obtain the equations of motion
Let us draw AD parallel to OC. From the graph, we observe that
BC = BD + DC = BD + OA
Substituting	BC	=	v and OA = u,
we get	v	=	BD + u
or BD	=	v – u	(8.8)
From the velocity-time graph (Fig. 8.8), the acceleration of the object is given by
8.5.2	Equation for position-time Relation
Let us consider that the object has travelled a distance s in time t under uniform acceleration a. In Fig. 8.8, the distance travelled by the object is obtained by the area enclosed within OABC under the velocity-time graph AB.
Thus, the distance s travelled by the object is given by
s	= area OABC (which is a trapezium)
= area of the rectangle OADC + area of the triangle ABD
8.5.3	Equation for position–velocity relation
From the velocity-time graph shown in Fig. 8.8, the distance s travelled by the object in time t, moving under uniform acceleration a is given by the area enclosed within the trapezium OABC under the graph. That is,
s = area of the trapezium OABC
Example 8.5 A train starting from rest attains a velocity of 72 km h –1 in 5 minutes. Assuming that the acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for attaining this velocity
Example 8.6 A car accelerates uniformly rom 18 km h–1 to 36 km h–1 in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.
Example 8.6 A car accelerates uniformly from 18 km h–1 to 36 km h–1 in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.
We are given that
u	=	18 km h–1 = 5 m s–1
v	=	36 km h–1 = 10 m s–1 and
t	=	5 s .
The acceleration of the car is 1 m s–2 and the distance covered is 37.5 m.
Example 8.7 The brakes applied to a car produce an acceleration of 6 m s -2 in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time.
Thus, the car will move 12 m before it stops after the application of brakes. Can you now appreciate why drivers are cautioned to maintain some distance between vehicles while travelling on the road?
1.	A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.
2.	A train is travelling at a speed of 90 km h–1. Brakes are applied so as to produce a uniform acceleration of – 0.5 m s-2. Find how far the train will go before it is brought to rest.
3.	A trolley, while going down an inclined plane, has an acceleration of 2 cm s-2. What will be its velocity 3 s after the start?
4.	A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 s after start?
5.	A stone is thrown in a vertically upward direction with a velocity of 5 m s-1. If the acceleration of the stone during its motion is 10 m s–2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
8.6	Uniform Circular Motion
When the velocity of an object changes, we say that the object is accelerating. The change in the velocity could be due to change in its magnitude or the direction of the motion or both. Can you think of an example when an object does not change its magnitude of velocity but only its direction of motion?
Let us consider an example of the motion of a body along a closed path. Fig 8.9 (a) shows the path of an athlete along a rectangular track ABCD. Let us assume that the athlete runs at a uniform speed on the straight parts AB, BC, CD and DA of the track. In order to keep himself on track, he quickly changes his speed at the corners. How many times will the athlete have to change his direction of motion, while he completes one round? It is clear that to move in a rectangular track once, he has to change his direction of motion four times.
Fig. 8.9: The motion of an athlete along closed tracks of different shapes.
Now, suppose instead of a rectangular track, the athlete is running along a hexagonal shaped path ABCDEF, as shown in Fig. 8.9(b). In this situation, the athlete will have to change his direction six times while he completes one round. What if the track was not a hexagon but a regular octagon, with eight equal sides as shown by ABCDEFGH in Fig. 8.9(c)? It is observed that as the number of sides of the track increases the athelete has to take turns more and more often. What would happen to the shape of the track as we go on increasing the number of sides indefinitely? If you do this you will notice that the shape of the track approaches the shape of a circle and the length of each of the sides will decrease to a point. If the athlete moves with a velocity of constant magnitude along the circular path, the only change in his velocity is due to the change in the direction of motion. The motion of the athlete moving along a circular path is, therefore, an example of an accelerated motion.
We know that the circumference of a circle of radius r is given by . If the athlete takes t seconds to go once around the circular path of radius r, the speed v is given by
When an object moves in a circular path with uniform speed, its motion is called uniform circular motion.
Activity 8.11
•	Take a piece of thread and tie a small piece of stone at one of its ends. Move the stone to describe a circular path with constant speed by holding the thread at the other end, as shown in Fig. 8.10.
Fig. 8.10:	A stone describing a circular path with a velocity of constant magnitude.
•	Now, let the stone go by releasing the thread.
•	Can you tell the direction in which the stone moves after it is released?
•	By repeating the activity for a few times and releasing the stone at different positions of the circular path, check whether the direction in which the stone moves remains the same or not.
