The acceleration of a motorcycle is given by `a_x(t)=At-Bt^2`, where `A=1.50ms^-3` and `B=0.120ms^-4`. The motorcycle is at rest at the origin at time `t=0`.
a. Find its position and velocity as funcitons of time.
b. Calculate the maximum velocity it attains.

Explanation

a. When `v_0=0` and `x_0=0`,
`v_x=int_0^t(At-Bt^2)dt=A/2t^2-B/3t^2`
`=(0.75ms^-3)t^3-(0.04ms^-4)t^3`
`x=int_0^t(A/2t^2-B/3t^3)dt=A/6t^3-B/12t^4`
`=(0.25ms^-3)t^3-(0.010ms^-4)t^4`
b. For the velocity to be a maximum, the acceleration must be zero, this occurs at `t=0` and `t=A/B=12.5s`. At `t=0` the velocity is a minimum, and `t=12.5s` the velocity is
`v_x=(0.75ms^-3)(12.5s)^2-(0.040ms^-4)(12.5s)^3`
`=39.1ms^-1`