# An object has a mass of 3 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?

May 5, 2017

The average speed is $= 163.3 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 3 k g$

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

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

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

$\frac{1}{2} m {u}_{2}^{2} = 72000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{3} \cdot 12000 = 8000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{3} \cdot 72000 = 48000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 8000\right)$ and $\left(5 , 48000\right)$

The equation of the line is

${v}^{2} - 8000 = \frac{48000 - 8000}{5} t$

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

So,

v=sqrt((8000t+8000)

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

$\left(5 - 0\right) \overline{v} = {\int}_{0}^{4} \sqrt{\left(8000 t + 8000\right)} \mathrm{dt}$

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

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

$= {48000}^{\frac{3}{2}} / 12000 - {8000}^{\frac{3}{2}} / 12000$

$= 816.7$

So,

$\overline{v} = \frac{816.7}{5} = 163.3 m {s}^{-} 1$