If you carefully note, on being released the stone moves along a straight line tangential to the circular path. This is because once the stone is released, it continues to move along the direction it has been moving at that instant. This shows that the direction of motion changed at every point when the stone was moving along the circular path.
When an athlete throws a hammer or a discus in a sports meet, he/she holds the hammer or the discus in his/her hand and gives it a circular motion by rotating his/her own body. Once released in the desired direction, the hammer or discus moves in the direction in which it was moving at the time it was released, just like the piece of stone in the activity described above. There are many more familiar examples of objects moving under uniform circular motion, such as the motion of the moon and the earth, a satellite in a circular orbit around the earth, a cyclist on a circular track at constant speed and so on.
What you have learnt
•	Motion is a change of position; it can be described in terms of the distance moved or the displacement.
•	The motion of an object could be uniform or non-uniform depending on whether its velocity is constant or changing.
•	The speed of an object is the distance covered per unit time, and velocity is the displacement per unit time.
•	The acceleration of an object is the change in velocity per unit time.
•	Uniform and non-uniform motions of objects can be shown through graphs.
•	The motion of an object moving at uniform acceleration can be described with the help of the following equations, namely
v = u + at
s = ut + ½ at2
2as = v2 – u2
where u is initial velocity of the object, which moves with uniform acceleration a for time t, v is its final velocity and s is the distance it travelled in time t.
•	If an object moves in a circular path with uniform speed, its motion is called uniform circular motion.
1.	An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
2.	Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph's average speeds and velocities in jogging (a) from A to B and (b) from A to C?
3.	Abdul, while driving to school, computes the average speed for his trip to be 20 km h–1. On his return trip along the same route, there is less traffic and the average speed is 30 km h–1. What is the average speed for Abdul's trip?
4.	A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s–2 for 8.0 s. How far does the boat travel during this time?
5.	A driver of a car travelling at 52 km h–1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h–1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
6.	Fig 8.11 shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions:
Fig. 8.11
(a)	Which of the three is travelling the fastest?
(b)	Are all three ever at the same point on the road?
(c)	How far has C travelled when B passes A?
(d)	How far has B travelled by the time it passes C?
7.	A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, with what velocity will it strike the ground? After what time will it strike the ground?
8.	The speed-time graph for a car is shown is Fig. 8.12.
(a)	Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
(b)	Which part of the graph represents uniform motion of the car?
9.	State which of the following situations are possible and give an example for each of these:
(a)	an object with a constant acceleration but with zero velocity
(b)	an object moving with an acceleration but with uniform speed.
(c)	an object moving in a certain direction with an acceleration in the perpendicular direction.
10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
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and dine with me? We can talk it over there."
her at once. "I have an engagement myself."
beforehand in his offer and that Carrie was about to go away.
"Come in later. I may know of something."
who had not caught all the details of the game he had played.
person to whom she was referred was not to be found.


text:
Reviews A-F
Reviews G-L
Reviews M-S
Reviews T-Z
Without the Walls
5/5 Rated Pubs
The Four Towers Challenge
Pub of the Year
York Beer News Feed
York Homebrew Club
Cawood Riverside Beer Festival
Indie Beer Yorkshire
-Evil Eye-
42 Stonegate, York, YO1 8AS.
https://www.evileyelounge.com/
Look & Feel:
Beer Garden:
The Evil Eye is another popular Stonegate destination, hidden behind a small off-licence which specialises predominantly in spirits. The inside is very vibrant with a kind of Middle Eastern shabby chic feel. There is a large bar which stretches down the main length of the back room which then opens out into a conservatory space and a very pleasant covered courtyard area with some fabulous views of the Minster. Unfortunately this is all that is really appealing about the Evil Eye as generally it's pretty filthy, rowdy and takes ages to get served. This is probably due to the huge amounts of cocktails on the menu, which I'm assured are some of the best in York. They do have a good selection of bottled beers but I'm not sure it's worth the effort to get to the bar and deal with the grumpy staff, especially given the high prices. There used to be an internet café upstairs by day but this space has now been transformed into a specialist Gin shop, so don't expect extra breathing space on the first floor. Access is okay but limited to the ground floor, that's if you can get through the crowds and the narrow rooms. The toilets leave much to be desired, as you can expect they get a good hammering every weekend. The Evil Eye is certainly popular but feels very overrated given there's much nicer places on Stonegate. Maybe it's just not our scene, either way it is not a place we would regularly return to. Best left to those who love their cocktails, as I can imagine if they can do that right then that's what keeps pulling people back in to overlook its other faults.
