# What is the average speed of an object that is moving at 18 m/s at t=0 and accelerates at a rate of a(t) =4t-8 on t in [0,2]?

Jun 28, 2016

${v}_{a} = 3.5 \text{ } \frac{m}{s}$

#### Explanation:

$\text{Strategy...}$
$\diamond \text{ find v(t) function}$
$\diamond \text{ use average value theorem}$

$\diamond \text{ --> } v \left(t\right) = \int a \left(t\right) d t$

$v \left(t\right) = \int \left(4 t - 8\right) d t$

$v \left(t\right) = 4 \cdot \frac{1}{2} \cdot {t}^{2} - 8 \cdot t$

$v \left(t\right) = 2 {t}^{2} - 8 t + C$

$\text{for "t=0" ; " v=18" "m/s" ; ""then } C = 18$

$v \left(t\right) = 2 {t}^{2} - 8 t + 18$

$\diamond \text{ -->} {v}_{a} = \frac{1}{2 - 0} {\int}_{0}^{2} v \left(t\right) d t$

${v}_{a} = \frac{1}{2} {\int}_{0}^{2} \left(2 {t}^{2} - 8 t + 18\right) d t$

${v}_{a} = \frac{1}{2} \left[| 2 \cdot \frac{1}{3} \cdot {t}^{3} - 8 \cdot \frac{1}{2} \cdot {t}^{2} + 18 \cdot t {|}_{0}^{2}\right]$

${v}_{a} = \frac{1}{2} \left[| \frac{2}{3} {t}^{3} - 4 {t}^{2} + 18 t {|}_{0}^{2}\right]$

${v}_{a} = \frac{1}{2} \left[\left(\frac{2}{3} \cdot {2}^{3} - 8 \cdot {2}^{2} + 18 \cdot 2\right) - \left(0\right)\right]$

${v}_{a} = \frac{1}{2} \left[\frac{16}{3} - 32 + 36\right]$

$v = a = \frac{1}{2} \left[\frac{16}{3} + 4\right] = \frac{1}{2} \left[\frac{16 + 12}{4}\right] = \frac{1}{2} \left[\frac{28}{4}\right] = \frac{1}{2} \cdot 7$

${v}_{a} = 3.5 \text{ } \frac{m}{s}$