# An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 144 KJ to  640KJ over t in [0, 3 s]. What is the average speed of the object?

May 14, 2017

#### Answer:

The average speed is $= 434.68 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 4 k g$

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

$\frac{1}{2} m {u}_{1}^{2} = 144000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 640000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{4} \cdot 144000 = 72000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{4} \cdot 640000 = 320000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 72000\right)$ and $\left(3 , 320000\right)$

The equation of the line is

${v}^{2} - 72000 = \frac{320000 - 72000}{3} t$

${v}^{2} = 82666.67 t + 72000$

So,

v=sqrt((82666.67t+72000)

We need to calculate the average value of $v$ over $t \in \left[0 , 3\right]$

$\left(3 - 0\right) \overline{v} = {\int}_{0}^{3} \sqrt{\left(82666.67 t + 72000\right)} \mathrm{dt}$

3 barv=[((82666.67t+72000)^(3/2)/(3/2*82666.67)]_0^3

$= \left({\left(82666.67 \cdot 3 + 72000\right)}^{\frac{3}{2}} / \left(124000\right)\right) - \left({\left(82666.67 \cdot 0 + 72000\right)}^{\frac{3}{2}} / \left(124000\right)\right)$

$= {320000}^{\frac{3}{2}} / 124000 - {72000}^{\frac{3}{2}} / 124000$

$= 1304.03$

So,

$\overline{v} = \frac{1304.03}{3} = 434.68 m {s}^{-} 1$