# An object has a mass of 9 kg. The object's kinetic energy uniformly changes from 54 KJ to  45 KJ over t in [0, 4 s]. What is the average speed of the object?

Dec 2, 2017

The average speed is $= 104.8 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

The mass is $m = 9 k g$

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

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

The initial kinetic energy is $\frac{1}{2} m {u}_{1}^{2} = 54000 J$

The final kinetic energy is $\frac{1}{2} m {u}_{2}^{2} = 45000 J$

Therefore,

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

and,

${u}_{2}^{2} = \frac{2}{9} \cdot 45000 = 10000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 12000\right)$ and $\left(4 , 10000\right)$

The equation of the line is

${v}^{2} - 12000 = \frac{10000 - 12000}{4} t$

${v}^{2} = - 500 t + 12000$

So,

$v = \sqrt{- 500 t + 12000}$

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

$\left(4 - 0\right) \overline{v} = {\int}_{0}^{4} \left(\sqrt{- 500 t + 12000}\right) \mathrm{dt}$

4 barv=[-500t+12000)^(3/2)/(3/2*-500))] _( 0) ^ (4)

$= \left({\left(- 500 \cdot 4 + 12000\right)}^{\frac{3}{2}} / \left(- 750\right)\right) - \left({\left(- 500 \cdot 0 + 12000\right)}^{\frac{3}{2}} / \left(- 750\right)\right)$

$= {12000}^{\frac{3}{2}} / 750 - {10000}^{\frac{3}{2}} / 750$

$= 419.4$

So,

$\overline{v} = \frac{419.4}{4} = 104.8 m {s}^{-} 1$

The average speed is $= 104.8 m {s}^{-} 1$