An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 144 KJ to  120KJ over t in [0, 2 s]. What is the average speed of the object?

Aug 14, 2017

The average speed is $= 256.8 m {s}^{-} 1$

Explanation:

The kinetic energy is

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

The mass is $= 4 k g$

The initial velocity is $= {u}_{1} m {s}^{-} 1$

The final velocity is $= {u}_{2} m {s}^{-} 1$

The initial kinetic energy is $\frac{1}{2} m {u}_{1}^{2} = 144000 J$

The final kinetic energy is $\frac{1}{2} m {u}_{2}^{2} = 120000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{4} \cdot 144000 = 72000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{4} \cdot 120000 = 60000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 72000\right)$ and $\left(2 , 60000\right)$

The equation of the line is

${v}^{2} - 72000 = \frac{60000 - 72000}{2} t$

${v}^{2} = - 6000 t + 72000$

So,

v=sqrt((-6000t+72000)

We need to calculate the average value of $v$ over $t \in \left[0 , 2\right]$

$\left(2 - 0\right) \overline{v} = {\int}_{0}^{2} \left(\sqrt{- 6000 t + 72000}\right) \mathrm{dt}$

$2 \overline{v} = {\left[\left({\left(- 6000 t + 72000\right)}^{\frac{3}{2}} / \left(\frac{3}{2} \cdot - 6000\right)\right)\right]}_{0}^{2}$

$= \left({\left(- 6000 \cdot 2 + 72000\right)}^{\frac{3}{2}} / \left(- 9000\right)\right) - \left({\left(- 6000 \cdot 0 + 72000\right)}^{\frac{3}{2}} / \left(- 9000\right)\right)$

$= {72000}^{\frac{3}{2}} / 9000 - {60000}^{\frac{3}{2}} / 9000$

$= 513.6$

So,

$\overline{v} = \frac{513.6}{2} = 256.8 m {s}^{-} 1$

The average speed is $= 256.8 m {s}^{-} 1$