(f) We arrived at the on the right (shown above) by adding 20t to the expression. x(t) (g) Examining the slopes of the graphs become zero, it is clear that the shift causes the = 0 point to correspond v to a larger value of (the top of the second curve shown in part (e) is higher than that of the first). x 10. Recognizing that the gap between the trains is closing at a constant rate of 60 km/h, the total time which elapses before they crash is = (60 km)/(60 km/h) = 1.0 h. During this time, the bird travels a distance of = = (60 t x vt km/h)(1.0 h) = 60 km. 12. We use Eq. 2-4. to solve the problem. (a) The velocity of the particle is Thus, at = 1 s, the velocity is = (–12 + (6)(1)) = m/s. t v –6 (b) 0, it is moving in the negative direction at = 1 s. v  x t (c) At = 1 s, the is |v| = 6 m/s. t speed (d) For 0 2 s, |v| decreases until it vanishes. For 2 3 s, |v| increases from zero to the value it had in part (c).  t   t  Then, |v| is larger than that value for  3 s. t (e) Yes, smoothly changes from negative values (consider the = 1 result) to positive (note that as  + , v t t we have + One can check that = 0 when v  ). v (f) No. In fact, from = + 6t, we know that 0 for 2 s. v v  t  –12 The velocity is given by = 2ct 3bt2 = (6.0 m/s2)t (6.0 m/s3)t2. v – – (f) Plugging in = 1 s, we obtain t (g) Similarly, (h) (i) The acceleration is given by = = 2c 6b = 6.0 m/s2 (12.0 m/s3)t. a dv/dt – – Plugging in = 1 s, we obtain t (k) (l)
(f) We arrived at the on the right (shown above) by adding 20t to the x(t) expression. (g) Examining the slopes of the graphs become zero, it is clear that the shift causes the v = 0 point to correspond to a larger value of x (the top of the second curve shown in part (e) is higher than that of the first). 10. Recognizing that the gap between the trains is closing at a constant rate of 60 km/h, the total time which elapses before they crash is t = (60 km)/(60 km/h) = 1.0 h. During this time, the bird travels a distance of x = vt = (60 km/h)(1.0 h) = 60 km. 12. We use Eq. 2-4. to solve the problem. (a) The velocity of the particle is Thus, at t = 1 s, the velocity is v = (–12 + (6)(1)) = –6 m/s. (b) v  0, it is moving in the negative x direction at t = 1 s. (c) At t = 1 s, the speed is |v| = 6 m/s. (d) For 0  t  2 s, |v| decreases until it vanishes. For 2  t  3 s, |v| increases from zero to the value it had in part (c). Then, |v| is larger than that value for t  3 s. (e) Yes, v smoothly changes from negative values (consider the t = 1 result) to positive (note that as t  + , we have v  + ). One can check that v = 0 when (f) No. In fact, from v = –12 + 6t, we know that v  0 for t  2 s. The velocity is given by v = 2ct – 3bt2 = (6.0 m/s2)t – (6.0 m/s3)t2. (f) Plugging in t = 1 s, we obtain (g) Similarly, (h) (i) The acceleration is given by a = dv/dt = 2c – 6b = 6.0 m/s2 – (12.0 m/s3)t. Plugging in t = 1 s, we obtain (k) (l)
APC%20Ch%202%20answers%20Randall.docx