# An object has a mass of 9 kg. The object's kinetic energy uniformly changes from 81 KJ to  18 KJ over t in [0, 4 s]. What is the average speed of the object?

Aug 30, 2017

The average speed is $= 103.0 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

The mass is $= 9 k g$

The initial velocity is $= {u}_{1} m {s}^{-} 1$

The final velocity is $= {u}_{2} m {s}^{-} 1$

The initial kinetic energy is $\frac{1}{2} m {u}_{1}^{2} = 81000 J$

The final kinetic energy is $\frac{1}{2} m {u}_{2}^{2} = 18000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{9} \cdot 81000 = 18000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{9} \cdot 18000 = 4000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 18000\right)$ and $\left(4 , 4000\right)$

The equation of the line is

${v}^{2} - 18000 = \frac{4000 - 18000}{4} t$

${v}^{2} = - 3500 t + 18000$

So,

v=sqrt((-3500t+18000)

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

$\left(4 - 0\right) \overline{v} = {\int}_{0}^{4} \left(\sqrt{- 3500 t + 18000}\right) \mathrm{dt}$

$4 \overline{v} = {\left[\left({\left(- 3500 t + 18000\right)}^{\frac{3}{2}} / \left(- \frac{3}{2} \cdot 3500\right)\right)\right]}_{0}^{4}$

$= \left({\left(- 3500 \cdot 4 + 18000\right)}^{\frac{3}{2}} / \left(- 5250\right)\right) - \left({\left(- 3500 \cdot 0 + 18000\right)}^{\frac{3}{2}} / \left(- 5250\right)\right)$

$= {18000}^{\frac{3}{2}} / 5250 - {4000}^{\frac{3}{2}} / 5250$

$= 411.8$

So,

$\overline{v} = \frac{411.8}{4} = 103.0 m {s}^{-} 1$

The average speed is $= 103.0 m {s}^{-} 1$