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

Apr 20, 2018

The average speed is $= 2 m {s}^{-} 1$

#### Explanation:

The speed is the integral of the acceleration

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

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

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

Plugging in the initial conditions

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

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

$C = 0$

The average speed on the interval $\left[0 , 2\right]$ is

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

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

$= \left(\frac{8}{3} - \frac{8}{3} + 4\right) - \left(0\right)$

$2 \overline{v} = 4$

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

Apr 20, 2018

Given the acceleration of the object at t th instant as

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

$\implies \frac{\mathrm{dv} \left(t\right)}{\mathrm{dt}} = 2 {t}^{2} - 2 t + 2$

$\implies v \left(t\right) = \int \left(2 {t}^{2} - 2 t + 2\right) \mathrm{dt} + c$, where c is integration constant.

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

Given that object is still at $t = 0$, we get $v \left(0\right) = 0$

Hence $v \left(0\right) = 0 = \frac{2}{3} \times 0 - \frac{2}{2} \times 0 + 2 \times 0 + c$

So $c = 0$
Hence
$v \left(t\right) = \frac{2}{3} {t}^{3} - {t}^{2} + 2 t$

$\implies \frac{\mathrm{ds} \left(t\right)}{\mathrm{dt}} = \frac{2}{3} {t}^{3} - {t}^{2} + 2 t$

So displacement in the time interval $t \in \left[0 , 2\right]$
$\implies s \left(t\right) = {\int}_{0}^{2} \left(\frac{2}{3} {t}^{3} - {t}^{2} + 2 t\right) \mathrm{dt}$

$\implies s \left(t\right) = {\left[\frac{2}{12} {t}^{4} - \frac{1}{3} {t}^{3} + \frac{2}{2} {t}^{2}\right]}_{0}^{2}$

$\implies s \left(t\right) = \left[\frac{2}{12} \cdot {2}^{4} - \frac{1}{3} \cdot {2}^{3} + {2}^{2}\right]$

$\implies s \left(t\right) = \left[\frac{8}{3} - \frac{8}{3} + 4\right] = 4$unit

Hence average velocity $\overline{v} = \text{displacement"/"time interval} = \frac{4}{2} = 2$ unit/s