# What is the average speed, on t in [0,5], of an object that is moving at 12 m/s at t=0 and accelerates at a rate of a(t) =5-3t on t in [0,4]?

Mar 18, 2016

$\overline{v} = 14 m / s$

#### Explanation:

The acceleration of an object is simply the derivative of its velocity with respect to time, so the velocity of an object is the integral of the acceleration with respect to time, in this case

$v \left(t\right) = \int a \left(t\right) \mathrm{dt} = \int \left(5 - 3 t\right) \mathrm{dt} = 5 t - \frac{3}{2} {t}^{2} + c$

Where we have added the arbitrary constant $c$. To find the value of $c$ we simply plug in $t = 0$ since we know the velocity at that time is $v \left(0\right) = 12 m / s$.

$v \left(t\right) = 5 t - \frac{3}{2} {t}^{2} + 12$

To get the average velocity, the simplest thing to do is to find the distance traveled in the specified time, which is the integral of the velocity between the limits:

$d = {\int}_{0}^{4} \left(5 t - \frac{3}{2} {t}^{2} + 12\right) \mathrm{dt} = {\left[\frac{5}{2} {t}^{2} - \frac{1}{2} {t}^{3} + 12 t\right]}_{0}^{4} = 56 m$

The average velocity is then the total distance over the total time:

$\overline{v} = \frac{56 m}{4 s} = 14 m / s$