# An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 12 KJ to  72KJ over t in [0,5s]. What is the average speed of the object?

Jan 13, 2018

The average speed is $= 141.5 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

The mass is $m = 4 k g$

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

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

The initial kinetic energy is $\frac{1}{2} m {u}_{1}^{2} = 12000 J$

The final kinetic energy is $\frac{1}{2} m {u}_{2}^{2} = 72000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{4} \cdot 12000 = 6000 {m}^{2} {s}^{-} 2$

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 $\left(0 , 6000\right)$ and $\left(5 , 36000\right)$

The equation of the line is

${v}^{2} - 6000 = \frac{36000 - 6000}{5} t$

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

So,

$v = \sqrt{6000 t + 6000}$

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

$\left(5 - 0\right) \overline{v} = {\int}_{0}^{5} \left(\sqrt{6000 t + 6000}\right) \mathrm{dt}$

$5 \overline{v} = {\left(6000 t + 6000\right)}^{\frac{3}{2}} / \left(\frac{3}{2} \cdot 6000\right) {|}_{0}^{5}$

$= \left({\left(6000 \cdot 5 + 6000\right)}^{\frac{3}{2}} / \left(9000\right)\right) - \left({\left(6000 \cdot 0 + 6000\right)}^{\frac{3}{2}} / \left(9000\right)\right)$

$= {36000}^{\frac{3}{2}} / 9000 - {6000}^{\frac{3}{2}} / 9000$

$= 707.31$

So,

$\overline{v} = \frac{707.31}{5} = 141.5 m {s}^{-} 1$

The average speed is $= 141.5 m {s}^{-} 1$

Jan 13, 2018

Using equation of kinetic energy, i.e E =$\left(\frac{1}{2}\right) m \left({v}^{2}\right)$
Now,using v=u+at(all symbols are bearing their conventional meaning) we get a is 22.45 $\frac{m}{{s}^{2}}$
Hence putting it in ${v}^{2} = {u}^{2} + 2 a s$ (all symbols are bearing their conventional meaning)
So, average speed = (total distance covered/total time) = $\left(\frac{668.152}{5}\right)$ i.e 133.63 m/s