What is the average speed, on t in [0,5], of an object that is moving at 7 m/s at t=0 and accelerates at a rate of a(t) =t^2-2t on t in [0,4]?

Dec 5, 2017

The average speed is $= 8.27 m {s}^{-} 1$

Explanation:

The speed is the integral of the acceleration

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

Then,

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

Plugging in the initial conditions at $t = 0$

$v \left(0\right) = 7 = 0 - 0 + C$

$C = 7$

Therefore,

$v \left(t\right) = \frac{1}{3} {t}^{3} - {t}^{2} + 7$ in the interval $\left[0 , 4\right]$

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

$v \left(4\right) = \frac{1}{3} \cdot {4}^{3} - {4}^{2} + 8 = \frac{64}{3} - 8 = 13.33 m {s}^{-} 1$

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

$5 \overline{v} = {\int}_{0}^{4} \left(\frac{1}{3} {t}^{3} - {t}^{2} + 7\right) \mathrm{dt} + 13.33 \cdot 1$

$= {\left[\frac{1}{12} {t}^{4} - \frac{1}{3} {t}^{3} + 7 t\right]}_{0}^{4} + 13.33$

$= \left(\frac{1}{12} \cdot {4}^{4} - {4}^{3} / 3 + 7 \cdot 4\right) + 13.33$

$= 41.33$

The average speed is

$\overline{v} = \frac{41.33}{5} = 8.27 m {s}^{-} 1$