# An object has a mass of 3 kg. The object's kinetic energy uniformly changes from 48 KJ to  32KJ over t in [0,6s]. What is the average speed of the object?

May 7, 2017

#### Answer:

The average speed is $= 163 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 3 k g$

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

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

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

$\frac{1}{2} m {u}_{2}^{2} = 32000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{3} \cdot 48000 = 32000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{3} \cdot 32000 = 21333.3 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 32000\right)$ and $\left(6 , 21333.3\right)$

The equation of the line is

${v}^{2} - 32000 = \frac{21333.3 - 32000}{6} t$

${v}^{2} = - 1777.8 t + 32000$

So,

v=sqrt((-1777.8t+32000)

We need to calculate the average value of $v$ over $t \in \left[0 , 6\right]$

$\left(6 - 0\right) \overline{v} = {\int}_{0}^{6} \sqrt{\left(- 1777.8 t + 32000\right)} \mathrm{dt}$

6 barv=[((-1777.8t+32000)^(3/2)/(-3/2*1777.8)]_0^6

$= \left({\left(- 1777.8 \cdot 6 + 32000\right)}^{\frac{3}{2}} / \left(- 2666.7\right)\right) - \left({\left(- 1777.8 \cdot 0 + 32000\right)}^{\frac{3}{2}} / \left(- 2666.7\right)\right)$

$= {32000}^{\frac{3}{2}} / 2666.7 - {21333.3}^{\frac{3}{2}} / 2666.7$

$= 978.1$

So,

$\overline{v} = \frac{978.1}{6} = 163 m {s}^{-} 1$