# An object has a mass of 9 kg. The object's kinetic energy uniformly changes from 135 KJ to  36KJ over t in [0, 6 s]. What is the average speed of the object?

Apr 19, 2016

I don't produce any number as the result, but here is how you should approach.

#### Explanation:

$K E = \frac{1}{2} m {v}^{2}$

Hence, $v = \sqrt{\frac{2 K E}{m}}$

We know $K E = {r}_{k} \cdot t + c$ where ${r}_{k} = 99 K J {s}^{- 1}$ and c = $36 K J$

So the rate of change of velocity ${r}_{v}$ is related to the rate of change of kinetic energy ${r}_{k}$ as:

$v = \sqrt{\frac{2 {r}_{k} \cdot t + 2 c}{m}}$

now, average velocity should be defined as:

${v}_{\text{avg}} = \frac{{\int}_{0}^{t} v \mathrm{dt}}{t} = \frac{1}{5} {\int}_{0}^{5} \sqrt{\frac{2 {r}_{k} \cdot t + 2 c}{m}} \mathrm{dt}$