# An object has a mass of 6 kg. The object's kinetic energy uniformly changes from 18 KJ to  4KJ over t in [0, 9 s]. What is the average speed of the object?

Mar 16, 2017

The average speed is $= 59.38 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

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

$\frac{1}{2} m {u}_{1}^{2} = 18000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 4000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{6} \cdot 18000 = 6000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{6} \cdot 4000 = 1333.3 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 6000\right)$ and $\left(9 , 1333.3\right)$

The equation of the line is

${v}^{2} - 6000 = \frac{1333.3 - 6000}{9} t$

${v}^{2} = - 518.5 t + 6000$

So,

v=sqrt((-518.5t+6000)

We need to calculate the average value of $v$ over $t \in \left[0 , 9\right]$

$\left(9 - 0\right) \overline{v} = {\int}_{0}^{9} \sqrt{- 518.5 t + 6000} \mathrm{dt}$

$9 \overline{v} = {\left[{\left(- 518.5 t + 6000\right)}^{\frac{3}{2}} / \left(\frac{3}{2} \cdot - 518.5\right)\right]}_{0}^{9}$

$= \left({\left(- 518.5 \cdot 9 + 6000\right)}^{\frac{3}{2}} / \left(- 777.8\right)\right) - \left({\left(- 518.5 \cdot 0 + 6000\right)}^{\frac{3}{2}} / \left(- 777.8\right)\right)$

$= {1333.5}^{\frac{3}{2}} / - 777.8 - {6000}^{\frac{3}{2}} / - 777.8$

$= \frac{1}{778.5} \left({6000}^{\frac{3}{2}} - {1333.5}^{\frac{3}{2}}\right)$

$= 534.4$

So,

$\overline{v} = \frac{534.4}{9} = 59.38 m {s}^{-} 1$