# 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]?

Jun 11, 2017

$1 \frac{m}{s}$

#### Explanation:

NOTE: I am using the given a(t) as the acceleration term. Dimensionally, that doesn’t look quite right $\left(\frac{m}{{s}^{2}}\right)$?? 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
${v}_{f} = {v}_{i} + a \cdot t$
where
$a$ = acceleration $\left(\frac{m}{{s}^{2}}\right)$
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:

${v}_{a} = \frac{{v}_{f} + {v}_{i}}{2} = \frac{- 23 + 25}{2} = 1 \frac{m}{s}$

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.