# An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 128 KJ to  48 KJ over t in [0, 5 s]. What is the average speed of the object?

Jun 12, 2017

The average speed is $= 207.9 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

The mass is $= 4 k g$

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

The initial kinetic energy is $\frac{1}{2} m {u}_{1}^{2} = 128000 J$

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

The final kinetic energy is $\frac{1}{2} m {u}_{2}^{2} = 48000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{4} \cdot 128000 = 64000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{4} \cdot 48000 = 24000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 64000\right)$ and $\left(5 , 24000\right)$

The equation of the line is

${v}^{2} - 64000 = \frac{24000 - 64000}{5} t$

${v}^{2} = - 8000 t + 64000$

So,

v=sqrt((-8000t+64000)

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

$\left(5 - 0\right) \overline{v} = {\int}_{0}^{5} \sqrt{\left(- 8000 t + 64000\right)} \mathrm{dt}$

5 barv=[((-8000t+64000)^(3/2)/(3/2*8000)]_0^5

$= \left({\left(- 8000 \cdot 5 + 64000\right)}^{\frac{3}{2}} / \left(12000\right)\right) - \left({\left(- 8000 \cdot 0 + 64000\right)}^{\frac{3}{2}} / \left(12000\right)\right)$

$= {64000}^{\frac{3}{2}} / 12000 - {24000}^{\frac{3}{2}} / 12000$

$= 1039.4$

So,

$\overline{v} = \frac{1039.4}{5} = 207.9 m {s}^{-} 1$

The average speed is $= 207.9 m {s}^{-} 1$