# An object has a mass of 3 kg. The object's kinetic energy uniformly changes from 75 KJ to 42 KJ over t in [0, 15 s]. What is the average speed of the object?

Mar 26, 2017

The average speed is $= 181.3 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

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

$\frac{1}{2} m {u}_{1}^{2} = 75000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 42000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{3} \cdot 75000 = 50000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{3} \cdot 42000 = 28000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 50000\right)$ and $\left(15 , 28000\right)$

The equation of the line is

${v}^{2} - 50000 = \frac{28000 - 50000}{15} t$

${v}^{2} = - 1466.7 t + 50000$

So,

v=sqrt((-1466.7t+50000)

We need to calculate the average value of $v$ over $t \in \left[0 , 15\right]$

(15-0)bar v=int_0^15sqrt((-1466.7t+50000)dt

15 barv=[((-1466.7t+50000)^(3/2)/(3/2*-1466.7)]_0^15

$= \left({\left(- 1466.7 \cdot 15 + 50000\right)}^{\frac{3}{2}} / \left(- 2200\right)\right) - \left({\left(- 1466.7 \cdot 0 + 50000\right)}^{\frac{3}{2}} / \left(- 2200\right)\right)$

$= {30000}^{\frac{3}{2}} / - 2200 - {50000}^{\frac{3}{2}} / - 2200$

$= {50000}^{\frac{3}{2}} / 2200 - {30000}^{\frac{3}{2}} / 2200$

$= 2720.1$

So,

$\overline{v} = \frac{2720.1}{15} = 181.3 m {s}^{-} 1$