# An object has a mass of 6 kg. The object's kinetic energy uniformly changes from 128 KJ to  720 KJ over t in [0, 3 s]. What is the average speed of the object?

May 21, 2017

#### Answer:

The average speed is $= 367.4 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 6 k g$

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

$\frac{1}{2} m {u}_{1}^{2} = 128000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 720000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{6} \cdot 128000 = 42666.7 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{6} \cdot 720000 = 240000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 42666.7\right)$ and $\left(3 , 240000\right)$

The equation of the line is

${v}^{2} - 42666.7 = \frac{240000 - 42666.7}{3} t$

${v}^{2} = 65777.8 t + 42666.7$

So,

v=sqrt((65777.8t+42666.7)

We need to calculate the average value of $v$ over $t \in \left[0 , 3\right]$

$\left(3 - 0\right) \overline{v} = {\int}_{0}^{8} \sqrt{\left(65777.8 t + 42666.7\right)} \mathrm{dt}$

3 barv=[((65777.8t+42666.7)^(3/2)/(3/2*65666.7)]_0^3

$= \left({\left(65777.8 \cdot 3 + 42666.7\right)}^{\frac{3}{2}} / \left(98666.7\right)\right) - \left({\left(65777.8 \cdot 0 + 42666.7\right)}^{\frac{3}{2}} / \left(98666.7\right)\right)$

$= {240000}^{\frac{3}{2}} / 98666.7 - {42666.7}^{\frac{3}{2}} / 98666.7$

$= 1102.3$

So,

$\overline{v} = \frac{1103.2}{3} = 367.4 m {s}^{-} 1$