An object has a mass of 6 kg. The object's kinetic energy uniformly changes from 540 KJ to  36 KJ over t in [0, 4 s]. What is the average speed of the object?

Apr 10, 2017

The average speed is $= 297.8 m {s}^{-} 1$

Explanation:

The kinetic energy is

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

mass is $= 6 k g$

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

$\frac{1}{2} m {u}_{1}^{2} = 540000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 36000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{6} \cdot 540000 = 180000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{6} \cdot 36000 = 12000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 180000\right)$ and $\left(4 , 12000\right)$

The equation of the line is

${v}^{2} - 180000 = \frac{12000 - 180000}{4} t$

${v}^{2} = - 42000 t + 180000$

So,

v=sqrt((-42000t+180000)

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

$\left(4 - 0\right) \overline{v} = {\int}_{0}^{4} \sqrt{\left(- 42000 t + 180000\right)} \mathrm{dt}$

4 barv=[((-42000t+180000)^(3/2)/(-3/2*42000)]_0^4

$= \left({\left(- 42000 \cdot 4 + 180000\right)}^{\frac{3}{2}} / \left(- 63000\right)\right) - \left({\left(- 42000 \cdot 0 + 180000\right)}^{\frac{3}{2}} / \left(- 63000\right)\right)$

$= {180000}^{\frac{3}{2}} / 63000 - {12000}^{\frac{3}{2}} / 63000$

$= 1191.3$

So,

$\overline{v} = \frac{1191.3}{4} = 297.8 m {s}^{-} 1$