Question

A particle moves according to the law of motion s(t) = t^{3}-8t^{2}+2t, where t is measured in seconds and s in feet.

(a) Find the velocity at time t.
(b) What is the velocity after 3 seconds?
(c) When is the particle at rest?

Answer 1

a)
`v(t) = 3t^(2) - 16t + 2`

b) The velocity after 3 seconds is -3m/s.

c)
`t = 0.13s` and
`t = 5.2s`.

Explanation:

The position is given by the following equation.

`s(t) = t^(3) - 8t^(2) + 2t`

(a) Find the velocity at time t.

The velocity is the derivative of position. So:

`v(t) = s^(\prime)(t) = 3t^(2) - 16t + 2`.

(b) What is the velocity after 3 seconds?

This is v(3).

`v(t) = 3t^(2) - 16t + 2`

`v(3) = 3*(3)^(2) - 16*(3) + 2 = -19`

The velocity after 3 seconds is -3m/s.

(c) When is the particle at rest?

This is when
`v(t) = 0`.

So:

`v(t) = 3t^(2) - 16t + 2`

`3t^(2) - 16t + 2 = 0`

This is when
`t = 0.13s` and
`t = 5.2s`.