An object has a mass of 12 kg. The object's kinetic energy uniformly changes from 96 KJ to  160 KJ over t in [0, 5 s]. What is the average speed of the object?

May 22, 2017

The average speed is $= 145.7 m {s}^{-} 1$

Explanation:

The kinetic energy is

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

mass is $= 12 k g$

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

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

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

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

Therefore,

${u}_{1}^{2} = \frac{2}{12} \cdot 96000 = 16000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{12} \cdot 160000 = 26666.7 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 16000\right)$ and $\left(5 , 26666.7\right)$

The equation of the line is

${v}^{2} - 16000 = \frac{26666.7 - 16000}{5} t$

${v}^{2} = 2133.3 t + 16000$

So,

v=sqrt((2133.3t+16000)

We need to calculate the average value of $v$ over $t \in \left[0 , 5\right]$

$\left(5 - 0\right) \overline{v} = {\int}_{0}^{5} \sqrt{\left(2133.3 t + 16000\right)} \mathrm{dt}$

5 barv=[((2133.3t+16000)^(3/2)/(3/2*2133.3)]_0^5

$= \left({\left(2133.3 \cdot 5 + 16000\right)}^{\frac{3}{2}} / \left(3200\right)\right) - \left({\left(213303 \cdot 0 + 16000\right)}^{\frac{3}{2}} / \left(3200\right)\right)$

$= {26666.7}^{\frac{3}{2}} / 3200 - {16000}^{\frac{3}{2}} / 3200$

$= 728.37$

So,

$\overline{v} = \frac{728.37}{5} = 145.7 m {s}^{-} 1$