# What is the average speed, on t in [0,5], of an object that is moving at 5 m/s at t=0 and accelerates at a rate of a(t) =t^2-2t on t in [0,4]?

Nov 16, 2017

#### Answer:

The average speed is $= 6.07 m {s}^{-} 1$

#### Explanation:

The speed is the integral of the acceleration

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

Then,

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

Plugging in the initial conditions at $t = 0$

$v \left(0\right) = 5 = 0 - 0 + C$

$C = 5$

Therefore,

$v \left(t\right) = {t}^{3} / 3 - {t}^{2} + 5$ in the interval $\left[0 , 4\right]$

After that $t > 4$, assume that the speed is constant

$v \left(4\right) = {4}^{3} / 3 - {4}^{2} + 5 = \frac{64}{3} - 16 + 5 = 10.33 m {s}^{-} 1$

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

$5 \overline{v} = {\int}_{0}^{4} \left({t}^{3} / 3 - {t}^{2} + 5\right) \mathrm{dt} + 10.33 \cdot 1$

$= {\left[{t}^{4} / 12 - {t}^{3} / 3 + 5 t\right]}_{0}^{4} + 10.33$

$= \left({4}^{2} / 12 - {4}^{3} / 3 + 5 \cdot 4\right) + 10.33$

$= 30.33$

The average speed is

$\overline{v} = \frac{30.33}{5} = 6.07 m {s}^{-} 1$