An object has a mass of 6 kg. The object's kinetic energy uniformly changes from 48 KJ to  225 KJ over t in [0, 8 s]. What is the average speed of the object?

Apr 7, 2017

The average speed is $= 209.2 m {s}^{-} 1$

Explanation:

The kinetic energy is

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

mass is $= 6 k g$

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

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

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

$\frac{1}{2} m {u}_{2}^{2} = 225000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{6} \cdot 48000 = 16000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{6} \cdot 225000 = 75000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 16000\right)$ and $\left(8 , 75000\right)$

The equation of the line is

${v}^{2} - 16000 = \frac{75000 - 16000}{8} t$

${v}^{2} = 7375 t + 16000$

So,

v=sqrt((7575t+16000)

We need to calculate the average value of $v$ over $t \in \left[0 , 8\right]$

$\left(8 - 0\right) \overline{v} = {\int}_{0}^{8} \sqrt{\left(7375 t + 16000\right)} \mathrm{dt}$

8 barv=[((7375t+16000)^(3/2)/(3/2*7375)]_0^8

$= \left({\left(7375 \cdot 8 + 16000\right)}^{\frac{3}{2}} / \left(11062.5\right)\right) - \left({\left(7375 \cdot 0 + 16000\right)}^{\frac{3}{2}} / \left(11062.5\right)\right)$

$= {75000}^{\frac{3}{2}} / 11062.5 - {16000}^{\frac{3}{2}} / 11062.5$

$= 1673.7$

So,

$\overline{v} = \frac{1673.7}{8} = 209.2 m {s}^{-} 1$