# An object has a mass of 6 kg. The object's kinetic energy uniformly changes from 96 KJ to  108 KJ over t in [0, 6 s]. What is the average speed of the object?

Mar 17, 2017

The average speed is $= 184.4 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

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

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

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

$\frac{1}{2} m {u}_{2}^{2} = 108000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{6} \cdot 96000 = 32000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{6} \cdot 108000 = 36000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 32000\right)$ and $\left(6 , 36000\right)$

The equation of the line is

${v}^{2} - 32000 = \frac{36000 - 32000}{6} t$

${v}^{2} = 666.7 t + 32000$

So,

v=sqrt((666.7t+32000)

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

(6-0)bar v=int_0^6sqrt((666.7t+32000)dt

6 barv=[((666.7t+32000)^(3/2)/(3/2*666.7)]_0^6

$= \left({\left(666.7 \cdot 6 + 32000\right)}^{\frac{3}{2}} / \left(1000\right)\right) - \left({\left(666.7 \cdot 0 + 32000\right)}^{\frac{3}{2}} / \left(1000\right)\right)$

$= {36000}^{\frac{3}{2}} / 1000 - {32000}^{\frac{3}{2}} / 1000$

$= 1106.2$

So,

$\overline{v} = \frac{1106.2}{6} = 184.4 m {s}^{-} 1$