# An object has a mass of 8 kg. The object's kinetic energy uniformly changes from 24 KJ to  42KJ over t in [0, 9 s]. What is the average speed of the object?

Apr 21, 2017

The average speed is $90.5 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 8 k g$

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

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

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

$\frac{1}{2} m {u}_{2}^{2} = 42000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{8} \cdot 24000 = 6000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{8} \cdot 42000 = 10500 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 6000\right)$ and $\left(9 , 10500\right)$

The equation of the line is

${v}^{2} - 6000 = \frac{10500 - 6000}{9} t$

${v}^{2} = 500 t + 6000$

So,

v=sqrt((500t+6000)

We need to calculate the average value of $v$ over $t \in \left[0 , 9\right]$

$\left(9 - 0\right) \overline{v} = {\int}_{0}^{9} \sqrt{\left(500 t + 6000\right)} \mathrm{dt}$

9 barv=[((500t+6000)^(3/2)/(3/2*500)]_0^9

$= \left({\left(500 \cdot 9 + 6000\right)}^{\frac{3}{2}} / \left(750\right)\right) - \left({\left(500 \cdot 0 + 6000\right)}^{\frac{3}{2}} / \left(750\right)\right)$

$= {10500}^{\frac{3}{2}} / 750 - {6000}^{\frac{3}{2}} / 750$

$= 814.9$

So,

$\overline{v} = \frac{814.9}{9} = 90.5 m {s}^{-} 1$