# An object has a mass of 5 kg. The object's kinetic energy uniformly changes from 150 KJ to  16KJ over t in [0,8s]. What is the average speed of the object?

Jun 5, 2017

The average speed is $= 176.4 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

The mass is $= 5 k g$

The initial velocity is $= {u}_{1}$

The initial kinetic energy is $\frac{1}{2} m {u}_{1}^{2} = 150000 J$

The final velocity is $= {u}_{2}$

The final kinetic energy is $\frac{1}{2} m {u}_{2}^{2} = 16000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{5} \cdot 150000 = 60000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{5} \cdot 16000 = 6400 {m}^{2} {s}^{-} 2$

The graph of ${v}^{2} = f \left(t\right)$ is a straight line

The points are $\left(0 , 60000\right)$ and $\left(8 , 6400\right)$

The equation of the line is

${v}^{2} - 60000 = \frac{6400 - 60000}{8} t$

${v}^{2} = - 6700 t + 60000$

So,

v=sqrt((-6700t+60000)

We need to calculate the average value of $v$ over $t \in \left[0 , 8\right]$

$\left(8 - 0\right) \overline{v} = {\int}_{0}^{8} \sqrt{\left(- 6700 t + 60000\right)} \mathrm{dt}$

8 barv=[((-6700t+60000)^(3/2)/(-3/2*6700)]_0^8

$= \left({\left(- 6700 \cdot 8 + 60000\right)}^{\frac{3}{2}} / \left(- 10050\right)\right) - \left({\left(- 6700 \cdot 0 + 60000\right)}^{\frac{3}{2}} / \left(- 10050\right)\right)$

$= {60000}^{\frac{3}{2}} / 10050 - {6400}^{\frac{3}{2}} / 10050$

$= 1411.4$

So,

$\overline{v} = \frac{1411.4}{8} = 176.4 m {s}^{-} 1$

The average speed is $= 176.4 m {s}^{-} 1$