# What is the average speed of an object that is moving at 6 m/s at t=0 and accelerates at a rate of a(t) =t^2-5t+3 on t in [0,3]?

Jul 2, 2017

The average speed is $= 5.25 m {s}^{-} 1$

#### Explanation:

The speed is the integral of the acceleration

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

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

$= \frac{1}{3} {t}^{3} - \frac{5}{2} {t}^{2} + 3 t + C$

Plugging in the initial conditions,

$v \left(0\right) = 6$

$v \left(0\right) = 0 - 0 + 0 + C = 6$

$\implies$, $C = 6$

Therefore,

$v \left(t\right) = \frac{1}{3} {t}^{3} - \frac{5}{2} {t}^{2} + 3 t + 6$

Let the average speed be $\overline{v}$

Then,

$\left(3 - 0\right) \overline{v} = {\int}_{0}^{3} \left(\frac{1}{3} {t}^{3} - \frac{5}{2} {t}^{2} + 3 t + 6\right) \mathrm{dt}$

$= {\left[\frac{1}{12} {t}^{4} - \frac{5}{6} {t}^{3} + \frac{3}{2} {t}^{2} + 6 t\right]}_{0}^{3}$

$= \left(\frac{1}{12} \cdot {3}^{4} - \frac{5}{6} \cdot {3}^{3} + \frac{3}{2} \cdot {3}^{2} + 6 \cdot 3\right) - \left(0\right)$

$= 15.75$

$\overline{v} = \frac{15.75}{3} = 5.25 m {s}^{-} 1$