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redpajama_set_name: RedPajamaCommonCrawl


text:
Home Personal Injury Pennsylvania Clinton County
Clinton County Personal Injury Lawyers
Find Clinton County Personal Injury Attorneys by City
Mackeyville
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Robert Englert
Clinton County, PA Personal Injury Attorney with 14 years of experience
(888) 973-3529	333 North Vesper Street
Lock Haven,	PA 17745
Personal Injury, Insurance Claims, Medical Malpractice and Products Liability
Pennsylvania State University - University Park and University of Pennsylvania Law School
Andrew Ritchie Rehmeyer
Centre County, PA Personal Injury Lawyer with 14 years of experience
(814) 343-9860	1317 North Atherton Street
State College,	PA 16803
Personal Injury, Civil Rights, Criminal Defense and DUI & DWI
"Andrew Rehmeyer is a partner at the law firm of Rehmeyer & Allatt. In 2012 and again 2013, Andrew was recognized as a Pennsylvania Super Lawyer Rising Star. Before becoming a partner at Rehmeyer & Allatt, Andrew Rehmeyer gained exponential and valuable experience by working at a mid-sized litigation firm in the Philadelphia area representing insurance companies, individuals and law enforcement personnel in a variety of complex civil litigation matters. Andrew Rehmeyer has obtained favorable verdicts on behalf of his clients including a multi-million dollar lawsuit involving a wrongful death action, insurance bad faith claims, breach of...
Julian G. Allatt
Centre County, PA Personal Injury Attorney with 13 years of experience
Julian Allatt is a dedicated criminal defense attorney and partner at Rehmeyer & Allatt, a law firm in State College, PA. He is experienced in representing clients charged with minor summary offenses, DUI and minor drug possession cases, to violent felonies including armed robbery, burglary, aggravated assault, attempted murder and related offenses. Additionally, Julian has extensive experience representing clients charged with the possession and distribution of controlled substances in prosecutions commenced by county district attorney's offices as well as the Attorney General of Pennsylvania. He is a highly experienced and respected courtroom advocate who prides himself in his ability...
Seth T. Mosebey
Juniata County, PA Personal Injury Lawyer with 14 years of experience
(717) 535-2100	621 West Main Street
Thompsontown,	PA 17094
Personal Injury, Estate Planning, Probate and Real Estate
Michael D. Suders
Union County, PA Personal Injury Attorney with 36 years of experience
(570) 966-3360	47 E Chestnut St
Mifflinburg,	PA 17844
Free ConsultationPersonal Injury, Business, Estate Planning and Insurance Claims
Clifford Rieders
Lycoming County, PA Personal Injury Attorney
(570) 323-8711	161 W Third Street
Williamsport,	PA 17701
Mr. Rieders is a partner in the Williamsport law firm of Rieders, Travis, Humphrey, Waters & Dohrmann. He graduated in 1970 with a Bachelor of Arts, Cum Laude, Phi Beta Kappa, Coat of Arms Society and Academic Scholarships from New York University and received his Juris Doctor from Georgetown University in 1973. In Washington D. C. he worked for the Maritime Administration, U.S. Dept. of Commerce and for defense counsel Henry Rothblatt in the celebrated Watergate case. He served as a law clerk to the Honorable Malcolm Muir, United States District Court, Middle District of Pennsylvania. ...
David B. Ross
Potter County, PA Personal Injury Attorney with 19 years of experience
(814) 274-8612	One East Third Street
Coudersport,	PA 16915
At Ross & Ross we are committed to helping you. We practice primarily in Potter County, Mckean County, Tioga county, Elk County and surrounding Pennsylvania Counties in the areas of Personal injury and serious accidents, oil and gas, family law, divorce custody, adoption, support, bankruptcy, and social security law. Stop by our web site or call us to make an appointment at our Coudersport, PA office. We care about our clients. Let us help you. www.rossandrosslaw.com (814) 274-8612 If your county is not listed, give us a call, we handle serious accident cases in multiple states. We even have...
Mr. Gerald E. Lynch III
Lycoming County, PA Personal Injury Lawyer
(570) 321-8090	339 Market Street
Personal Injury, Criminal Defense, Family and Immigration
Daryl A. Yount
Union County, PA Personal Injury Attorney
New Berlin,	PA 17855
Personal Injury, Construction, Elder and Real Estate
Matthew J Lager
McKean County, PA Personal Injury Lawyer with 8 years of experience
(814) 887-6009	203 West Main St, Suite B
Smethport,	PA 16749
Free ConsultationPersonal Injury, Social Security Disability and Workers' Comp
Matt is a lifelong Pennsylvania resident, having lived in Bucks, Centre, and Erie counties. As an undergraduate, he attended The Pennsylvania State University's main campus, where he received a Bachelor of Science degree in biology. Afterwards, he went on to earn his Juris Doctor degree from The Pennsylvania State University, The Dickinson School of Law, where he graduated in the top third of his class. While at Penn State Law, Matt was a member of the Rural Economic Development Clinic and was recognized as a Public Interest Advocate for providing free legal assistance to small businesses in rural...
