# What is the average speed of an object that is not moving at t=0 and accelerates at a rate of a(t) =6t-9 on t in [3, 5]?

Jan 3, 2016

Take the differential definition of acceleration, derive a formula connecting speed and time, find the two speeds and estimate the average.

${u}_{a v} = 15$

#### Explanation:

The definition of acceleration:

$a = \frac{\mathrm{du}}{\mathrm{dt}}$

$a \cdot \mathrm{dt} = \mathrm{du}$

${\int}_{0}^{t} a \left(t\right) \mathrm{dt} = {\int}_{0}^{u} \mathrm{du}$

${\int}_{0}^{t} \left(6 t - 9\right) \mathrm{dt} = {\int}_{0}^{u} \mathrm{du}$

${\int}_{0}^{t} \left(6 t \cdot \mathrm{dt}\right) - {\int}_{0}^{t} 9 \mathrm{dt} = {\int}_{0}^{u} \mathrm{du}$

$6 {\int}_{0}^{t} \left(t \cdot \mathrm{dt}\right) - 9 {\int}_{0}^{t} \mathrm{dt} = {\int}_{0}^{u} \mathrm{du}$

$6 \cdot {\left[{t}^{2} / 2\right]}_{0}^{t} - 9 \cdot {\left[t\right]}_{0}^{t} = {\left[u\right]}_{0}^{u}$

$6 \cdot \left({t}^{2} / 2 - {0}^{2} / 2\right) - 9 \cdot \left(t - 0\right) = \left(u - 0\right)$

$3 {t}^{2} - 9 t = u$

$u \left(t\right) = 3 {t}^{2} - 9 t$

So the speed at $t = 3$ and $t = 5$:

$u \left(3\right) = 3 \cdot {3}^{2} - 9 \cdot 3 = 0$

$u \left(5\right) = 30$

The average speed for $t \in \left[3 , 5\right]$:

${u}_{a v} = \frac{u \left(3\right) + u \left(5\right)}{2}$

${u}_{a v} = \frac{0 + 30}{2}$

${u}_{a v} = 15$