# What is the average speed of an object that is still at t=0 and accelerates at a rate of a(t) = 12-3t from t in [0, 4]?

Aug 3, 2017

The average speed is $= 16 m {s}^{-} 1$

#### Explanation:

The speed is the integral of the acceleration

$a \left(t\right) = 12 - 3 t$

$v \left(t\right) = \int \left(12 - 3 t\right) \mathrm{dt} = 12 t - \frac{3}{2} {t}^{2} + C$

Plugging in the initial conditions, $v \left(0\right) = 0$

Therefore,

$0 = 0 - 0 + C$

So,

$v \left(t\right) = 12 t - \frac{3}{2} {t}^{2}$

The average speed is

$4 \overline{v} = {\int}_{0}^{4} \left(12 t - \frac{3}{2} {t}^{2}\right) \mathrm{dt} = {\left[\frac{12}{2} {t}^{2} - \frac{1}{2} {t}^{3}\right]}_{0}^{4}$

$= \left(96 - 32\right) - \left(0\right) = 64$

$\overline{v} = \frac{64}{4} = 16 m {s}^{-} 1$