An object has a mass of 12 kg. The object's kinetic energy uniformly changes from 254 KJ to  24 KJ over t in [0, 5 s]. What is the average speed of the object?

Jan 9, 2018

Answer:

The average speed is $= 147.1 m {s}^{-} 1$

Explanation:

The kinetic energy is

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

The mass is $m = 12 k g$

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

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

The initial kinetic energy is $\frac{1}{2} m {u}_{1}^{2} = 254000 J$

The final kinetic energy is $\frac{1}{2} m {u}_{2}^{2} = 24000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{12} \cdot 254000 = 42333.3 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{12} \cdot 24000 = 4000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 42333.3\right)$ and $\left(5 , 4000\right)$

The equation of the line is

${v}^{2} - 42333.3 = \frac{4000 - 42333.3}{5} t$

${v}^{2} = - 7666.7 t + 42333.3$

So,

$v = \sqrt{- 7666.7 t + 42333.3}$

We need to calculate the average value of $v$ over $t \in \left[0 , 5\right]$

$\left(5 - 0\right) \overline{v} = {\int}_{0}^{5} \left(\sqrt{- 7666.7 t + 42333.3}\right) \mathrm{dt}$

$5 \overline{v} = {\left(- 7666.7 t + 42333.3\right)}^{\frac{3}{2}} / \left(\frac{3}{2} \cdot - 7666.7\right) {|}_{0}^{5}$

$= \left({\left(- 7666.7 \cdot 5 + 42333.3\right)}^{\frac{3}{2}} / \left(- 11500\right)\right) - \left({\left(- 7666.7 \cdot 0 + 42333.3\right)}^{\frac{3}{2}} / \left(- 11500\right)\right)$

$= {42333.3}^{\frac{3}{2}} / 11500 - {4000}^{\frac{3}{2}} / 11500$

$= 735.4$

So,

$\overline{v} = \frac{735.4}{5} = 147.1 m {s}^{-} 1$

The average speed is $= 147.1 m {s}^{-} 1$