# An object has a mass of 2 kg. The object's kinetic energy uniformly changes from 32 KJ to 80 KJ over t in [0, 4 s]. What is the average speed of the object?

Aug 16, 2017

The average speed is $= 234.8 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

The mass is $= 2 k g$

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

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

The initial kinetic energy is $\frac{1}{2} m {u}_{1}^{2} = 32000 J$

The final kinetic energy is $\frac{1}{2} m {u}_{2}^{2} = 80000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{2} \cdot 32000 = 32000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{2} \cdot 80000 = 80000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 32000\right)$ and $\left(4 , 80000\right)$

The equation of the line is

${v}^{2} - 32000 = \frac{80000 - 32000}{4} t$

${v}^{2} = 12000 t + 32000$

So,

v=sqrt((12000t+32000)

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

$\left(4 - 0\right) \overline{v} = {\int}_{0}^{4} \left(\sqrt{12000 t + 32000}\right) \mathrm{dt}$

$4 \overline{v} = {\left[\left({\left(12000 t + 32000\right)}^{\frac{3}{2}} / \left(\frac{3}{2} \cdot 12000\right)\right)\right]}_{0}^{4}$

$= \left({\left(12000 \cdot 4 + 32000\right)}^{\frac{3}{2}} / \left(18000\right)\right) - \left({\left(12000 \cdot 0 + 32000\right)}^{\frac{3}{2}} / \left(18000\right)\right)$

$= {80000}^{\frac{3}{2}} / 18000 - {32000}^{\frac{3}{2}} / 18000$

$= 939.1$

So,

$\overline{v} = \frac{939.1}{4} = 234.8 m {s}^{-} 1$

The average speed is $= 234.8 m {s}^{-} 1$