Whitney S Graham
Dauphin County, PA Personal Injury Lawyer with 9 years of experience
(800) 218-0808	4711 Locust Lane
Harrisburg,	PA 17109
Personal Injury and Social Security Disability
Temple University Beasley School of Law
Whitney S. Graham is a partner at Graham & Mauer, P.C. She earned her J.D. from Temple University's Beasley School of Law, where she was a member of the Moot Court Honor Society. Ms. Graham competed in Moot Court competitions in Philadelphia and Washington D.C. and was a semi-finalist in the Samuel Polsky Moot Court Competition. Ms. Graham was also a member of the Rubin Public Interest Law Honor Society for her law-related volunteer activities.
Ms. Graham has focused her career on litigation since her first summer in law school when she had the privilege to work an an intern for...
Jeffrey Bryant Engle
Dauphin County, PA Personal Injury Lawyer with 25 years of experience
(717) 545-3032	2205 Forest Hills Drive
Free ConsultationPersonal Injury and Criminal Defense
My office was founded in Millersburg, Dauphin County in 1951 by Attorney Allen Shaffer. Since I joined the Firm in 1999, I took over as managing partner in 2001 and later as sole shareholder in 2006. The office has expanded from a two lawyer firm to a four attorney firm and moved to Harrisburg. We handle a variety of legal areas, including Family Law, Divorce, Child Custody, Criminal Law, Municipal Law, and Educational Law.
Lackawanna County, PA Personal Injury Attorney with 40 years of experience
(888) 536-2426	116 N Washington Ave
Scranton,	PA 18503
Free ConsultationPersonal Injury, Education and Workers' Comp
Daniel Munley
Lackawanna County, PA Personal Injury Lawyer with 24 years of experience
(570) 865-4699	227 Penn Ave
For over 20 years, Daniel Munley has been fighting for his clients in personal injury cases throughout northeast Pennsylvania. He has been recognized as one of the Top 100 Trial Lawyers in PA by the National Trial Lawyers Association, listed in "Best Lawyers in America," and is a board-certified Civil Trial Advocate by the National Board of Trial Advocacy. Dan is a past chairperson of the American Association of Justice's Trucking Litigation Group and is a current executive board member of the AAJ's Trucking Litigation Group. Daniel holds a Juris Doctor degree from the University of Tulsa College of Law...
Paula Silverstein
Lancaster County, PA Personal Injury Attorney
East Petersburg,	PA 17520
Personal Injury, Criminal Defense and Family
Paula S. Silverstein has two decades of experience representing clients with their personal injury, family law and domestic matters and general civil litigation. If you are living in the Central Pennsylvania area including York and Lancaster and have been injured in a car acccident, facing divorce or any domestic issue, been convicted of a crime or any related law matter, contact Paula S. Silverstein today. Attorney Silverstein in married to her law partner, Bernard Ilkhanoff. They have been in their law practice together since 1999. Attorney Silverstein was an Assistant City Solicitor for the City of Philadelphia, then worked for...
Douglas V. Stoehr
Blair County, PA Personal Injury Attorney
(814) 946-4100	3600 5th Avenue
Altoona,	PA 16602
Personal Injury, Insurance Claims and Nursing Home
Attorney Stoehr operates a solo law practice in Altoona, Pennsylvania, with 21 years of experience. After working at a larger law firm where he became a partner, Attorney Stoehr decided to become a solo practitioner in 2004. He is a former claims representative which gives him great insight into how insurance companies settle claims. Attorney Stoehr represents injured people in Blair and the surrounding eight counties handling automobile cases, claims against nursing homes and hospitals, and general personal injury litigation. He is a member of the Million Dollar Advocates Forum, selected by Super Lawyers in 2012, 2013,...
