# An object has a mass of 3 kg. The object's kinetic energy uniformly changes from 12 KJ to  72KJ over t in [0,6s]. What is the average speed of the object?

Apr 3, 2018

The average speed is $= 163.3 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

The mass is $m = 3 k g$

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

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

The initial kinetic energy is $\frac{1}{2} m {u}_{1}^{2} = 12000 J$

The final kinetic energy is $\frac{1}{2} m {u}_{2}^{2} = 72000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{3} \cdot 12000 = 8000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{3} \cdot 72000 = 48000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 8000\right)$ and $\left(6 , 48000\right)$

The equation of the line is

${v}^{2} - 8000 = \frac{48000 - 8000}{6} t$

${v}^{2} = 6666.7 t + 8000$

So,

$v = \sqrt{6666.7 t + 8000}$

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

$\left(6 - 0\right) \overline{v} = {\int}_{0}^{6} \left(\sqrt{6666.7 t + 8000}\right) \mathrm{dt}$

$6 \overline{v} = {\left[{\left(6666.7 t + 8000\right)}^{\frac{3}{2}} / \left(\frac{3}{2} \cdot 6666.7\right)\right]}_{0}^{6}$

$= \left({\left(6666.7 \cdot 6 + 8000\right)}^{\frac{3}{2}} / \left(10000\right)\right) - \left({\left(6666.7 \cdot 0 + 8000\right)}^{\frac{3}{2}} / \left(10000\right)\right)$

$= {48000}^{\frac{3}{2}} / 10000 - {8000}^{\frac{3}{2}} / 10000$

$= 980.07$

So,

$\overline{v} = \frac{980.07}{6} = 163.3 m {s}^{-} 1$

The average speed is $= 163.3 m {s}^{-} 1$