# An object has a mass of 3 kg. The object's kinetic energy uniformly changes from 75 KJ to 135 KJ over t in [0, 9 s]. What is the average speed of the object?

Apr 22, 2016

${v}_{a v e} = 237 \frac{m}{s}$

#### Explanation:

For simplicity, Let: y = the kinetic energy

The kinetic energy changes by
$\setminus \Delta y = \left(135 - 75\right) K J = 60 K J$

over a time period of 9 seconds.
The rate of change of the kinetic energy, which I'm calling $y$, is

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{60 K J}{9 s} = 6.7 \frac{K J}{s}$

Therefore the kinetic energy as a function of time is

${y}_{\left(t\right)} = \left(6.7 \frac{K J}{s}\right) t + 75 K J$

This follows from ${y}_{\left(t\right)} = {y}_{\left(t = 0\right)} + \setminus {\int}_{0}^{t} \frac{\mathrm{dy}}{\mathrm{dt}} \mathrm{dt}$,

where I used that at $t = 0$, $\setminus \quad {y}_{\left(t = 0\right)} = 75 K J$.

We know that Kinetic Energy $= y = \frac{1}{2} m {v}^{2}$, Therefore

${v}_{\left(t\right)} = \setminus \sqrt{\frac{2 y}{m}}$.

The average speed is

${v}_{a v e} = \frac{1}{9} \setminus {\int}_{0}^{9} {v}_{\left(t\right)} \mathrm{dt} = \frac{1}{9 s} \setminus \sqrt{\frac{2}{m}} \setminus {\int}_{0}^{9} \setminus \sqrt{y \setminus \quad} \mathrm{dt}$

We could substitute $y$ and integrate in terms of $t$. But it looks like it would be easier to integrate over $y$. The endpoints of the integral correspond to the initial and final kinetic energies and we have that $\mathrm{dt} = \left(\frac{1}{6.7} \frac{s}{K J}\right) \mathrm{dy}$

${v}_{a v e} = \frac{1}{9 s} \setminus \sqrt{\frac{2}{m}} \left(\frac{1}{6.7} \frac{s}{K J}\right) \setminus {\int}_{{y}_{i}}^{{y}_{f}} {y}^{\frac{1}{2}} \mathrm{dy}$

Note that $\left(9 s\right) \left(6.7 \frac{K J}{s}\right) = {y}_{f} - {y}_{i}$, because that's how we found the rate of change of the kinetic energy.

${v}_{a v e} = \frac{1}{\left({y}_{f} - {y}_{i}\right)} \setminus \sqrt{\frac{2}{m}} \frac{2}{3} \left[{y}_{f}^{\frac{3}{2}} - {y}_{i}^{\frac{3}{2}}\right]$

Be sure to use ${y}_{i}$ and ${y}_{f}$ in Joules, the answer is
${v}_{a v e} = 237 \frac{m}{s}$