# What is the average speed of an object that is still at t=0 and accelerates at a rate of a(t) = t/4+2 from t in [0, 6]?

Mar 18, 2016

${v}_{a} = 7 , 5 \text{ } \frac{m}{s}$

#### Explanation:

$\text{1- Find displacement at interval (0,6)}$
$\text{2-use the formula :} {v}_{a} = \frac{\Delta x}{\Delta t}$

$v \left(t\right) = \int a \left(t\right) d t \text{ ; } x = {\int}_{a}^{b} v \left(t\right) d t$

$v \left(t\right) = \int \left(\frac{t}{4} + 2\right) d t = \frac{1}{4} \cdot \frac{1}{2} {t}^{2} + 2 t + C$
$v \left(t\right) = \frac{1}{8} {t}^{2} + 2 t + C \text{ for t=0 ; v=0 ; C=0}$
$v \left(t\right) = \frac{1}{8} {t}^{2} + 2 t$

$\Delta x = {\int}_{0}^{6} \left(\frac{1}{8} {t}^{2} + 2 t\right) d t$

$\Delta x = | \frac{1}{8} \cdot \frac{1}{3} \cdot {t}^{3} + 2 \cdot \frac{1}{2} \cdot {t}^{2} {|}_{0}^{6}$
$\Delta x = \left(\frac{1}{24} \cdot {6}^{3} + {6}^{2}\right) - \left(0 + 0\right)$

$\Delta x = 9 + 36$
$\Delta x = 45$

${v}_{a} = \frac{45}{6 - 0}$
${v}_{a} = \frac{45}{6}$
${v}_{a} = 7 , 5 \text{ } \frac{m}{s}$