An object has a mass of 8 kg. The object's kinetic energy uniformly changes from 240 KJ to  640KJ over t in [0, 3 s]. What is the average speed of the object?

Jun 2, 2017

The average speed is $= 328.7 m {s}^{-} 1$

Explanation:

The kinetic energy is

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

mass is $= 8 k g$

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

$\frac{1}{2} m {u}_{1}^{2} = 240000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 640000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{8} \cdot 240000 = 60000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{8} \cdot 640000 = 160000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 60000\right)$ and $\left(3 , 160000\right)$

The equation of the line is

${v}^{2} - 60000 = \frac{160000 - 60000}{3} t$

${v}^{2} = 33333.3 t + 60000$

So,

v=sqrt((33333.3t+60000)

We need to calculate the average value of $v$ over $t \in \left[0 , 3\right]$

$\left(3 - 0\right) \overline{v} = {\int}_{0}^{3} \sqrt{\left(33333.3 t + 60000\right)} \mathrm{dt}$

3 barv=[((33333.3t+60000)^(3/2)/(3/2*33333.3)]_0^3

$= \left({\left(33333.3 \cdot 3 + 60000\right)}^{\frac{3}{2}} / \left(50000\right)\right) - \left({\left(33333.3 \cdot 0 + 60000\right)}^{\frac{3}{2}} / \left(50000\right)\right)$

$= {160000}^{\frac{3}{2}} / 50000 - {60000}^{\frac{3}{2}} / 50000$

$= 986.1$

So,

$\overline{v} = \frac{986.1}{3} = 328.7 m {s}^{-} 1$