# 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) =t-1 on t in [0,2]?

May 22, 2016

${v}_{a} = \frac{8}{3} \text{ } \frac{m}{s}$

#### Explanation:

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

$v \left(t\right) = \int \left(t - 1\right) d t$

$v \left(t\right) = {t}^{2} / 2 - t + C$

$\text{for t=0 ; v=3 ; C=3}$

$v \left(t\right) = \frac{1}{2} {t}^{2} - t + 3$

$\text{1-Solution using ' the average value' theorem :}$

${v}_{a} = \frac{1}{2 - 0} {\int}_{0}^{2} \left(\frac{1}{2} {t}^{2} - t + 3\right)$

${v}_{a} = \frac{1}{2} \left[| \frac{1}{2} \cdot {t}^{3} / 3 - \frac{1}{2} {t}^{2} + 3 t {|}_{0}^{2}\right]$

${v}_{a} = \frac{1}{2} \left[\left(\frac{1}{2} \cdot {2}^{3} / 3 - \frac{1}{2} \cdot {2}^{2} + 3 \cdot 2\right) - 0\right]$

${v}_{a} = \frac{1}{2} \left[\frac{4}{3} - 2 + 6\right]$

${v}_{a} = \frac{1}{2} \left[\frac{4}{3} + 4\right]$

${v}_{a} = \frac{1}{2} \left[\frac{16}{3}\right]$

${v}_{a} = \frac{16}{6}$

${v}_{a} = \frac{8}{3} \text{ } \frac{m}{s}$

$\text{2-Solution using definition of average velocity}$

${v}_{a} = \frac{\Delta x}{\Delta t} = \left(\text{displacement")/("time}\right)$

$\Delta x = {\int}_{0}^{2} v \left(t\right) d t = {\int}_{0}^{2} \left(\frac{1}{2} {t}^{2} - t + 3\right) d x$

$\Delta x = \left[| \frac{1}{2} \cdot \frac{1}{3} \cdot {t}^{3} - \frac{1}{2} {t}^{2} - 3 t {|}_{0}^{2}\right] = \left[\left(\frac{1}{2} \cdot \frac{1}{3} \cdot {2}^{3} - \frac{1}{2} \cdot {2}^{2} + 3 \cdot 2\right) - 0\right]$

$\Delta x = \frac{4}{3} - 2 + 6 \text{ ; "Delta x=4/3+4" ; "Delta x=16/3" } m$

${v}_{a} = \frac{\frac{16}{3}}{2}$

${v}_{a} = \frac{16}{6}$

${v}_{a} = \frac{8}{3} \text{ } \frac{m}{s}$