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.

Answer 1

Answer:

a) $v(t) = 3t^{2} - 16t + 2$

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

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

Step-by-step 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 $t = 5.2s$.