# An object has a mass of 2 kg. The object's kinetic energy uniformly changes from 48 KJ to  54KJ over t in [0,12s]. What is the average speed of the object?

May 6, 2017

The average speed is $= 159.7 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 2 k g$

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

$\frac{1}{2} m {u}_{1}^{2} = 48000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 54000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{4} \cdot 48000 = 24000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{4} \cdot 54000 = 27000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 24000\right)$ and $\left(12 , 27000\right)$

The equation of the line is

${v}^{2} - 24000 = \frac{27000 - 24000}{12} t$

${v}^{2} = 250 t + 24000$

So,

v=sqrt((250t+24000)

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

$\left(12 - 0\right) \overline{v} = {\int}_{0}^{12} \sqrt{\left(250 t + 24000\right)} \mathrm{dt}$

12 barv=[((250t+24000)^(3/2)/(3/2*250)]_0^12

$= \left({\left(250 \cdot 12 + 24000\right)}^{\frac{3}{2}} / \left(375\right)\right) - \left({\left(250 \cdot 0 + 24000\right)}^{\frac{3}{2}} / \left(375\right)\right)$

$= {27000}^{\frac{3}{2}} / 375 - {24000}^{\frac{3}{2}} / 375$

$= 1916$

So,

$\overline{v} = \frac{1916}{12} = 159.7 m {s}^{-} 1$