# An object has a mass of 5 kg. The object's kinetic energy uniformly changes from 15 KJ to  64KJ over t in [0,12s]. What is the average speed of the object?

Apr 24, 2017

The average speed is $= 123.5 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 5 k g$

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

$\frac{1}{2} m {u}_{1}^{2} = 15000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 64000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{5} \cdot 15000 = 6000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{5} \cdot 64000 = 25600 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 6000\right)$ and $\left(12 , 25600\right)$

The equation of the line is

${v}^{2} - 6000 = \frac{25600 - 6000}{12} t$

${v}^{2} = 1633.3 t + 6000$

So,

v=sqrt((1633.3t+6000)

We need to calculate the average value of $v$ over $t \in \left[0 , 12\right]$

$\left(12 - 0\right) \overline{v} = {\int}_{0}^{12} \sqrt{\left(1633.3 t + 6000\right)} \mathrm{dt}$

12 barv=[((1633.3t+6000)^(3/2)/(3/2*1633.3)]_0^12

$= \left({\left(1633.3 \cdot 12 + 6000\right)}^{\frac{3}{2}} / \left(2450\right)\right) - \left({\left(1633.3 \cdot 0 + 6000\right)}^{\frac{3}{2}} / \left(2450\right)\right)$

$= {25600}^{\frac{3}{2}} / 2450 - {6000}^{\frac{3}{2}} / 2450$

$= 1482.1$

So,

$\overline{v} = \frac{1482.1}{12} = 123.5 m {s}^{-} 1$