# An object has a mass of 5 kg. The object's kinetic energy uniformly changes from 55 KJ to  24 KJ over t in [0,4s]. What is the average speed of the object?

Apr 20, 2017

The average speed is $= 124.9 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 5 k g$

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

$\frac{1}{2} m {u}_{1}^{2} = 55000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 24000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{5} \cdot 55000 = 22000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{5} \cdot 24000 = 9600 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 22000\right)$ and $\left(4 , 9600\right)$

The equation of the line is

${v}^{2} - 22000 = \frac{9600 - 22000}{4} t$

${v}^{2} = - 3100 t + 22000$

So,

v=sqrt((-3100t+22000)

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

$\left(4 - 0\right) \overline{v} = {\int}_{0}^{4} \sqrt{\left(- 3100 t + 22000\right)} \mathrm{dt}$

4 barv=[((-3100t+22000)^(3/2)/(-3/2*3100)]_0^4

$= \left({\left(- 3100 \cdot 4 + 22000\right)}^{\frac{3}{2}} / \left(- 4650\right)\right) - \left({\left(- 3100 \cdot 0 + 22000\right)}^{\frac{3}{2}} / \left(- 4650\right)\right)$

$= {22000}^{\frac{3}{2}} / 4650 - {9600}^{\frac{3}{2}} / 4650$

$= 499.5$

So,

$\overline{v} = \frac{499.5}{4} = 124.9 m {s}^{-} 1$