# What is the average speed of an object that is moving at 8 m/s at t=0 and accelerates at a rate of a(t) =5-2t on t in [0,3]?

Feb 8, 2016

12.5

#### Explanation:

Integrate the acceleration to get the velocity as a function of time.

$v \left(t\right) = v \left(0\right) + {\int}_{0}^{t} a \left(\tau\right) d \tau$

$= 8 + {\int}_{0}^{t} \left(5 - 2 \tau\right) d \tau$

$= 8 + {\left[5 \tau - {\tau}^{2}\right]}_{0}^{t}$

$= 8 + 5 t - {t}^{2}$

Integrate the velocity to get the displacement as a function of time.

$x \left(t\right) - x \left(0\right) = {\int}_{0}^{t} v \left(\tau\right) d \tau$

$= {\int}_{0}^{t} \left(8 + 5 \tau - {\tau}^{2}\right) d \tau$

$= {\left[8 \tau + \frac{5}{2} {\tau}^{2} - \frac{1}{3} {\tau}^{3}\right]}_{0}^{t}$

$= 8 t + \frac{5}{2} {t}^{2} - \frac{1}{3} {t}^{3}$

From this, we can calculate the displacement from $t = 0$ to $t = 3$.

$x \left(3\right) - x \left(0\right) = 8 \left(3\right) + \frac{5}{2} {\left(3\right)}^{2} - \frac{1}{3} {\left(3\right)}^{3} = 37.5$

Divide this by the time (3 seconds) to get the average velocity.

$\overline{v} = \frac{37.5}{3 - 0} = 12.5$