# An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 64 KJ to  160 KJ over t in [0, 12 s]. What is the average speed of the object?

Apr 24, 2017

The average speed is $= 234.8 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 4 k g$

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

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

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

$\frac{1}{2} m {u}_{2}^{2} = 160000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{4} \cdot 64000 = 32000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{4} \cdot 160000 = 80000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 32000\right)$ and $\left(12 , 80000\right)$

The equation of the line is

${v}^{2} - 32000 = \frac{80000 - 32000}{12} t$

${v}^{2} = 4000 t + 32000$

So,

v=sqrt((4000t+32000)

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

$\left(12 - 0\right) \overline{v} = {\int}_{0}^{12} \sqrt{\left(4000 t + 32000\right)} \mathrm{dt}$

12 barv=[((4000t+32000)^(3/2)/(3/2*4000)]_0^12

$= \left({\left(4000 \cdot 12 + 32000\right)}^{\frac{3}{2}} / \left(6000\right)\right) - \left({\left(4000 \cdot 0 + 32000\right)}^{\frac{3}{2}} / \left(6000\right)\right)$

$= {80000}^{\frac{3}{2}} / 6000 - {32000}^{\frac{3}{2}} / 6000$

$= 2817.2$

So,

$\overline{v} = \frac{2817.2}{12} = 234.8 m {s}^{-} 1$