# What is the average speed of an object that is not moving at t=0 and accelerates at a rate of a(t) =5t-10 on t in [3, 5]?

May 11, 2016

${v}_{a} = 51.25 \frac{m}{s}$

#### Explanation:

$\frac{d v}{d t} = a \left(t\right)$

$d v = a \left(t\right) \cdot d t$

$\int d v = \int a \left(t\right) \cdot d t$

$v = \int \left(5 t - 10\right) \cdot d t$

$v = 5 \cdot {t}^{2} / 2 - 10 \cdot t + C$

$t = 0 \rightarrow v = 0 \rightarrow c = 0$

$v = \frac{5}{2} {t}^{2} - 10 t$

${v}_{a} = \frac{1}{5 - 3} {\int}_{3}^{5} | \left(\frac{5}{2} {t}^{2} - 10 t\right) | d t$

${v}_{a} = \frac{1}{2} \left[| \frac{5}{2} \cdot {t}^{3} / 3 - 10 \cdot {t}^{2} / 2 {|}_{3}^{5}\right]$

${v}_{a} = \frac{1}{2} \left[\left(\frac{5}{2} \cdot {5}^{3} / 3 - 5 \cdot {5}^{2}\right) - \left(\frac{5}{2} \cdot {3}^{3} / 3 - 5 \cdot {3}^{2}\right)\right]$

${v}_{a} = \frac{1}{2} \left[\left(\frac{625}{6} - 125\right) - \left(\frac{45}{2} - 45\right)\right]$

${v}_{a} = \frac{1}{2} \left[\left(625 - 750\right) - \left(- \frac{45}{2}\right)\right]$

${v}_{a} = \frac{1}{2} \left[- 125 + \frac{45}{2}\right]$

${v}_{a} = | - 51.25 |$

${v}_{a} = 51.25 \frac{m}{s}$