# An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 16 KJ to  66KJ over t in [0, 6 s]. What is the average speed of the object?

Jun 15, 2018

The average speed is $= 140.8 m {s}^{-} 1$

#### 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 $\frac{1}{2} m {u}_{1}^{2} = 16000 J$

The final kinetic energy is $\frac{1}{2} m {u}_{2}^{2} = 66000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{4} \cdot 16000 = 8000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{4} \cdot 66000 = 33000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 8000\right)$ and $\left(6 , 33000\right)$

The equation of the line is

${v}^{2} - 8000 = \frac{33000 - 8000}{6} t$

${v}^{2} = 4166.7 t + 8000$

So,

$v = \sqrt{4166.7 t + 8000}$

We need to calculate the average value of $v$ over $t \in \left[0 , 6\right]$

$\left(6 - 0\right) \overline{v} = {\int}_{0}^{6} \left(\sqrt{4166.7 t + 8000}\right) \mathrm{dt}$

$6 \overline{v} = {\left[{\left(4166.7 t + 8000\right)}^{\frac{3}{2}} / \left(\frac{3}{2} \cdot 4166.7\right)\right]}_{0}^{6}$

$= \left({\left(4166.7 \cdot 6 + 8000\right)}^{\frac{3}{2}} / \left(6250\right)\right) - \left({\left(4166.7 \cdot 0 + 8000\right)}^{\frac{3}{2}} / \left(6250\right)\right)$

$= {33000}^{\frac{3}{2}} / 6250 - {8000}^{\frac{3}{2}} / 6250$

$= 844.67$

So,

$\overline{v} = \frac{844.67}{6} = 140.8 m {s}^{-} 1$

The average speed is $= 140.8 m {s}^{-} 1$