# An object has a mass of 5 kg. The object's kinetic energy uniformly changes from 15 KJ to  42KJ over t in [0,8s]. What is the average speed of the object?

May 14, 2017

The average speed is $= 105.73 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 5 k g$

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

$\frac{1}{2} m {u}_{1}^{2} = 15000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 42000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{5} \cdot 15000 = 6000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{5} \cdot 42000 = 16800 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 6000\right)$ and $\left(8 , 16800\right)$

The equation of the line is

${v}^{2} - 6000 = \frac{16800 - 6000}{8} t$

${v}^{2} = 1350 t + 6000$

So,

v=sqrt((1350t+6000)

We need to calculate the average value of $v$ over $t \in \left[0 , 8\right]$

$\left(8 - 0\right) \overline{v} = {\int}_{0}^{8} \sqrt{\left(3200 t + 10666.67\right)} \mathrm{dt}$

8 barv=[((1350t+6000)^(3/2)/(3/2*1350)]_0^8

$= \left({\left(1350 \cdot 8 + 6000\right)}^{\frac{3}{2}} / \left(2025\right)\right) - \left({\left(1350 \cdot 0 + 6000\right)}^{\frac{3}{2}} / \left(2025\right)\right)$

$= {16800}^{\frac{3}{2}} / 2025 - {6000}^{\frac{3}{2}} / 2025$

$= 845.81$

So,

$\overline{v} = \frac{845.81}{8} = 105.73 m {s}^{-} 1$