# 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) =5t^2-2t on t in [0,3]?

Jun 11, 2017

The average speed is $=$

#### Explanation:

The speed is the integral of the acceleration

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

$v \left(t\right) = \int \left(5 {t}^{2} - 2 t\right) \mathrm{dt}$

$= \frac{5}{3} {t}^{3} - {t}^{2} + C$

Plugging in the initial conditions

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

Therefore,

$v \left(t\right) = \frac{5}{3} {t}^{3} - {t}^{2} + 8$

When $t = 3$,

$v \left(3\right) = 45 - 9 + 8 = 44 m {s}^{-} 2$

When $t \in \left[3 , 5\right]$, the speed is $= 44 m {s}^{-} 1$

The average speed is $= \overline{v}$

$5 \overline{v} = {\int}_{0}^{3} \left(\frac{5}{3} {t}^{3} - {t}^{2} + 8\right) \mathrm{dt} + 2 \cdot 44$

$5 \overline{v} = {\left[\frac{5}{12} {t}^{4} - \frac{1}{3} {t}^{3} + 8 t\right]}_{0}^{3} + 88$

$5 \overline{v} = \frac{135}{4} - 9 + 24 + 88 = 136.75$

$\overline{v} = \frac{136.75}{5} = 27.35 m {s}^{-} 1$