# An object has a mass of 12 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?

Mar 15, 2017

The average speed is $= 89.16 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

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

$\frac{1}{2} m {u}_{1}^{2} = 81000 J$

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

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

Therefore,

${u}_{1}^{2} = \frac{2}{12} \cdot 81000 = 13500 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{12} \cdot 18000 = 3000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 13500\right)$ and $\left(4 , 3000\right)$

The equation of the line is

${v}^{2} - 13500 = \frac{3000 - 13500}{4} t$

${v}^{2} = - 2625 t + 13500$

So,

v=sqrt((-2625t+13500)

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

$\left(4 - 0\right) \overline{v} = {\int}_{0}^{5} \sqrt{- 2625 t + 13500} \mathrm{dt}$

$4 \overline{v} = {\left[{\left(- 2625 t + 13500\right)}^{\frac{3}{2}} / \left(\frac{3}{2} \cdot - 2625\right)\right]}_{0}^{4}$

$= \left({\left(- 2625 \cdot 4 + 13500\right)}^{\frac{3}{2}} / \left(- 3937.5\right)\right) - \left({\left(- 2625 \cdot 0 + 13500\right)}^{\frac{3}{2}} / \left(- 3937.5\right)\right)$

$= {3000}^{\frac{3}{2}} / - 3937.5 - {13500}^{\frac{3}{2}} / - 3937.5$

=1/3937.5(13500^(3/2)-3000^(3/2)))

$= 356.63$

So,

$\overline{v} = \frac{356.63}{4} = 89.16 m {s}^{-} 1$