# An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 104 KJ to  64KJ over t in [0,5s]. What is the average speed of the object?

Apr 5, 2017

The average speed is $= 204.4 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 4 k g$

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

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

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

$\frac{1}{2} m {u}_{2}^{2} = 64000 J$

Therefore,

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

and,

${u}_{2}^{2} = \frac{2}{4} \cdot 64000 = 32000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 52000\right)$ and $\left(5 , 32000\right)$

The equation of the line is

${v}^{2} - 52000 = \frac{32000 - 52000}{5} t$

${v}^{2} = - 4000 t + 52000$

So,

v=sqrt((-4000t+52000)

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

$\left(5 - 0\right) \overline{v} = {\int}_{0}^{5} \sqrt{\left(- 4000 t + 52000\right)} \mathrm{dt}$

5 barv=[((-4000t+52000)^(3/2)/(-3/2*4000)]_0^5

$= \left({\left(- 4000 \cdot 5 + 52000\right)}^{\frac{3}{2}} / \left(- 6000\right)\right) - \left({\left(- 4000 \cdot 0 + 52000\right)}^{\frac{3}{2}} / \left(- 6000\right)\right)$

$= - {32000}^{\frac{3}{2}} / 6000 + {52000}^{\frac{3}{2}} / 6000$

$= 1022.2$

So,

$\overline{v} = \frac{1022.2}{5} = 204.4 m {s}^{-} 1$