# An object has a mass of 3 kg. The object's kinetic energy uniformly changes from 96 KJ to 240 KJ over t in [0, 9 s]. What is the average speed of the object?

May 13, 2017

The average speed is $= 332.01 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 3 k g$

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

$\frac{1}{2} m {u}_{1}^{2} = 96000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 240000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{3} \cdot 96000 = 64000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{3} \cdot 240000 = 160000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 64000\right)$ and $\left(9 , 160000\right)$

The equation of the line is

${v}^{2} - 64000 = \frac{160000 - 64000}{9} t$

${v}^{2} = 10666.67 t + 64000$

So,

v=sqrt((10666.67t+64000)

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

$\left(9 - 0\right) \overline{v} = {\int}_{0}^{9} \sqrt{\left(10666.67 t + 64000\right)} \mathrm{dt}$

9 barv=[((10666.67t+64000)^(3/2)/(3/2*10666.67)]_0^9

$= \left({\left(10666.67 \cdot 9 + 64000\right)}^{\frac{3}{2}} / \left(16000\right)\right) - \left({\left(10666.67 \cdot 0 + 64000\right)}^{\frac{3}{2}} / \left(16000\right)\right)$

$= {160000}^{\frac{3}{2}} / 16000 - {64000}^{\frac{3}{2}} / 16000$

$= 2988.07$

So,

$\overline{v} = \frac{2988.07}{9} = 332.01 m {s}^{-} 1$