An object has a mass of 9 kg. The object's kinetic energy uniformly changes from 135 KJ to  36KJ over t in [0, 4 s]. What is the average speed of the object?

Mar 14, 2017

The average speed is $= 135.8 m {s}^{-} 1$

Explanation:

The kinetic energy is

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

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

$\frac{1}{2} m {u}_{1}^{2} = 135000 J$

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

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

Therefore,

${u}_{1}^{2} = \frac{2}{9} \cdot 135000 = 30000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{9} \cdot 36000 = 8000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 30000\right)$ and $\left(4 , 8000\right)$

The equation of the line is

${v}^{2} - 30000 = \frac{8000 - 30000}{4} t$

${v}^{2} = - 5500 t + 30000$

So,

v=sqrt((-5500t+30000)

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

$\left(4 - 0\right) \overline{v} = {\int}_{0}^{5} \sqrt{- 5500 t + 30000} \mathrm{dt}$

$4 \overline{v} = {\left[{\left(- 5500 t + 30000\right)}^{\frac{3}{2}} / \left(\frac{3}{2} \cdot - 5500\right)\right]}_{0}^{4}$

$= \left({\left(- 5500 \cdot 4 + 30000\right)}^{\frac{3}{2}} / \left(- 8250\right)\right) - \left({\left(- 5500 \cdot 0 + 30000\right)}^{\frac{3}{2}} / \left(- 8250\right)\right)$

$= {8000}^{\frac{3}{2}} / - 8250 - {30000}^{\frac{3}{2}} / - 8250$

=1/8250(30000^(3/2)-8000^(3/2)))

$= 543.1$

So,

$\overline{v} = \frac{543.1}{4} = 135.8 m {s}^{-} 1$