Question

1. A hamster moving along the x axis has a position given by x = (11t – 7.0t³) m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero?

Answer 1

x = 11t - 7t^3

v = dx/dt

    = 11 - 21^2

when v = 0

11 - 21t^2 = 0

t = 0.73 s

a = dv/dt = -42t

= -42*0.73 = -30.4 m/s2

so the manitude of acceleration is 30.4 m/s2