# An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 84 KJ to  124KJ over t in [0, 6 s]. What is the average speed of the object?

Mar 27, 2017

The average speed is $= 227.7 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 4 k g$

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

$\frac{1}{2} m {u}_{1}^{2} = 84000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 124000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{4} \cdot 84000 = 42000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{4} \cdot 124000 = 62000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 42000\right)$ and $\left(6 , 62000\right)$

The equation of the line is

${v}^{2} - 42000 = \frac{62000 - 42000}{6} t$

${v}^{2} = 3333.3 t + 42000$

So,

v=sqrt((3333.3t+42000)

We need to calculate the average value of $v$ over $t \in \left[0 , 6\right]$

(6-0)bar v=int_0^6sqrt((3333.3t+42000)dt

6 barv=[((3333.3t+42000)^(3/2)/(3/2*3333.3)]_0^3

$= \left({\left(3333.3 \cdot 6 + 42000\right)}^{\frac{3}{2}} / \left(5000\right)\right) - \left({\left(3333.3 \cdot 0 + 42000\right)}^{\frac{3}{2}} / \left(5000\right)\right)$

$= {62000}^{\frac{3}{2}} / 5000 - {42000}^{\frac{3}{2}} / 5000$

$= 1366.1$

So,

$\overline{v} = \frac{1366.1}{6} = 227.7 m {s}^{-} 1$