# An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 42 KJ to  320 KJ over t in [0, 12 s]. What is the average speed of the object?

May 6, 2017

The average speed is $= 292.4 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 4 k g$

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

$\frac{1}{2} m {u}_{1}^{2} = 42000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 320000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{4} \cdot 42000 = 21000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{4} \cdot 320000 = 160000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 21000\right)$ and $\left(12 , 160000\right)$

The equation of the line is

${v}^{2} - 21000 = \frac{160000 - 21000}{12} t$

${v}^{2} = 11583.3 t + 21000$

So,

v=sqrt((11583.3t+21000)

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

$\left(12 - 0\right) \overline{v} = {\int}_{0}^{12} \sqrt{\left(11583.3 t + 21000\right)} \mathrm{dt}$

12 barv=[((11583.3t+21000)^(3/2)/(3/2*11583.3)]_0^12

$= \left({\left(11583.3 \cdot 12 + 21000\right)}^{\frac{3}{2}} / \left(17375\right)\right) - \left({\left(1583.3 \cdot 0 + 21000\right)}^{\frac{3}{2}} / \left(17375\right)\right)$

$= {160000}^{\frac{3}{2}} / 17375 - {21000}^{\frac{3}{2}} / 17375$

$= 3508.3$

So,

$\overline{v} = \frac{3508.3}{12} = 292.4 m {s}^{-} 1$