# 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 [0,3]?

Jan 16, 2018

The average speed is $= 4 m {s}^{-} 1$

#### Explanation:

The speed is the integral of the acceleration

$a \left(t\right) = 2 {t}^{2} - 1$

Then,

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

Plugging in the initial conditions at $t = 0$

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

$C = 1$

Therefore,

$v \left(t\right) = \frac{2}{3} {t}^{3} - t + 1$ in the interval $\left[0 , 3\right]$

After that $t > 3$, assume that the speed is constant

$v \left(3\right) = \frac{2}{3} \cdot {3}^{3} - 3 + 1 = 4 m {s}^{-} 1$

The average speed on the interval $\left[0 , 5\right]$ is

$5 \overline{v} = {\int}_{0}^{3} \left(\frac{2}{3} {t}^{3} - t + 1\right) \mathrm{dt} + \left(4 \cdot 2\right)$

$= {\left[\frac{1}{6} {t}^{4} - \frac{1}{2} {t}^{2} + t\right]}_{0}^{3} + 8$

$= \left(\frac{1}{6} \cdot {3}^{4} - \frac{1}{2} \cdot {3}^{2} + 3\right) + 8$

$= 20$

The average speed is

$\overline{v} = \frac{20}{5} = 4 m {s}^{-} 1$