# An object has a mass of 2 kg. The object's kinetic energy uniformly changes from 35 KJ to 18 KJ over t in [0, 15 s]. What is the average speed of the object?

Jan 13, 2018

The average speed is $= 162.1 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

The mass is $m = 2 k g$

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

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

The initial kinetic energy is $\frac{1}{2} m {u}_{1}^{2} = 35000 J$

The final kinetic energy is $\frac{1}{2} m {u}_{2}^{2} = 18000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{2} \cdot 35000 = 35000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{2} \cdot 18000 = 18000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 35000\right)$ and $\left(15 , 18000\right)$

The equation of the line is

${v}^{2} - 35000 = \frac{18000 - 35000}{15} t$

${v}^{2} = - 1133.3 t + 35000$

So,

$v = \sqrt{- 1133.3 t + 35000}$

We need to calculate the average value of $v$ over $t \in \left[0 , 15\right]$

$\left(15 - 0\right) \overline{v} = {\int}_{0}^{15} \left(\sqrt{- 1133.3 t + 35000}\right) \mathrm{dt}$

$15 \overline{v} = {\left(- 1133.3 t + 35000\right)}^{\frac{3}{2}} / \left(\frac{3}{2} \cdot - 1133.3\right) {|}_{0}^{15}$

$= \left({\left(- 1133.3 \cdot 15 + 35000\right)}^{\frac{3}{2}} / \left(- 1700\right)\right) - \left({\left(- 1133.3 \cdot 0 + 35000\right)}^{\frac{3}{2}} / \left(- 1700\right)\right)$

$= {35000}^{\frac{3}{2}} / 1700 - {18000}^{\frac{3}{2}} / 1700$

$= 2431.1$

So,

$\overline{v} = \frac{2431.1}{15} = 162.1 m {s}^{-} 1$

The average speed is $= 162.1 m {s}^{-} 1$