# What is the average speed of an object that is moving at 8 m/s at t=0 and accelerates at a rate of a(t) =2-t^2 on t in [0,2]?

Apr 8, 2017

#### Answer:

The average speed is $= 9.33 m {s}^{-} 1$

#### Explanation:

The velocity is the integral of the acceleration.

$a \left(t\right) = 2 - {t}^{2}$

$v \left(t\right) = \int \left(2 - {t}^{2}\right) \mathrm{dt} = 2 t - \frac{1}{3} {t}^{3} + C$

Plugging in the initial conditions

$v \left(0\right) = 8 = 0 - 0 + C$

So,

$C = 8$

$v \left(t\right) = 2 t - \frac{1}{3} {t}^{3} + 8$

We can calculate the mean speed

$\left(2 - 0\right) \overline{v} = {\int}_{0}^{2} \left(2 t - \frac{1}{3} {t}^{3} + 8\right) \mathrm{dt}$

$= {\left[\frac{2}{2} {t}^{2} - \frac{1}{12} {t}^{4} + 8 t\right]}_{0}^{2}$

$= \left(4 - \frac{4}{3} + 16\right) - \left(0\right)$

$= \frac{56}{3}$

Therefore,

$\overline{v} = \frac{1}{2} \cdot \frac{56}{3} = \frac{28}{3} = 9.33 m {s}^{-} 1$