# An object has a mass of 2 kg. The object's kinetic energy uniformly changes from 18 KJ to  64KJ over t in [0,12s]. What is the average speed of the object?

Mar 15, 2017

The average speed of the object is $= 199.6 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

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

$\frac{1}{2} m {u}_{1}^{2} = 18000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 64000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{2} \cdot 18000 = 18000 {m}^{2} {s}^{-} 2$

and,

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

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

The points are $\left(0 , 18000\right)$ and $\left(12 , 64000\right)$

The equation of the line is

${v}^{2} - 18000 = \frac{64000 - 18000}{12} t$

${v}^{2} = 3833.3 t + 18000$

So,

v=sqrt((3833.3t+18000)

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

(12-0)bar v=int_0^12sqrt((3833.3t+18000)dt

12 barv=[((3833.3t+18000)^(3/2)/(3/2*3833.3)]_0^12

$= \left({\left(3833.3 \cdot 12 + 18000\right)}^{\frac{3}{2}} / \left(5750\right)\right) - \left({\left(3833.3 \cdot 0 + 18000\right)}^{\frac{3}{2}} / \left(5750\right)\right)$

$= {63999.6}^{\frac{3}{2}} / 5750 - {18000}^{\frac{3}{2}} / 5750$

$= 2395.8$

So,

$\overline{v} = \frac{2395.8}{12} = 199.6 m {s}^{-} 1$