# An object has a mass of 12 kg. The object's kinetic energy uniformly changes from 64 KJ to  160 KJ over t in [0, 5 s]. What is the average speed of the object?

May 14, 2017

The average speed is $= 91.79 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 12 k g$

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

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

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

$\frac{1}{2} m {u}_{2}^{2} = 160000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{12} \cdot 64000 = 10666.67 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{12} \cdot 160000 = 26666.67 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 10666.67\right)$ and $\left(5 , 26666.67\right)$

The equation of the line is

${v}^{2} - 10666.67 = \frac{26666.67 - 10666.67}{5} t$

${v}^{2} = 3200 t + 10666.67$

So,

v=sqrt((3200t+10666.67)

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

$\left(5 - 0\right) \overline{v} = {\int}_{0}^{5} \sqrt{\left(3200 t + 10666.67\right)} \mathrm{dt}$

5 barv=[((3200t+10666.67)^(3/2)/(3/2*3200)]_0^5

$= \left({\left(3200 \cdot 5 + 10666.67\right)}^{\frac{3}{2}} / \left(4800\right)\right) - \left({\left(3200 \cdot 0 + 10666.67\right)}^{\frac{3}{2}} / \left(4800\right)\right)$

$= {26666.67}^{\frac{3}{2}} / 4800 - {10666.67}^{\frac{3}{2}} / 4800$

$= 458.96$

So,

$\overline{v} = \frac{458.96}{5} = 91.79 m {s}^{-} 1$