The automobile is originally at rest s = 0. if it then starts to increase its speed at v # = (0.05t 2) ft> s2, where t is in seconds, determine the magnitudes of its velocity and acceleration at s = 550 ft.

Ans : dv/dt is the acceleration. In order to find the speed, integrate v = ⫠0.05 t^2 dt = 0.05/3 t^3 + v0 = 0.05/3 t^3 since v0 = 0 Similarly, space s is given by s = ⫠v dt = ⫠0.05/3 t^3 dt = 0.05/12 t^4 Then, for s = 550 ft, 0.05/12 t^4 = 550 t^4 = 550*12/0.05 = 132000 s t = (4th)root 132000 = 19.1 s Accordingly, v = 0.05/3 t^3 = 115 ft/s a = 0.05 t^2 = 18.2 ft/s²

The answer is b, i got it right on a test. hope it ; )

Either 2 or 3 i think