# What is the average speed of an object that is not moving at t=0 and accelerates at a rate of a(t) =4-t on t in [2,3]?

Dec 14, 2015

Remember how one calculates average value of any quantity:
$\setminus \left\langle f \setminus\right\rangle = \setminus \frac{1}{{t}_{2} - {t}_{1}} \setminus {\int}_{{t}_{1}}^{{t}_{2}} f \left(t\right) \mathrm{dt}$

We need to calculate $\setminus \left\langle v \setminus\right\rangle$, where $v$ is the velocity, for that we will need to first find the velocity it self, but we do know that
$\setminus \frac{\mathrm{dv}}{\mathrm{dt}} = a$, where a is the acceleration or in other words:
$v \left({t}_{2}\right) - v \left({t}_{1}\right) = \setminus {\int}_{{t}_{1}}^{{t}_{2}} a \left(t\right) \mathrm{dt}$

We have been given:
$a \left(t\right) = 4 - t$ and $v \left(0\right) = 0$, lets us plug these in the above formula, with ${t}_{1} = 0$, ${t}_{2} = t$
We get,
$v \left(t\right) = \setminus {\int}_{0}^{t} \left(4 - t\right) \mathrm{dt}$
$v \left(t\right) = \left(4 t - \setminus \frac{{t}^{2}}{2}\right)$
Now for its average value we use the first formula used
$\setminus \left\langle v \setminus\right\rangle = \setminus \frac{1}{3 - 2} \setminus {\int}_{2}^{3} \left(4 t - \setminus \frac{{t}^{2}}{2}\right) \mathrm{dt}$
$\setminus \left\langle v \setminus\right\rangle = \setminus \frac{1}{3 - 2} \left(2 {t}^{2} - \setminus \frac{{t}^{3}}{6}\right)$ with limits 2 and 3.
Inserting limits
$\setminus \left\langle v \setminus\right\rangle = \left(50 - \frac{19}{6}\right)$
$\setminus \left\langle v \setminus\right\rangle = \left(\frac{281}{6}\right) = 46.8 u n i t \frac{s}{\sec}$