# What is the average speed of an object that is not moving at t=0 and accelerates at a rate of a(t) =10-t on t in [6, 10]?

Sep 20, 2016

Given
$\text{Acceleration} = a \left(t\right) = 10 - t$

$\frac{d \left(v \left(t\right)\right)}{\mathrm{dt}} = 10 - t$

Integrating we get

$v \left(t\right) = \int \left(10 - t\right) \mathrm{dt} = 10 t - \frac{1}{2} {t}^{2} + c$

It is also given that the object is not moving at $t = 0$,

$\text{so } v \left(0\right) = 0 \implies c = 0$

$\therefore v \left(t\right) = 10 t - \frac{1}{2} {t}^{2}$

$\implies \frac{d \left(s \left(t\right)\right)}{\mathrm{dt}} = 10 t - \frac{1}{2} {t}^{2}$

Integrating for $t \in \left[6 , 10\right]$ we get the distance traversed

$s = {\int}_{6}^{10} \left(10 t - \frac{1}{2} {t}^{2}\right) \mathrm{dt}$

$= {\left[10 \times {t}^{2} / 2 - \frac{1}{6} {t}^{3}\right]}_{6}^{10}$

$= 5 \times {10}^{2} - \frac{1}{6} \times {10}^{3} - 5 \times {6}^{2} + \frac{1}{6} \times {6}^{3}$

$= 536 - 180 - \frac{500}{3}$

$= 356 - 166.67 = 189.33$

$\frac{\text{Average speed"=s/(Deltat)=189.33/4~~47.33"unit}}{s}$