# An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 16 KJ to  96KJ over t in [0, 6 s]. What is the average speed of the object?

Apr 27, 2017

163.345

#### Explanation:

Here, let's assume that uniform change means that the graph of Kinetic energy vs. time is a straight line.

Let y coordinate be kinetic energy, and x coordinate be time. As at x=0, y=16000, so y-intercept of line is 16000.

Slope of the line is $\frac{96000 - 16000}{6 - 0} = \frac{40000}{3}$

Equation of line is
$y = \left(\frac{40000}{3}\right) x + 16000$

$K . E . = \left(\frac{40000}{3}\right) t + 16000$

$\frac{1}{2} m {v}^{2} = \left(\frac{40000}{3}\right) t + 16000$
Putting value of m as 4 and simplifying:

${v}^{2} = \left(\frac{20000}{3}\right) t + 8000$
$v = \sqrt{\left(\frac{20000}{3}\right) t + 8000}$

Putting $v = \frac{\mathrm{ds}}{\mathrm{dt}}$, taking dt to R.H.S and integrating from t=0 to t=6, we get

$s = {\int}_{0}^{6} \left(\sqrt{\left(\frac{20000}{3}\right) t + 8000}\right) \mathrm{dt}$

Upon integrating (which I leave up to you), $s \approx 980.07$

We know that average speed = total distance / total time.

So Avg. Speed $\approx \frac{980.07}{6}$

Avg. Speed $\approx 163.345 \frac{m}{s}$