Lawrence J. Moran
Personal Injury, Civil Rights and Criminal Defense
Michael J. O'Connor
Schuylkill County, PA Personal Injury Lawyer with 33 years of experience
(800) 518-4529	608 W Oak St
PO Box #201
Frackville,	PA 17931
Free ConsultationPersonal Injury, Elder, Estate Planning and Products Liability
James R. Ronca
Dauphin County, PA Personal Injury Attorney with 44 years of experience
(866) 735-2792	Olde Liberty Square
4807 Jonestown Road Suite 148
Free ConsultationPersonal Injury, Construction, Medical Malpractice and Products Liability
James Ronca focuses his practice on complex automobile and commercial vehicle accident lawsuits, pharmaceutical and medical device cases, construction accident lawsuits and more. Mr. Ronca was named one of the Top 10 Lawyers in Pennsylvania by Super Lawyers® in 2012 and 2014.
Robert J. Munley
538 Spruce St
Atty. Robert J. Munley practices Criminal and D.U.I. defense in Lackawanna, Monroe, Luzerne, Pike, Wayne, Susquehanna and all other counties in Northeast Pennsylvania. Atty. Munley has successfully represented thousands of individuals in criminal court throughout his career. He represents individuals charged with D.U.I., theft, assault, rape and sex crimes, computer crimes, drug delivery and possession, and all types of crimes. Atty. Munley has been a Certified Criminal Trial Specialist by the National Board of Trial Advocacy since 2003. He is also certified by the Pennsylvania Supreme Court to represent individuals facing the death penalty. Atty. Munley's objectives for every case...
Michael Edwin Kosik
(888) 476-0807	209 State Street
Michael E. Kosik grew up in Northeastern Pennsylvania and attended Wilkes University where he graduated cum laude with a degree in Economics and Political Science. He relocated to Central Pennsylvania to attend the Dickinson School of Law in Carlisle. He has been a trial lawyer handling trials throughout Pennsylvania since 1982. Since 1984, Mike has handled only cases on behalf of victims of automobile accidents, dangerous products, premise liability and other injury claims.
Timothy Kraeer
Chester County, PA Personal Injury Attorney with 7 years of experience
(610) 410-0099	158 West Gay Street
West Chester,	PA 19380
Free ConsultationPersonal Injury, Criminal Defense, DUI & DWI and Workers' Comp
Tim comes to Fink Law Offices with experience in civil litigation, personal injury, workers' compensation and criminal defense. He attended law school at Widener University School of Law in Wilmington, DE. Tim graduated with honors and finished in the top twenty percent of his class while earning Dean's List Honors and Certificates of Achievement in several subjects.
Tim grew up in West Chester attending SS. Simon & Jude and then Bishop Shanahan High School in Downingtown. He attended Western Michigan University in Kalamazoo, Michigan where he graduated summa cum laude while earning a Bachelor of Science degree in Aviation Flight Science....
Brian Lenahan
Free ConsultationPersonal Injury, Insurance Claims, Products Liability and Workers' Comp
J. T. Herber III
Schuylkill County, PA Personal Injury Attorney with 16 years of experience
(570) 682-8174	9 Park Lane
Hegins,	PA 17938
Wake Forest University School of Law and Lafayette College
J.T. Herber, III primarily practices in the fields of criminal defense and civil litigation, including personal injury. His criminal defense practice includes the representation of clients facing felony and misdemeanor charges, as well as serious traffic violations in the Courts of Common Pleas throughout central and eastern Pennsylvania. Attorney Herber previously served as an Assistant District Attorney in Schuylkill County. In civil cases, J.T. represents plaintiffs and defendants in a wide range of civil disputes, including personal injury and wrongful death cases, as well as real estate, commercial, construction, landlord-tenant, and collections matters. Despite being...
Adam Eugene Barnett
McKean County, PA Personal Injury Lawyer with 14 years of experience
(800) 999-0750	203 W Main Street, Suite B
Born and raised in Erie, Adam went on to obtain his bachelor's degree in Business Administration from Clarion University, where he graduated summa cum laude. After college, he went on to earn his Doctor of Jurisprudence from the West Virginia University College of Law in Morgantown, West Virginia. While in law school, Adam was a zealous advocate in West Virginia University's Clinical Law Program and West Virginia University's Immigration Clinic, where he fought for immigrants' rights. Adam has focused his practice on Car Accidents, Personal Injury, Workers' Compensation, and Social Security Disability law. Adam has earned a great deal...
Jonathan Comitz
Luzerne County, PA Personal Injury Attorney
(855) 589-9458	46 Public Square
Wilkes-Barre,	PA 18701
Free ConsultationPersonal Injury, Business, Family and Medical Malpractice
Born and raised in northeastern Pennsylvania, I grew up in the Heights section of Wilkes-Barre, am a graduate of G.A.R. Memorial High School, and received my B.A. from Juniata College in Huntingdon and my J.D. from Widener University School of Law in Harrisburg. After finishing law school, it was important to me to return to the Wilkes-Barre area to begin my career as an attorney. As you can see, family is extremely important to me. My wife Kelly is also from the Wilkes-Barre/Scranton area and it was important to us that we raise our children, Samuel, Carolyn, and John in close...
