# What is the average speed of an object that is still at t=0 and accelerates at a rate of a(t) = 16-t^2 from t in [0, 4]?

Apr 28, 2016

$26 \frac{2}{3}$

#### Explanation:

given,

$a \left(t\right) = 16 - {t}^{2}$

$\implies v \left(t\right) = {\int}_{o}^{t} \left(16 - {t}^{2}\right) \mathrm{dt} = 16 t - \frac{1}{3} {t}^{3} ,$ as v(0) = 0;

$\text{avg. speed} = \frac{{\int}_{o}^{4} \left(16 t - \frac{1}{3} {t}^{3}\right) \mathrm{dt}}{{\int}_{0}^{4} \mathrm{dt}} = \frac{1}{4} \left(8 {\left[{t}^{2}\right]}_{0}^{4} - \frac{1}{12} {\left[{t}^{4}\right]}_{0}^{4}\right) = \left(32 - \frac{16}{3}\right) = 26 \frac{2}{3}$