# An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 104 KJ to  72KJ over t in [0,5s]. What is the average speed of the object?

Mar 15, 2017

The average speed is $= 209.5 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

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

$\frac{1}{2} m {u}_{1}^{2} = 104000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 72000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{4} \cdot 104000 = 52000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{4} \cdot 72000 = 36000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 52000\right)$ and $\left(5 , 36000\right)$

The equation of the line is

${v}^{2} - 52000 = \frac{36000 - 52000}{5} t$

${v}^{2} = - 3200 t + 52000$

So,

v=sqrt((-3200t+52000)

We need to calculate the average value of $v$ over $t \in \left[0 , 5\right]$

$\left(5 - 0\right) \overline{v} = {\int}_{0}^{5} \sqrt{- 3200 t + 52000} \mathrm{dt}$

$5 \overline{v} = {\left[{\left(- 3200 t + 52000\right)}^{\frac{3}{2}} / \left(\frac{3}{2} \cdot - 3200\right)\right]}_{0}^{5}$

$= \left({\left(- 3200 \cdot 5 + 52000\right)}^{\frac{3}{2}} / \left(- 48000\right)\right) - \left({\left(- 3200 \cdot 0 + 52000\right)}^{\frac{3}{2}} / \left(- 48000\right)\right)$

$= {36000}^{\frac{3}{2}} / - 4800 - {52000}^{\frac{3}{2}} / - 4800$

$= \frac{1}{4800} \left({52000}^{\frac{3}{2}} - {36000}^{\frac{3}{2}}\right)$

$= 1047.4$

So,

$\overline{v} = \frac{1047.4}{5} = 209.5 m {s}^{-} 1$