Thomas L. Kearney IV
York County, PA Personal Injury Attorney with 11 years of experience
(717) 668-8159	160 E. Market Street
York,	PA 17401
Free ConsultationPersonal Injury, Criminal Defense, DUI & DWI and Traffic Tickets
Thomas L. Kearney, IV (T.L.), is a Magna Cum Laude graduate of the University of Pittsburgh where he received a B.S. in both Business and Psychology. T.L. graduated Cum Laude from Widener University School of Law in Wilmington, Delaware where he received his Juris Doctorate degree. While attending Widener, T.L. served on the Executive Board, as an Editor, for the Delaware Journal of Corporate Law and was the Vice President of the entire Student Bar Association. He was also selected for an externship with the Honorable Justice Richard Barclay Surrick of the United States District Court for the Eastern District...
James R. Moyles
Cumberland County, PA Personal Injury Lawyer with 42 years of experience
(717) 233-5400	20 Erford Road
Lemoyne,	PA 17043
Free ConsultationPersonal Injury, Medical Malpractice, Nursing Home and Products Liability
With more than 30 years of legal experience, Pittsburgh personal injury lawyer, Jim Moyles, has the extensive personal injury knowledge and polished litigation skills to obtain maximum compensation for his clients. During his notable career, Jim has obtained for several clients settlements of more than $1 million. Jim founded The Moyles Law Firm in Pittsburgh in 1999. The law firm now includes a second office in Harrisburg and focuses exclusively on personal injury settlements and litigation. Recognized by Super Lawyers, Harrisburg personal injury attorney, Jim Moyles, will do his best to represent you in litigation to keep you on...
Free ConsultationPersonal Injury and Insurance Claims
As an attorney at Schmidt Kramer, P.C., I specialize in personal injury law. Specifically I work with motor vehicle accidents and insurances cases in the Harrisburg, PA area. I'm an active member of the Pennsylvania Association for Justice and had the pleasure of serving as its 2012-2013 president.
Bernard Ilkhanoff
Lancaster County, PA Personal Injury Lawyer with 25 years of experience
Personal Injury, Criminal Defense, Family and Products Liability
My interests lie in representing those whose disputes require a skilled lawyer in the courtroom. As a graduate of Temple University School of Law's famed trial advocacy program, I acquired the necessary skills, techniques and experience to serve my clients well, both pre-trial and at trial. I thrive in and am comfortable in competitive situations and parlay that into good results for my clients.
Daniel E. Cummins
(570) 319-5899	610 Morgan Highway
Clarks Summit,	PA 18411
Pennsylvania State University - Dickinson School of Law and University of London (England), Faculty of Laws
Daniel E. Cummins, Esquire is an AV-rated partner in the Clarks Summit (Scranton area), Pennsylvania law firm of of Cummins Law with over 22 years of litigation experience in Northeastern Pennsylvania, including insurance defense work in the Poconos area and the surrounding counties.
Attorney Cummins has focused his practice on defending motor vehicle accident liability cases and UIM/UM arbitration matters along with premises liability and products liability cases. Mr. Cummins is also routinely summoned by his clients to handle matters involving insurance coverage questions and insurance subrogation claims.
Mr. Cummins has been selected for inclusion in...
John R. Lenahan
Free ConsultationPersonal Injury, Nursing Home and Products Liability
Timothy G. Lenahan
James Stuczynski
James J. Stuczynski is a lifelong resident of the Erie area. After graduating from Cathedral Prep in 1971, he attended and graduated from Alliance College in 1975, summa cum laude. He furthered his education at the University of Pittsburgh School of Law, graduating in 1979 and returning to Erie to accept a position with Ely & Bernard. In 1985, he became a partner in Bernard & Stuczynski and has continued to practice with its successor firm, Bernard Stuczynski & Barnett. After a practice which handled a wide variety of legal claims, Attorney Stuczynski has specialized in handling insurance and...
Jeremy Donham
(717) 881-7855	256 E Market St
Free ConsultationPersonal Injury and Employment
Widener University Commonwealth Law School and West Virginia University
Jeremy Donham has a practice focused in the field of Employment Disability Discrimination and Workplace Privacy litigation. Mr. Donham also assists individuals who have been injured by the negligence of others. Mr. Donham is admitted to practice in all Pennsylvania and West Virginia State Courts. He is also admitted to all Federal District Courts in Pennsylvania and the Third Circuit Court of Appeals. Mr. Donham has access to other State and Federal Courts through local collaborative counsel arrangements and by special admission.
