# An object has a mass of 12 kg. The object's kinetic energy uniformly changes from 81 KJ to  180 KJ over t in [0, 4 s]. What is the average speed of the object?

Apr 22, 2017

The average speed is $= 146.6 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 12 k g$

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

$\frac{1}{2} m {u}_{1}^{2} = 81000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 180000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{12} \cdot 81000 = 13500 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{12} \cdot 180000 = 30000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 13500\right)$ and $\left(4 , 30000\right)$

The equation of the line is

${v}^{2} - 13500 = \frac{30000 - 13500}{4} t$

${v}^{2} = 4125 t + 13500$

So,

v=sqrt((4125t+13500)

We need to calculate the average value of $v$ over $t \in \left[0 , 4\right]$

$\left(4 - 0\right) \overline{v} = {\int}_{0}^{4} \sqrt{\left(4125 t + 13500\right)} \mathrm{dt}$

4 barv=[((4125t+13500)^(3/2)/(3/2*4125)]_0^4

$= \left({\left(4125 \cdot 4 + 13500\right)}^{\frac{3}{2}} / \left(6187.5\right)\right) - \left({\left(4125 \cdot 0 + 13500\right)}^{\frac{3}{2}} / \left(6187.5\right)\right)$

$= {30000}^{\frac{3}{2}} / 6187.5 - {13500}^{\frac{3}{2}} / 6187.5$

$= 586.3$

So,

$\overline{v} = \frac{586.3}{4} = 146.6 m {s}^{-} 1$