What is the average speed of an object that is not moving at t=0 and accelerates at a rate of a(t) =8-t^2/4 on t in [2,5]?

Jun 17, 2016

v_a=28.67" "("unit")/s

Explanation:

"$\text{we can calculate using the average value theorem}$
${v}_{a} = \frac{1}{5 - 2} {\int}_{2}^{5} v \left(t\right) \cdot d t$

$\text{v(t) represents the speed function with respect to time}$

$\text{first we must find the function v(t)}$

v(t)=int a(t)*d t; " so " a(t)=8-t^2/4

$v \left(t\right) = \int \left(8 - {t}^{2} / 4\right) d t$

$v \left(t\right) = 8 t - \frac{1}{4} \cdot \frac{1}{3} \cdot {t}^{3} + C$

$\text{C=0 because it isn't moving at t=0}$

$v \left(t\right) = 8 t - \frac{1}{12} {t}^{3} + 0$

$\text{now we can calculate the average speed of the object}$

${v}_{a} = \frac{1}{5 - 2} {\int}_{2}^{5} \left(8 t - \frac{1}{12} \cdot {t}^{3}\right) d t$

${v}_{a} = \frac{1}{3} \left[| 8 \cdot \frac{1}{2} \cdot {t}^{2} - \frac{1}{12} \cdot \frac{1}{4} \cdot {t}^{4} {|}_{2}^{5}\right]$

${v}_{a} = \frac{1}{3} \left[| 4 {t}^{2} - \frac{1}{48} {t}^{4} {|}_{2}^{5}\right]$

${v}_{a} = \frac{1}{3} \left[\left(4 \cdot {5}^{2} - \frac{1}{48} \cdot {5}^{4}\right) - \left(4 \cdot {2}^{2} - \frac{1}{48} \cdot {2}^{4}\right)\right]$

${v}_{a} = \frac{1}{3} \left[\left(100 - \frac{625}{48}\right) - \left(16 - \frac{16}{48}\right)\right]$

${v}_{a} = \frac{1}{3} \left[\frac{4800 - 625}{48} - \frac{47}{48}\right]$

${v}_{a} = \frac{1}{3} \left[\frac{4175 - 47}{48}\right]$

${v}_{a} = \frac{1}{3} \left[\frac{4128}{48}\right]$

${v}_{a} = \frac{4128}{144}$

v_a=28.67" "("unit")/s