# What is the average speed of an object that is still at t=0 and accelerates at a rate of a(t) = 36-t^2 from t in [2, 6]?

##### 1 Answer
Jan 1, 2018

$117.3 m {s}^{-} 1$

#### Explanation:

To get the average speed, we must sum the velocity and divide by the time traveled. We have a continuous function so we must integrate to sum the velocity. So average velocity is given by:

${v}_{a v g} = \frac{1}{\Delta t} {\int}_{{t}_{i}}^{{t}_{f}} v \left(t\right) \mathrm{dt}$

Now, we have been given the acceleration which means we must first integrate to find the formula for velocity, so:

$v \left(t\right) = \int a \left(t\right) \mathrm{dt} = \int 36 - {t}^{2} \mathrm{dt} = 36 t - \frac{1}{3} {t}^{3} + C$

Using the fact that the object is still at at $t = 0$ then:

$v \left(0\right) = 36 \left(0\right) - \frac{1}{3} {\left(0\right)}^{3} + C = 0 \to C = 0$

So we have for the velocity:

$v \left(t\right) = 36 t - \frac{1}{3} {t}^{3}$

Now we can use this to get the average velocity, in our case the initial time ${t}_{i} = 2 s$ and the final time ${t}_{f} = 6 s$ so $\Delta t = 6 - 2 = 4 s$

Now putting that into the expression for the average velocity:

$\frac{1}{\Delta t} {\int}_{{t}_{i}}^{{t}_{f}} v \left(t\right) \mathrm{dt} = \frac{1}{4} {\int}_{2}^{6} 36 t - \frac{1}{3} {t}^{3} \mathrm{dt}$

$= \frac{1}{4} {\left[18 {t}^{2} - \frac{1}{12} {t}^{4}\right]}_{2}^{6}$

$= \frac{1}{4} \left(\left\{18 {\left(6\right)}^{2} - \frac{1}{12} {\left(6\right)}^{4}\right\} - \left\{18 {\left(2\right)}^{2} - \frac{1}{12} {\left(2\right)}^{4}\right\}\right)$

$= \frac{1408}{12} \approx 117.3 m {s}^{-} 1$