# What is the average speed of an object that is not moving at t=0 and accelerates at a rate of a(t) =8-t on t in [2,5]?

Nov 25, 2017

The average speed is $= 21.5 m {s}^{-} 1$

#### Explanation:

The speed is the integral of the acceleration

$a \left(t\right) = 8 - t$

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

Plugging in the initial conditions

$v \left(0\right) = 0$

so,

$v \left(0\right) = 8 \cdot 0 - \frac{1}{2} \cdot O + C = 0$

$C = 0$

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

The average speed on the interval $t \in \left[2 , 5\right]$ is

$\left(5 - 2\right) \overline{v} = {\int}_{2}^{5} \left(8 t - \frac{1}{2} {t}^{2}\right) \mathrm{dt}$

$3 \overline{v} = {\left[\frac{8}{2} {t}^{2} - \frac{1}{6} {t}^{3}\right]}_{2}^{5}$

$3 \overline{v} = \left(4 \cdot 25 - \frac{1}{6} \cdot 125\right) - \left(4 \cdot 4 - \frac{1}{6} \cdot 8\right)$

$3 \overline{v} = 84 - \frac{125}{6} + \frac{4}{3} = 64.5$

$\overline{v} = \frac{64.5}{3} = 21.5 m {s}^{-} 1$