Question

What is the average speed of an object that is moving at `25 m/s` at `t=0` and accelerates at a rate of `a(t) =2-2t^2` on `t in [0,5]`?

Answer 1

`1 m/s`

Explanation:

NOTE: I am using the given a(t) as the acceleration term. Dimensionally, that doesn’t look quite right `(m/(s^2))`?? If it should be `a = 2 – 2*t^2` then the calculation of the final velocity will be different.

Average Velocity
`v_a = (v_f + v_i) / 2`   
where
`v_a` = average velocity (m/s)
`v_i` = initial velocity (m/s)
`v_f` = final velocity (m/s)

Final Velocity
`v_f = v_i + a*t`    
where
`a` = acceleration `(m/(s^2))`
t = time taken (s)

In this case:
`v_f = v_i + (2 – 2*t^2)` ;    `v_f = 25 + (2 – 2*5^2)`
`v_f = 25 + -48 = -23`
It is moving backwards, as indicated by the negative acceleration relative to the starting point. It initially had a forward direction, so it changed direction as well as velocity.

The average velocity is thus:

`v_a = (v_f + v_i) / 2 = (-23+ 25)/2 = 1 m/s`

This positive value just means that at time (5) it has not yet quite reached the location it was at at time (0). On average it only moved 1 meter over the time span and is now heading back.