# An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 12 KJ to  360 KJ over t in [0, 12 s]. What is the average speed of the object?

Aug 17, 2017

The average speed is $= 290.8 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

The mass is $= 4 k g$

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

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

The initial kinetic energy is $\frac{1}{2} m {u}_{1}^{2} = 12000 J$

The final kinetic energy is $\frac{1}{2} m {u}_{2}^{2} = 360000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{4} \cdot 12000 = 6000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{4} \cdot 360000 = 180000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 6000\right)$ and $\left(12 , 180000\right)$

The equation of the line is

${v}^{2} - 6000 = \frac{180000 - 6000}{12} t$

${v}^{2} = 14500 t + 6000$

So,

v=sqrt((14500t+6000)

We need to calculate the average value of $v$ over $t \in \left[0 , 12\right]$

$\left(12 - 0\right) \overline{v} = {\int}_{0}^{12} \left(\sqrt{14500 t + 6000}\right) \mathrm{dt}$

$12 \overline{v} = {\left[\left({\left(14500 t + 6000\right)}^{\frac{3}{2}} / \left(\frac{3}{2} \cdot 14500\right)\right)\right]}_{0}^{12}$

$= \left({\left(14500 \cdot 12 + 6000\right)}^{\frac{3}{2}} / \left(21750\right)\right) - \left({\left(14500 \cdot 0 + 6000\right)}^{\frac{3}{2}} / \left(21750\right)\right)$

$= {180000}^{\frac{3}{2}} / 21750 - {6000}^{\frac{3}{2}} / 21750$

$= 3489.8$

So,

$\overline{v} = \frac{3489.8}{12} = 290.8 m {s}^{-} 1$

The average speed is $= 290.8 m {s}^{-} 1$