# What is the average speed, on t in [0,5], of an object that is moving at 1 m/s at t=0 and accelerates at a rate of a(t) =2t^2+1 on t in [0,2]?

May 2, 2017

The average speed is $= 6.33 m {s}^{-} 1$

#### Explanation:

The speed is the integral of the acceleration

$v \left(t\right) = \int a \left(t\right) \mathrm{dt}$

$= \int \left(2 {t}^{2} + 1\right) \mathrm{dt}$

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

Plugging in the initial conditions

$v \left(0\right) = 0 + 0 + C = 1$

Therefore,

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

$v \left(2\right) = \frac{2}{3} \cdot {2}^{3} + 2 + 1 = \frac{25}{3}$

The average value is

$\left(5 - 0\right) \overline{v} = {\int}_{0}^{2} \left(\frac{2}{3} {t}^{3} + t + 1\right) \mathrm{dt} + \frac{25}{3} \cdot 3$

$5 \overline{v} = {\left[\frac{2}{3} \cdot \frac{1}{4} \cdot {t}^{4} + \frac{1}{2} {t}^{2} + t\right]}_{0}^{2} + 25$

$= \left(\frac{8}{3} + 2 + 2\right) - \left(0\right) + 25$

$= \frac{20}{3} + 25 = \frac{95}{3}$

$\overline{v} = \frac{95}{15} = 6.33 m {s}^{-} 1$