# An object has a mass of 6 kg. The object's kinetic energy uniformly changes from 15 KJ to  64KJ over t in [0,12s]. What is the average speed of the object?

Jul 19, 2017

The average speed is $= 107.3 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

The mass is $= 6 k g$

The initial velocity is $= {u}_{1} m {s}^{-} 1$

The final velocity is $= {u}_{2} m {s}^{-} 1$

The initial kinetic energy is $\frac{1}{2} m {u}_{1}^{2} = 15000 J$

The final kinetic energy is $\frac{1}{2} m {u}_{2}^{2} = 64000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{6} \cdot 15000 = 3000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{6} \cdot 64000 = 21333.3 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 3000\right)$ and $\left(12 , 21333.3\right)$

The equation of the line is

${v}^{2} - 3000 = \frac{21333.3 - 3000}{12} t$

${v}^{2} = 1527.8 t + 3000$

So,

v=sqrt((1527.8t+3000)

We need to calculate the average value of $v$ over $t \in \left[0 , 12\right]$

(12-0)bar v=int_0^12sqrt(1527.8t+3000))dt

12 barv=[((1527.8t+3000)^(3/2)/(3/2*1527.8)]_0^12

$= \left({\left(1527.8 \cdot 12 + 3000\right)}^{\frac{3}{2}} / \left(2291.7\right)\right) - \left({\left(1527.8 \cdot 0 + 3000\right)}^{\frac{3}{2}} / \left(2291.7\right)\right)$

$= {21333.6}^{\frac{3}{2}} / 2291.7 - {3000}^{\frac{3}{2}} / 2291.7$

$= 1288$

So,

$\overline{v} = \frac{1288}{12} = 107.3 m {s}^{-} 1$

The average speed is $= 107.3 m {s}^{-} 1$