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^2-t on t in [0,4]?

Jun 22, 2016

${v}_{a} = 5.67 \text{ } \frac{m}{s}$

Explanation:

$\text{we can solve either using the average value of a function,}$
$\text{or using definition of the average speed}$

$\text{solution using definition of average speed:}$
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$v \left(t\right) = \int a \left(t\right) \cdot d t$

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

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

$\text{So " t=0" ; "v=3 m/s" ; } C = 3$

$v \left(t\right) = {t}^{3} / 3 - {t}^{2} / 2 + 3 \text{ *}$

$\Delta s = {\int}_{0}^{4} v \left(t\right) \cdot d t$

$\Delta s = {\int}_{0}^{4} \left({t}^{3} / 3 - {t}^{2} / 2 + 3\right) d t$

$\Delta s = | \frac{1}{3} \cdot \frac{1}{4} \cdot {t}^{4} - \frac{1}{2} \cdot \frac{1}{3} \cdot {t}^{3} + 3 \cdot t {|}_{0}^{4}$

$\Delta s = \left(\frac{1}{12} \cdot {4}^{4} - \frac{1}{6} \cdot {4}^{3} + 3 \cdot 4\right) - \left(\frac{1}{12} \cdot 0 - \frac{1}{6} \cdot 0 + 3 \cdot 0\right)$

$\Delta s = \left(\frac{256}{12} - \frac{64}{6} + 12\right) - 0$

$\Delta s = \frac{256 - 128 + 144}{12} = \frac{272}{12} \text{ m , total distance covered(t=0;t=4)}$

v_a=("total distance covered")/("given time interval")" definition of average velocity"

${v}_{a} = \frac{\frac{272}{12}}{4 - 0} = \frac{272}{4 \cdot 12} = \frac{272}{48} = 5.67 \text{ } \frac{m}{s}$

$\text{solution using the average value theorem:}$
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$\text{using speed v(t) * function above}$
${v}_{a} = \frac{1}{4 - 0} {\int}_{0}^{4} v \left(t\right) \cdot d t$

${v}_{a} = \frac{1}{4} {\int}_{0}^{4} \left({t}^{3} / 3 - {t}^{2} / 2 + 3\right) \cdot d t$

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

${v}_{a} = \frac{1}{4} \left[\left(\frac{1}{12} \cdot {4}^{4} - \frac{1}{6} \cdot {4}^{3} + 3 \cdot 4\right) - \left(0\right)\right]$

${v}_{a} = \frac{1}{4} \left[\frac{272}{12}\right]$

${v}_{a} = 5.67 \text{ } \frac{m}{s}$