# An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 128 KJ to  48 KJ over t in [0, 12 s]. What is the average speed of the object?

Jun 1, 2017

#### Answer:

The average speed is $= 207.9 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 4 k g$

The initial velocity is $= {u}_{1}$

$\frac{1}{2} m {u}_{1}^{2} = 128000 J$

The final velocity is $= {u}_{2}$

$\frac{1}{2} m {u}_{2}^{2} = 48000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{4} \cdot 128000 = 64000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{4} \cdot 48000 = 24000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 64000\right)$ and $\left(12 , 24000\right)$

The equation of the line is

${v}^{2} - 64000 = \frac{24000 - 64000}{12} t$

${v}^{2} = - 3333.3 t + 64000$

So,

v=sqrt((-3333.3t+6400)

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

$\left(12 - 0\right) \overline{v} = {\int}_{0}^{12} \sqrt{\left(- 3333.3 t + 64000\right)} \mathrm{dt}$

12 barv=[((-3333.3t+64000)^(3/2)/(3/2*3333.3)]_0^12

$= \left({\left(- 3333.3 \cdot 12 + 64000\right)}^{\frac{3}{2}} / \left(5000\right)\right) - \left({\left(- 3333.3 \cdot 0 + 64000\right)}^{\frac{3}{2}} / \left(5000\right)\right)$

$= {64000}^{\frac{3}{2}} / 5000 - {24000}^{\frac{3}{2}} / 5000$

$= 2494.6$

So,

$\overline{v} = \frac{2494.6}{12} = 207.9 m {s}^{-} 1$