# What is the average speed of an object that is still at t=0 and accelerates at a rate of a(t) =12-2t from t in [3, 4]?

Oct 20, 2016

Given acceleration of the object at t th sec

$a \left(t\right) = 12 - 2 t$

$\implies \frac{\mathrm{dv} \left(t\right)}{\mathrm{dt}} = 12 - 2 t$

$\implies v \left(t\right) = \int \left(12 - 2 t\right) \mathrm{dt} + c$

$\implies v \left(t\right) = \left(12 t - \frac{2 {t}^{2}}{2}\right) + c$

$\text{Given at "t=0, v(0)=0" " So" } c = 0$

$\implies v \left(t\right) = 12 t - {t}^{2}$

$\implies \frac{\mathrm{ds} \left(t\right)}{\mathrm{dt}} = 12 t - {t}^{2}$

So the distance traversed during $t \in \left\mid 3 , 4 \right\mid$

$s = {\int}_{3}^{4} \left(12 t - {t}^{2}\right) \mathrm{dt}$

$= {\left[\frac{12 {t}^{2}}{2} - {t}^{3} / 3\right]}_{3}^{4}$

$= 6 \cdot {4}^{2} - {4}^{3} / 3 - 6 \cdot {3}^{2} + {3}^{3} / 3 = \frac{89}{3}$

So the average speed

${V}_{\text{average"="Distance covered"/"Time taken}} = \frac{\frac{89}{3}}{1} = \frac{89}{3}$