# An object has a mass of 6 kg. The object's kinetic energy uniformly changes from 120 KJ to  720 KJ over t in [0, 4 s]. What is the average speed of the object?

Jul 21, 2017

The average speed is $= 365.3 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

The mass is $= 6 k g$

The initial velocity is $= {u}_{1} m {s}^{-} 1$

The final velocity is $= {u}_{2} m {s}^{-} 1$

The initial kinetic energy is $\frac{1}{2} m {u}_{1}^{2} = 120000 J$

The final kinetic energy is $\frac{1}{2} m {u}_{2}^{2} = 720000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{6} \cdot 120000 = 40000 {m}^{2} {s}^{-} 2$

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 $\left(0 , 40000\right)$ and $\left(4 , 240000\right)$

The equation of the line is

${v}^{2} - 40000 = \frac{240000 - 40000}{4} t$

${v}^{2} = 50000 t + 40000$

So,

v=sqrt((50000t+40000)

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

$\left(4 - 0\right) \overline{v} = {\int}_{0}^{4} \left(\sqrt{50000 t + 40000}\right) \mathrm{dt}$

$4 \overline{v} = {\left[\left({\left(50000 t + 40000\right)}^{\frac{3}{2}} / \left(\frac{3}{2} \cdot 50000\right)\right)\right]}_{0}^{4}$

$= \left({\left(50000 \cdot 4 + 40000\right)}^{\frac{3}{2}} / \left(75000\right)\right) - \left({\left(50000 \cdot 0 + 40000\right)}^{\frac{3}{2}} / \left(75000\right)\right)$

$= {240000}^{\frac{3}{2}} / 75000 - {40000}^{\frac{3}{2}} / 75000$

$= 1461$

So,

$\overline{v} = \frac{1461}{4} = 365.3 m {s}^{-} 1$

The average speed is $= 365.3 m {s}^{-} 1$