# What is the average speed of an object that is still at t=0 and accelerates at a rate of a(t) = 3t^2-t+4 from t in [0,4]?

Jul 10, 2017

The average speed is $= 21.3 m {s}^{-} 1$

#### Explanation:

The speed is the integral of the acceleration.

$a \left(t\right) = 3 {t}^{2} - t + 4$

$v \left(t\right) = \int \left(3 {t}^{2} - t + 4\right) \mathrm{dt}$

$v \left(t\right) = {t}^{3} - \frac{1}{2} {t}^{2} + 4 t + C$

Plugging in the initial conditions, $t = 0$

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

Therefore,

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

The average speed is

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

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

$= \left(64 - \frac{32}{3} + 32\right) - \left(0\right)$

$= \frac{256}{3}$

$\overline{v} = \frac{64}{3} = 21.3 m {s}^{-} 1$