Question

A particle moves along the x-axis so that at time t its position is given by s(t) = (t + 3)(t −1)^3, t > 0. For what values of t is the velocity of the particle decreasing?

Answer 1

`0<t<1`

Explanation:

We want to know when the velocity is decreasing, which would mean the acceleration is less than 0.
Acceleration is the second derivative of position, so derive the equation twice.

(If you're comfortable using the product rule with powers, go straight into the derivation, otherwise simplify the equation first using algebra):
`s(t) = (t+3)(t^3-3t^2+3t-1)`
`s(t) = t^4-6t^2+8t-3`

Take the first derivative:
`v(t) = 4t^3-12t+8`

Take the second derivative:
`a(t) = 12t^2-12`

Set this acceleration function to < 0 and solve for `t` when `a(t)<0`:
`12t^2-12<0`
`12(t^2-1)<0`
`t^2<1`
`t<+-sqrt1`
`t<+-1`

In the problem statement, time is `t>0`, so the answer is
`0<t<1`