# An object has a mass of 1 kg. The object's kinetic energy uniformly changes from 48 KJ to 12 KJ over t in [0, 3 s]. What is the average speed of the object?

May 13, 2017

The average speed is $= 240.99 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 1 k g$

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

$\frac{1}{2} m {u}_{1}^{2} = 48000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 12000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{1} \cdot 48000 = 96000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{1} \cdot 120000 = 24000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 96000\right)$ and $\left(3 , 24000\right)$

The equation of the line is

${v}^{2} - 96000 = \frac{24000 - 96000}{3} t$

${v}^{2} = - 24000 t + 96000$

So,

v=sqrt((-24000t+96000)

We need to calculate the average value of $v$ over $t \in \left[0 , 3\right]$

$\left(3 - 0\right) \overline{v} = {\int}_{0}^{3} \sqrt{\left(- 24000 t + 96000\right)} \mathrm{dt}$

3 barv=[((-24000t+96000)^(3/2)/(-3/2*24000)]_0^3

$= \left({\left(- 24000 \cdot 3 + 96000\right)}^{\frac{3}{2}} / \left(- 36000\right)\right) - \left({\left(- 24000 \cdot 0 + 96000\right)}^{\frac{3}{2}} / \left(- 36000\right)\right)$

$= {96000}^{\frac{3}{2}} / 36000 - {24000}^{\frac{3}{2}} / 36000$

$= 722.96$

So,

$\overline{v} = \frac{722.96}{3} = 240.99 m {s}^{-} 1$