# An object has a mass of 12 kg. The object's kinetic energy uniformly changes from 64 KJ to  140 KJ over t in [0, 5 s]. What is the average speed of the object?

Mar 13, 2017

The average speed is $= 129.6 m {s}^{-} 1$

#### Explanation:

The kinetis energy is

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

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

$\frac{1}{2} m {u}_{1}^{2} = 64000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 140000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{12} \cdot 64000 = 10666.7 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{12} \cdot 140000 = 23333.3 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 1066.7\right)$ and $\left(5 , 23333.3\right)$

The equation of the line is

${v}^{2} - 10666.7 = \frac{23333.3 - 10666.7}{5} t$

${v}^{2} = \frac{12666.6}{5} t + 10666.7$

So,

$v = \sqrt{\frac{12666.6}{5} t + 10666.7}$

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

$\left(5 - 0\right) \overline{v} = {\int}_{0}^{5} \sqrt{\frac{12666.6}{5} t + 10666.7} \mathrm{dt}$

$5 \overline{v} = {\left[{\left(\frac{12666.6}{5} t + 10666.7\right)}^{\frac{3}{2}} / \left(\frac{3}{2} \cdot \frac{12666.6}{5}\right)\right]}_{0}^{5}$

$= \left({\left(\frac{12666.6}{5} \cdot 5 + 10666.7\right)}^{\frac{3}{2}} / \left(3800\right)\right) - \left({\left(\frac{12666.6}{5} \cdot 0 + 10666.7\right)}^{\frac{3}{2}} / \left(3800\right)\right)$

$= {23333.3}^{\frac{3}{2}} / 3800 - {10666.7}^{\frac{3}{2}} / 3800$

$= 648$

So,

$\overline{v} = \frac{648}{5} = 129.6 m {s}^{-} 1$