# An object has a mass of 8 kg. The object's kinetic energy uniformly changes from 150 KJ to  640KJ over t in [0, 3 s]. What is the average speed of the object?

Mar 26, 2017

The average speed is $= 308.8 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 8 k g$

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

$\frac{1}{2} m {u}_{1}^{2} = 150000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 640000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{8} \cdot 150000 = 37500 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{8} \cdot 640000 = 160000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 37500\right)$ and $\left(3 , 160000\right)$

The equation of the line is

${v}^{2} - 37500 = \frac{160000 - 37500}{3} t$

${v}^{2} = 40833.3 t + 37500$

So,

v=sqrt((40833.3t+37500)

We need to calculate the average value of $v$ over $t \in \left[0 , 3\right]$

(3-0)bar v=int_0^3sqrt((40833.3t+37500)dt

3 barv=[((40833.3t+37500)^(3/2)/(3/2*40833.3)]_0^3

$= \left({\left(40833.3 \cdot 3 + 37500\right)}^{\frac{3}{2}} / \left(61250\right)\right) - \left({\left(40833.3 \cdot 0 + 37500\right)}^{\frac{3}{2}} / \left(61250\right)\right)$

$= {160000}^{\frac{3}{2}} / 61250 - {37500}^{\frac{3}{2}} / 61250$

$= 926.3$

So,

$\overline{v} = \frac{926.3}{3} = 308.8 m {s}^{-} 1$