# An object has a mass of 5 kg. The object's kinetic energy uniformly changes from 36 KJ to 135 KJ over t in [0, 9 s]. What is the average speed of the object?

May 25, 2017

The average speed is $= 182.2 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 5 k g$

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

$\frac{1}{2} m {u}_{1}^{2} = 36000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 135000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{5} \cdot 36000 = 14400 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{5} \cdot 135000 = 54000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 14400\right)$ and $\left(9 , 54000\right)$

The equation of the line is

${v}^{2} - 14400 = \frac{54000 - 14400}{9} t$

${v}^{2} = 4400 t + 14400$

So,

v=sqrt((4400t+14400)

We need to calculate the average value of $v$ over $t \in \left[0 , 9\right]$

$\left(9 - 0\right) \overline{v} = {\int}_{0}^{9} \sqrt{\left(4400 t + 14400\right)} \mathrm{dt}$

9 barv=[((4400t+14400)^(3/2)/(3/2*4400)]_0^9

$= \left({\left(4400 \cdot 9 + 14400\right)}^{\frac{3}{2}} / \left(6600\right)\right) - \left({\left(4400 \cdot 0 + 14400\right)}^{\frac{3}{2}} / \left(6600\right)\right)$

$= {54000}^{\frac{3}{2}} / 6600 - {14400}^{\frac{3}{2}} / 6600$

$= 1639.5$

So,

$\overline{v} = \frac{1639.5}{9} = 182.2 m {s}^{-} 1$