# What is the average speed of an object that is still at t=0 and accelerates at a rate of a(t) = 6-t from t in [0, 2]?

Apr 6, 2017

The average speed is $= 5.33 m {s}^{-} 1$

#### Explanation:

The velocity is the integral of the acceleration

$a \left(t\right) = 6 - t$

$v \left(t\right) = \int \left(6 - t\right) \mathrm{dt}$

$= 6 t - {t}^{2} / 2 + C$

We apply the initial conditions to find $C$

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

Therefore, $C = 0$

So,

$v \left(t\right) = 6 t - {t}^{2} / 2$

We can calculate the average value of the velocity

$\left(2 - 0\right) \overline{v} = {\int}_{0}^{2} \left(6 t - {t}^{2} / 2\right) \mathrm{dt}$

$= {\left[3 {t}^{2} - {t}^{3} / 6\right]}_{0}^{2}$

$= \left(12 - \frac{4}{3}\right) - \left(0\right)$

$= \frac{32}{3}$

$= 10.67$

So,

$\overline{v} = \frac{10.67}{2} = 5.33 m {s}^{-} 1$