# An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 96 KJ to  270 KJ over t in [0, 12 s]. What is the average speed of the object?

May 5, 2017

The average speed is $= 299.5 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 4 k g$

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

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

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

$\frac{1}{2} m {u}_{2}^{2} = 270000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{4} \cdot 96000 = 48000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{4} \cdot 270000 = 135000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 48000\right)$ and $\left(12 , 135000\right)$

The equation of the line is

${v}^{2} - 48000 = \frac{135000 - 48000}{12} t$

${v}^{2} = 7250 t + 48000$

So,

v=sqrt((7250t+48000)

We need to calculate the average value of $v$ over $t \in \left[0 , 12\right]$

$\left(12 - 0\right) \overline{v} = {\int}_{0}^{12} \sqrt{\left(7250 t + 48000\right)} \mathrm{dt}$

12 barv=[((7250t+48000)^(3/2)/(3/2*7250)]_0^12

$= \left({\left(7250 \cdot 12 + 48000\right)}^{\frac{3}{2}} / \left(10875\right)\right) - \left({\left(7250 \cdot 0 + 48000\right)}^{\frac{3}{2}} / \left(10875\right)\right)$

$= {135000}^{\frac{3}{2}} / 10875 - {48000}^{\frac{3}{2}} / 10875$

$= 3594.1$

So,

$\overline{v} = \frac{3594.1}{12} = 299.5 m {s}^{-} 1$