# An object has a mass of 3 kg. The object's kinetic energy uniformly changes from 75 KJ to 48 KJ over t in [0, 9 s]. What is the average speed of the object?

May 5, 2017

#### Answer:

The average speed is $= 202.1 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 3 k g$

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

$\frac{1}{2} m {u}_{1}^{2} = 75000 J$

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

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

Therefore,

${u}_{1}^{2} = \frac{2}{3} \cdot 75000 = 50000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{3} \cdot 48000 = 32000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 50000\right)$ and $\left(9 , 32000\right)$

The equation of the line is

${v}^{2} - 50000 = \frac{32000 - 50000}{9} t$

${v}^{2} = - 2000 t + 50000$

So,

v=sqrt((-2000t+50000)

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

$\left(9 - 0\right) \overline{v} = {\int}_{0}^{9} \sqrt{\left(- 2000 t + 50000\right)} \mathrm{dt}$

9 barv=[((-2000t+50000)^(3/2)/(-3/2*2000)]_0^9

$= \left({\left(- 2000 \cdot 9 + 50000\right)}^{\frac{3}{2}} / \left(- 3000\right)\right) - \left({\left(- 2000 \cdot 0 + 50000\right)}^{\frac{3}{2}} / \left(- 3000\right)\right)$

$= {50000}^{\frac{3}{2}} / 3000 - {32000}^{\frac{3}{2}} / 3000$

$= 1818.7$

So,

$\overline{v} = \frac{1818.7}{9} = 202.1 m {s}^{-} 1$