# What is the average speed, on t in [0,5], of an object that is moving at 8 m/s at t=0 and accelerates at a rate of a(t) =5t^2-3t on t in [0,4]?

Nov 29, 2017

The average speed is $= 38.67 m {s}^{-} 1$

#### Explanation:

The speed is the integral of the acceleration

$a \left(t\right) = 5 {t}^{2} - 3 t$

Then,

$v \left(t\right) = \int \left(5 {t}^{2} - 3 t\right) \mathrm{dt} = \frac{5}{3} {t}^{3} - \frac{3}{2} {t}^{2} + C$

Plugging in the initial conditions at $t = 0$

$v \left(0\right) = 8 = 0 - 0 + C$

$C = 8$

Therefore,

$v \left(t\right) = \frac{5}{3} {t}^{3} - \frac{3}{2} {t}^{2} + 8$ in the interval $\left[0 , 4\right]$

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

$v \left(4\right) = \frac{5}{3} \cdot {4}^{3} - \frac{3}{2} \cdot {4}^{2} + 8 = \frac{320}{3} - 28 + 8 = 86.67 m {s}^{-} 1$

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

$5 \overline{v} = {\int}_{0}^{4} \left(\frac{5}{3} {t}^{3} - \frac{3}{2} {t}^{2} + 8\right) \mathrm{dt} + 86.67 \cdot 1$

$= {\left[\frac{5}{12} {t}^{4} - \frac{1}{2} {t}^{3} + 8 t\right]}_{0}^{4} + 86.67$

$= \left(\frac{5}{12} \cdot {4}^{2} - {4}^{3} / 2 + 8 \cdot 4\right) + 86.67$

$= 193.3$

The average speed is

$\overline{v} = \frac{193.3}{5} = 38.67 m {s}^{-} 1$