# An object has a mass of 9 kg. The object's kinetic energy uniformly changes from 54 KJ to  0 KJ over t in [0, 4 s]. What is the average speed of the object?

May 12, 2017

The average speed is $= 73.03 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 9 k g$

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

$\frac{1}{2} m {u}_{1}^{2} = 54000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 0 J$

Therefore,

${u}_{1}^{2} = \frac{2}{9} \cdot 54000 = 12000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{9} \cdot 0 = 0 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 12000\right)$ and $\left(4 , 0\right)$

The equation of the line is

${v}^{2} - 12000 = \frac{0 - 12000}{4} t$

${v}^{2} = - 3000 t + 12000$

So,

v=sqrt((-3000t+12000)

We need to calculate the average value of $v$ over $t \in \left[0 , 4\right]$

$\left(4 - 0\right) \overline{v} = {\int}_{0}^{4} \sqrt{\left(- 3000 t + 12000\right)} \mathrm{dt}$

4 barv=[((-3000t+12000)^(3/2)/(-3/2*3000)]_0^4

$= \left({\left(- 3000 \cdot 4 + 12000\right)}^{\frac{3}{2}} / \left(- 4500\right)\right) - \left({\left(- 3000 \cdot 0 + 12000\right)}^{\frac{3}{2}} / \left(- 4500\right)\right)$

$= {12000}^{\frac{3}{2}} / 4500 - {0}^{\frac{3}{2}} / 4500$

$= 292.12$

So,

$\overline{v} = \frac{292.12}{4} = 73.03 m {s}^{-} 1$