27 C1 3 C1 for a good diagram illustrating the journey M Distance travelled = (15 × (u + 3u)) + Drawing a velocity/time graph allows us to (10 × 3u) + (20 × 3u) illustrate the journey. The area under the = 30u + 30u +30u = 90u graph is the distance travelled. M1 for method of using diagram The assumption is that acceleration is M1 M1 for correct area equation at a steady rate when the motorbike A1 cao M1 speeds up and slows down. A1 for clear explanation of assumption. A1 A1 5 28 a x B1 2 B1 cao M = 0 b C2 C1 for explanation referring to a right angle triangle The circle has a radius of 5 and its C1 for explaining how using theorem helps Pythagoras’ centre is at (0, 0), halfway between D in finding the length 5 and E. Find the distance of F(–3, 4) from the origin. If it is 5 units from the origin, it is on the circumference of the circle. Using Pythagoras’ theorem: distance = = 5 So F(–3, 4) is on the circumference of the circle. c The tangent at F will be at right angles C3 C1 for explaining the gradient is tangent of the angle to the line joining the point to the origin. y mx c C1 for showing = + Gradient of the line is –. The product of the gradient of this line C1 for complete explanation showing how to work out and the tangent is Use this to work the equation of the line –1. out the gradient of the tangent. Substitute the gradient and the coordinates of the point into the general equation of a straight line (y = mx c) y-intercept c. + to find the 6
27 Drawing a velocity/time graph allows us to illustrate the journey. The area under the graph is the distance travelled. Distance travelled = (15 × (u + 3u)) + (10 × 3u) + (20 × 3u) = 30u + 30u +30u = 90u The assumption is that acceleration is at a steady rate when the motorbike speeds up and slows down. C1 M1 M1 A1 A1 3 C1 for a good diagram illustrating the journey M1 for method of using diagram M1 for correct area equation A1 cao A1 for clear explanation of assumption. M 5 28 a b c x = 0 The circle has a radius of 5 and its centre is at (0, 0), halfway between D and E. Find the distance of F(–3, 4) from the origin. If it is 5 units from the origin, it is on the circumference of the circle. Using Pythagoras’ theorem: distance = = 5 So F(–3, 4) is on the circumference of the circle. The tangent at F will be at right angles to the line joining the point to the origin. Gradient of the line is –. The product of the gradient of this line and the tangent is –1. Use this to work out the gradient of the tangent. Substitute the gradient and the coordinates of the point into the general equation of a straight line (y = mx + c) to find the y-intercept c. B1 C2 C3 2 B1 cao C1 for explanation referring to a right angle triangle C1 for explaining how using Pythagoras’ theorem helps in finding the length 5 C1 for explaining the gradient is tangent of the angle C1 for showing y = mx + c C1 for complete explanation showing how to work out the equation of the line M 6
Higher_Skills_Algebra_MarkScheme_Edexcel.docx