# What is the average speed, on t in [0,5], of an object that is moving at 8 m/s at t=0 and accelerates at a rate of a(t) =t^2-5 on t in [0,3]?

Apr 25, 2018

The average speed is $= 3.92 m {s}^{-} 1$

#### Explanation:

The speed is the integral of the acceleration

$a \left(t\right) = {t}^{2} - 5$

$v \left(t\right) = \frac{1}{3} {t}^{3} - 5 t + C$

Plugging in the initial conditions

$v \left(0\right) = 8 m {s}^{-} 1$

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

Therefore,

$v \left(t\right) = \frac{1}{3} {t}^{3} - 5 t + 8$

$v \left(3\right) = 9 - 15 + 8 = 2$

$v \left(4\right) = \frac{64}{3} - 20 + 8 = \frac{28}{3}$

The average speed is

$\left(5 - 0\right) \overline{v} = {\int}_{0}^{3} \left(\frac{1}{3} {t}^{3} - 5 t + 8\right) \mathrm{dt} + v \left(3\right) + v \left(4\right)$

$5 \overline{v} = {\left[\frac{1}{12} {t}^{4} - \frac{5}{2} {t}^{2} + 8 t\right]}_{0}^{3} + 2 + \frac{28}{3}$

$= \frac{27}{4} - \frac{45}{2} + 24 + \frac{34}{3}$

$= 19.58$

$\overline{v} = \frac{19.58}{5} = 3.92 m {s}^{-} 1$