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

##### 1 Answer
Mar 24, 2017

The average speed is $= 6.17 m {s}^{-} 1$

#### Explanation:

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

Velocity is

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

$= \frac{2}{3} {t}^{3} - \frac{1}{2} {t}^{2} - 4 t + C$

Initial conditions, $v \left(0\right) = 0$

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

So,

$C = 0$

Average speed is

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

$= {\left[\frac{2}{3} {t}^{3} - \frac{1}{2} {t}^{2} - 4 t\right]}_{2}^{3}$

$= \left(18 - \frac{9}{2} - 12\right) - \left(\frac{16}{3} - 2 - 8\right)$

$= 6 - \frac{9}{2} - \frac{16}{3} + 10$

$= 16 - \frac{59}{6}$

$= \frac{37}{6}$

The average speed is $= 6.17 m {s}^{-} 1$