# An object has a mass of 2 kg. The object's kinetic energy uniformly changes from 32 KJ to 96 KJ over t in [0, 4 s]. What is the average speed of the object?

Apr 24, 2017

The average speed is $= 250.2 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 2 k g$

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

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

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

$\frac{1}{2} m {u}_{2}^{2} = 96000 J$

Therefore,

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

and,

${u}_{2}^{2} = \frac{2}{2} \cdot 96000 = 96000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 32000\right)$ and $\left(4 , 96000\right)$

The equation of the line is

${v}^{2} - 32000 = \frac{96000 - 32000}{4} t$

${v}^{2} = 16000 t + 32000$

So,

v=sqrt((16000t+32000)

We need to calculate the average value of $v$ over $t \in \left[0 , 4\right]$

$\left(4 - 0\right) \overline{v} = {\int}_{0}^{4} \sqrt{\left(16000 t + 32000\right)} \mathrm{dt}$

4 barv=[((16000t+32000)^(3/2)/(3/2*16000)]_0^4

$= \left({\left(16000 \cdot 4 + 32000\right)}^{\frac{3}{2}} / \left(24000\right)\right) - \left({\left(16000 \cdot 0 + 32000\right)}^{\frac{3}{2}} / \left(24000\right)\right)$

$= {96000}^{\frac{3}{2}} / 24000 - {32000}^{\frac{3}{2}} / 24000$

$= 1000.8$

So,

$\overline{v} = \frac{1000.8}{4} = 250.2 m {s}^{-} 1$