# An object has a mass of 6 kg. The object's kinetic energy uniformly changes from 450 KJ to 129 KJ over t in [0, 8 s]. What is the average speed of the object?

Mar 30, 2017

The average speed is $= 306.4 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 6 k g$

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

$\frac{1}{2} m {u}_{1}^{2} = 450000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 129000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{6} \cdot 450000 = 150000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{6} \cdot 129000 = 43000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 150000\right)$ and $\left(8 , 43000\right)$

The equation of the line is

${v}^{2} - 150000 = \frac{43000 - 150000}{8} t$

${v}^{2} = - 13375 t + 150000$

So,

v=sqrt((-13375t+150000)

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

$\left(8 - 0\right) \overline{v} = {\int}_{0}^{8} \sqrt{\left(- 13375 t + 150000\right)} \mathrm{dt}$

8 barv=[((-13375t+150000)^(3/2)/(-3/2*13375)]_0^8

$= \left({\left(- 13375 \cdot 8 + 150000\right)}^{\frac{3}{2}} / \left(- 20062.5\right)\right) - \left({\left(- 13375 \cdot 0 + 150000\right)}^{\frac{3}{2}} / \left(- 20062.5\right)\right)$

$= {43000}^{\frac{3}{2}} / - 20062.5 - {150000}^{\frac{3}{2}} / - 20062.5$

$= 2451.2$

So,

$\overline{v} = \frac{22451.2}{8} = 306.4 m {s}^{-} 1$