Mr. Donham gives personal attention to his clients by providing technologically based client interaction portals where clients can view...
Gerard Kramer
Attorney Gerard Kramer is a 1982 graduate of Ursinus College and a 1985 graduate of Penn State University's Dickinson School of Law. Mr. Kramer has been admitted to practice law in both Pennsylvania and New Jersey. He is also admitted to the United States District Court for the Middle District, the United States District Court for the Eastern Court, and the Federal Court of New Jersey. He holds professional memberships with the Pennsylvania Bar Association, American Association for Justice, the Pennsylvania Association for Justice, and the Central Pennsylvania Trial Lawyers Association. Mr. Kramer was also selected for inclusion...
Gerard Michael Karam
(570) 348-0776	Suite 201 Bank Towers
Joseph Chapman
Free ConsultationPersonal Injury and Tax
A trial attorney since 2009, Attorney Joe Chapman has helped individuals navigate the insurance system since 1996. He was elected to the Board of Governors of Pennsylvania Association for Justice by his peers, and is the chair of PAJ's End Distracted Driving campaign.
Free ConsultationPersonal Injury, Medical Malpractice, Nursing Home and Social Security Disability
A life-long resident of Harrisburg, Attorney Charles E. Schmidt, Jr. graduated from Bishop McDevitt High School in 1965, the University of Notre Dame in 1969, and Dickinson School of Law in 1974. Between college and law school, he served in the U.S. Army. He was admitted to the Pennsylvania Bar in 1974 and the Federal Bar in 1979. Mr. Schmidt is a former panel chairman of the Pennsylvania State Supreme Court Disciplinary Committee and a former Special Master in Dauphin County (a lawyer who is appointed to act as a judge, taking sworn testimony in divorce cases and rendering...
Michael H. Roth
Personal Injury, Insurance Claims and Medical Malpractice
The Oyez Lawyer Directory contains lawyers who have claimed their profiles and are actively seeking clients. Find more Clinton County Personal Injury Lawyers in the Justia Legal Services and Lawyers Directory which includes profiles of more than one million lawyers licensed to practice in the United States, in addition to profiles of legal aid, pro bono and legal service organizations.


text:
Weather in the Netherlands
Severe weather in the Netherlands
Recreational weather
Weather-Extra
Aruba, Bonaire and Curaçao
Saint Martin, Sint Eustatius and Saba
Weather in Liège, BE (Weather Station)
Precipitation: -
intermittent drizzle with partly clear sky
Precipitation 0 l/m² 0.1 l/m² 0.9 l/m²
Wind gusts 28.5 km/h 29.4 km/h 30.8 km/h 31.9 km/h 36.5 km/h 41.6 km/h 42.3 km/h
Avg. wind speed 11.9 km/h 13.9 km/h 16.1 km/h 18.4 km/h 20.3 km/h 19.3 km/h 20.8 km/h
Snow depth 0 cm 0 cm 0 cm
Precipitation 0 l/m² 0.1 l/m² 0.9 l/m² 0 l/m²
Wind gusts 29.4 km/h 30.8 km/h 31.9 km/h 36.5 km/h 41.6 km/h 42.3 km/h 39.4 km/h 33.2 km/h
Avg. wind speed 13.9 km/h 16.1 km/h 18.4 km/h 20.3 km/h 19.3 km/h 20.8 km/h 19.3 km/h 14.6 km/h
Snow depth 0 cm 0 cm 0 cm 0 cm
Wind gusts 31.6 km/h 31 km/h 29.6 km/h 30.7 km/h 30.8 km/h 31.2 km/h 25.6 km/h 20.1 km/h
Avg. wind speed 15.3 km/h 14.6 km/h 14.1 km/h 15.1 km/h 15 km/h 14.4 km/h 9.7 km/h 7.1 km/h
Avg. wind speed 7.5 km/h 8.8 km/h 11.3 km/h 13.1 km/h 14.2 km/h 13.5 km/h 9.1 km/h 5.7 km/h
Avg. wind speed 7.5 km/h 7.8 km/h 9.1 km/h 10 km/h 11.5 km/h 13.3 km/h 12.6 km/h 9.1 km/h
Wednesday Thursday Friday
Data provided by weather station Grâce-Hollogne-Bierset, 178m
+0 km - Liège
+29 km - Liège
+38 km - Maastricht-Airport
+38 km - Spa
Weather in Liège, 19.07.2019
In the early morning it will be nearly cloudless. Before noon it will become cloudy, but mostly dry. The noon offers mainly sunny weather with a few clouds. Morning temperatures will be around 15 °C. During the day they will reach a maximum of 24 °C. Low temperatures at night will be around 18°C. Southwesterly winds are moderate to brisk.
