# An object has a mass of 6 kg. The object's kinetic energy uniformly changes from 66 KJ to  225 KJ over t in [0, 8 s]. What is the average speed of the object?

Jul 13, 2017

The average speed is $= 217.3 m {s}^{-} 1$

#### 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 $\frac{1}{2} m {u}_{1}^{2} = 66000 J$

The final kinetic energy is $\frac{1}{2} m {u}_{2}^{2} = 225000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{6} \cdot 66000 = 22000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{6} \cdot 225000 = 75000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 22000\right)$ and $\left(8 , 75000\right)$

The equation of the line is

${v}^{2} - 22000 = \frac{75000 - 22000}{8} t$

${v}^{2} = 6625 t + 22000$

So,

v=sqrt((6625t+22000)

We need to calculate the average value of $v$ over $t \in \left[0 , 8\right]$

(8-0)bar v=int_0^8sqrt(6625t+22000))dt

8 barv=[((6625t+22000)^(3/2)/(3/2*6625)]_0^8

$= \left({\left(6625 \cdot 8 + 22000\right)}^{\frac{3}{2}} / \left(9937.5\right)\right) - \left({\left(6625 \cdot 0 + 22000\right)}^{\frac{3}{2}} / \left(9937.5\right)\right)$

$= {75000}^{\frac{3}{2}} / 9937.5 - {22000}^{\frac{3}{2}} / 9937.5$

$= 1738.5$

So,

$\overline{v} = \frac{1738.5}{8} = 217.3 m {s}^{-} 1$

The average speed is $= 217.3 m {s}^{-} 1$