# An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 18 KJ to  48KJ over t in [0, 9 s]. What is the average speed of the object?

Oct 28, 2017

The average speed is $= 127.3 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

The mass is $m = 4 k g$

The initial velocity is $= {u}_{1} m {s}^{-} 1$

The final velocity is $= {u}_{2} m {s}^{-} 1$

The initial kinetic energy is $\frac{1}{2} m {u}_{1}^{2} = 18000 J$

The final kinetic energy is $\frac{1}{2} m {u}_{2}^{2} = 48000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{4} \cdot 18000 = 9000 {m}^{2} {s}^{-} 2$

and,

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

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

The points are $\left(0 , 9000\right)$ and $\left(9 , 24000\right)$

The equation of the line is

${v}^{2} - 9000 = \frac{24000 - 9000}{9} t$

${v}^{2} = 1666.7 t + 9000$

So,

v=sqrt((1666.7t+9000)

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

$\left(9 - 0\right) \overline{v} = {\int}_{0}^{12} \left(\sqrt{1666.7 t + 9000}\right) \mathrm{dt}$

$9 \overline{v} = {\left[\left({\left(1666.7 t + 9000\right)}^{\frac{3}{2}} / \left(\frac{3}{2} \cdot 1666.7\right)\right)\right]}_{0}^{9}$

$= \left({\left(1666.7 \cdot 9 + 9000\right)}^{\frac{3}{2}} / \left(2500\right)\right) - \left({\left(- 1666.7 \cdot 0 + 9000\right)}^{\frac{3}{2}} / \left(2500\right)\right)$

$= {24000}^{\frac{3}{2}} / 2500 - {9000}^{\frac{3}{2}} / 2500$

$= 1145.7$

So,

$\overline{v} = \frac{1145.7}{9} = 127.3 m {s}^{-} 1$

The average speed is $= 127.3 m {s}^{-} 1$