On Saturday late morning it will be sunny. The noon will bring variable clouds with isolated rain showers. In the late evening cloudless weather will dominate. Morning temperatures will be around 18 °C. During the day they will reach a maximum of 25 °C. Low temperatures at night will be around 15°C. Moderate to brisk winds will shift from the south to the southwest at noon.
+ Sunshine duration (min)


Board of Education – The Board of Education makes decisions about school policy, budget, and curriculum. The District 115 Board is the official governing body of Lake Forest High School and addresses educational, budget, and financial issues at its monthly meetings.
Education Committee – This committee provides oversight to all major District educational initiatives, new programs, and makes recommendations to the Board of Education on major topics or matters related to curriculum and instruction. The committee provides guidance to management on the accomplishment of educational strategic initiatives and improvement of student outcomes.
Finance/Operations Committee – This committee participates in the shaping of strategic financial directions and monitors all policies related to the financial administration of District 115. In addition, this committee contributes valuable input to the District's operational areas: facility management, transportation, and food service.
If you are interested, please indicate area(s) of interest and submit online or return the attached application to: Eileen Faller, Board Clerk, District Administrative Center, 300 S. Waukegan Rd., Lake Forest (Phone: 847-604-7420, Fax: 847-234-2372, Email: efaller@lfschools.net). We will contact and interview individuals as openings occur.
Thank you very much for your interest in serving Lake Forest High School District 115.


As residential home builders, we understand the value of your home investment. One of our priorities is to give you peace of mind and as stress-free an experience as possible.
Find answers to your questions about designing and building your custom home, remodeling and additions, or general contractor services with Butler Developments.
We have strong relationships with local financial institutions with which we can coordinate a meeting to help you receive a financing option. Please contact us to see how we can help. You may also visit our page describing financing options available through Butler Developments.
WHAT AWARDS HAS BUTLER DEVELOPMENTS RECEIVED?
Butler Developments is the recipient of the "Best of General contractor for the White Mountains" award. We are very grateful for the continued support and votes.
Photos of our previous work are displayed in our residential home builders gallery. We also encourage you to contact us to schedule a tour to view our current and past projects.
DO YOU OFFER SMART HOMES AND/OR ENERGY EFFICIENT HOMES?
Yes! Butler Developments takes pride in leading the way in new construction technology, energy efficiency, and environmental safety to make your home a 'smart' home.
CAN I CUSTOMIZE YOUR FLOOR PLANS?
Absolutely. Make your favorite Butler Developments floor plan entirely your own. We partner with a team of architects and draftsmen to assist in the design process of your new home, addition, or remodeling project.
No, our plans are not for sale. Butler Developments has engineered construction drawings that are unavailable outside our design process. The final plans are created to work with our unique process and techniques. Please keep in mind our plans are copyrighted and may not be reproduced or used without Butler Developments' consent.
HOW MUCH WILL MY CUSTOM HOME COST TO BUILD?
The first step to building a home or starting a construction project is understanding and approving the budget for making your dream a reality. We begin our Design Process by learning about your ideals and goals so that we can create a plan that fits within your financial expectations.
WHAT GEOGRAPHICAL AREA IS SERVED BY BUTLER DEVELOPMENTS?
Butler Developments services the state of Arizona and the entire United States through our Federal contract. But we are proud to be local residential home builders. We live, work, and play in the Show Low area and specialize in log cabin homes.
Please contact us if you have more questions, would like to schedule a tour of past and current construction projects, or want to schedule a consultation.


Boutiques + Brands + Birmingham—the height of Alabama shopping. For food, fashion, special events, films, and anytime-style inspiration in Alabama, visit the The Summit today.
Santa is available in The North Pole (next to the Cinema) daily from 12-6 PM (except Tuesdays, until 4 PM). Experiences with Santa are complimentary and professional photography is available for purchase Thursdays through Sundays. Walk-ins are always welcome during this time but will be served on a first-come, first-served basis between the scheduled appointments. Present your appointment confirmation (via print or mobile device) to the North Pole Greeter at Santa's Photo station upon arrival!
*Professional photography available + appointments. Credit and debit cards accepted (via square). No cash or check payments are accepted for Santa photography. Santa Photo fee of $20 includes the sitting fee, 1-4×6 print, and 1-digital file (with rights to print). You may purchase additional photos.