# An object has a mass of 2 kg. The object's kinetic energy uniformly changes from 64 KJ to 28 KJ over t in [0, 15 s]. What is the average speed of the object?

Jan 6, 2018

#### Answer:

The average speed is $= 319.6 m {s}^{-} 1$

#### 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 $\frac{1}{2} m {u}_{1}^{2} = 64000 J$

The final kinetic energy is $\frac{1}{2} m {u}_{2}^{2} = 28000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{2} \cdot 64000 = 64000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{2} \cdot 28000 = 28000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 64000\right)$ and $\left(15 , 28000\right)$

The equation of the line is

${v}^{2} - 64000 = \frac{28000 - 64000}{15} t$

${v}^{2} = - 2400 t + 64000$

So,

$v = \sqrt{- 2400 t + 64000}$

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

$\left(15 - 0\right) \overline{v} = {\int}_{0}^{15} \left(\sqrt{- 2400 t + 64000}\right) \mathrm{dt}$

$15 \overline{v} = {\left(- 2400 t + 64000\right)}^{\frac{3}{2}} / \left(\frac{3}{2} \cdot - 2400\right) {|}_{0}^{15}$

$= \left({\left(- 2400 \cdot 15 + 64000\right)}^{\frac{3}{2}} / \left(- 2400\right)\right) - \left({\left(- 2400 \cdot 0 + 64000\right)}^{\frac{3}{2}} / \left(- 2400\right)\right)$

$= {64000}^{\frac{3}{2}} / 2400 - {28000}^{\frac{3}{2}} / 2400$

$= 4794$

So,

$\overline{v} = \frac{4794}{15} = 319.6 m {s}^{-} 1$

The average speed is $= 319.6 m {s}^{-} 1$