# What is the average speed of an object that is moving at 3 m/s at t=0 and accelerates at a rate of a(t) =9-t^2/3 on t in [3,5]?

##### 1 Answer
Mar 30, 2017

$31 \frac{4}{9}$m/s

#### Explanation:

$a \left(t\right) = 9 - {t}^{2} / 3$

Speed, $v = \int a \left(t\right) = \int 9 - {t}^{2} / 3 = 9 t - {t}^{3} / 9 + c$

$t = 0 , v = 3$ m/s $\to c = 3$, therefore,
$v = 9 t - {t}^{3} / 9 + 3$

Displacement,
$s = \int v \left(t\right)$,

${s}_{3 , 5} = {\int}_{3}^{5} v \left(t\right) = {\int}_{3}^{5} 9 t - {t}^{3} / 9 + 3 = {\left[\frac{9}{2} {t}^{2} - {t}^{4} / 36 + 3 t\right]}_{3}^{5}$

${s}_{3 , 5} = \left[\frac{9}{2} {\left(5\right)}^{2} - {\left(5\right)}^{4} / 36 + 3 \left(5\right)\right] - \left[\frac{9}{2} {\left(3\right)}^{2} - {\left(3\right)}^{4} / 36 + 3 \left(3\right)\right]$

${s}_{3 , 5} = \left[\frac{225}{2} - \frac{625}{36} + 15\right] - \left[\frac{81}{2} - \frac{81}{36} + 9\right]$

${s}_{3 , 5} = \left[\frac{225}{2} - \frac{625}{36} + 15\right] - \left[\frac{81}{2} - \frac{81}{36} + 9\right]$

${s}_{3 , 5} = \frac{225}{2} - \frac{81}{2} - \frac{625}{36} + \frac{81}{36} + 15 - 9$

${s}_{3 , 5} = \frac{144}{2} - \frac{544}{36} + 6$

${s}_{3 , 5} = 72 - \frac{136}{9} + 6 = \frac{566}{9}$

Average speed = displacement/time $= \frac{566}{9} \div 2 = \frac{566}{18} = 31 \frac{4}{9}$m/s