# What is the average speed of an object that is moving at 2 m/s at t=0 and accelerates at a rate of a(t) =5-t on t in [0,3]?

Apr 20, 2016

Given acceleration $a \left(t\right) = 5 - t$
OR,$\frac{\mathrm{dv}}{\mathrm{dt}} = 5 - t$
,$\implies \mathrm{dv} = 5 \mathrm{dt} - t \mathrm{dt}$
,$\implies \int \mathrm{dv} = 5 \int \mathrm{dt} - \int t \mathrm{dt} + c$,where c= integration constant
$\implies v = 5 t - {t}^{2} / 2 + c$ ....(1)
Given at $t = 0 , v = 2 \frac{m}{s}$
So $c = 2$
and eq(1) becomes
$\implies v = 5 t - {t}^{2} / 2 + 2$

Now distance traversed during $t = 0 \to t = 3 s$

$s = {\int}_{0}^{3} v \mathrm{dt} = {\int}_{0}^{3} \left(5 t - {t}^{2} / 2 + 2\right) \mathrm{dt}$
$= {\left[5 {t}^{2} / 2 - {t}^{3} / 6 + 2 t\right]}_{0}^{3} = 5 \cdot {3}^{2} / 2 - {3}^{3} / 6 + 2 \cdot 3 = 24 m$
Hence Average speed
$= \frac{s}{t} = \frac{24}{3} = 8 \frac{m}{s}$