An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 18 KJ to  4KJ over t in [0, 9 s]. What is the average speed of the object?

Apr 4, 2017

The average speed of the object is $= 72.8 m {s}^{-} 1$

Explanation:

The kinetic energy is

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

mass is $= 4 k g$

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

$\frac{1}{2} m {u}_{1}^{2} = 18000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 4000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{4} \cdot 18000 = 9000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{4} \cdot 4000 = 2000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 9000\right)$ and $\left(9 , 2000\right)$

The equation of the line is

${v}^{2} - 9000 = \frac{2000 - 9000}{9} t$

${v}^{2} = - 777.8 t + 9000$

So,

v=sqrt((-777.8t+9000)

We need to calculate the average value of $v$ over $t \in \left[0 , 9\right]$

$\left(9 - 0\right) \overline{v} = {\int}_{0}^{9} \sqrt{\left(- 777.8 t + 9000\right)} \mathrm{dt}$

9 barv=[((-777.8t+9000)^(3/2)/(-3/2*777.8)]_0^9

$= \left({\left(- 777.8 \cdot 9 + 9000\right)}^{\frac{3}{2}} / \left(- 1166.7\right)\right) - \left({\left(- 777.8 \cdot 0 + 9000\right)}^{\frac{3}{2}} / \left(- 1166.7\right)\right)$

$= - {2000}^{\frac{3}{2}} / 1166.7 + {9000}^{\frac{3}{2}} / 1166.7$

$= 655.2$

So,

$\overline{v} = \frac{655.2}{9} = 72.8 m {s}^{-} 1$