# An object has a mass of 2 kg. The object's kinetic energy uniformly changes from 16 KJ to  36KJ over t in [0, 6 s]. What is the average speed of the object?

Apr 3, 2017

The average speed is $= 160.2 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 2 k g$

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

$\frac{1}{2} m {u}_{1}^{2} = 16000 J$

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

$\frac{1}{2} m {u}_{2}^{2} = 36000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{2} \cdot 16000 = 16000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{2} \cdot 36000 = 36000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 16000\right)$ and $\left(6 , 36000\right)$

The equation of the line is

${v}^{2} - 16000 = \frac{36000 - 16000}{6} t$

${v}^{2} = 3333.3 t + 16000$

So,

v=sqrt((3333.3t+16000)

We need to calculate the average value of $v$ over $t \in \left[0 , 6\right]$

$\left(6 - 0\right) \overline{v} = {\int}_{0}^{8} \sqrt{\left(3333.3 t + 16000\right)} \mathrm{dt}$

6 barv=[((3333.3t+16000)^(3/2)/(3/2*3333.3)]_0^6

$= \left({\left(3333.3 \cdot 6 + 16000\right)}^{\frac{3}{2}} / \left(5000\right)\right) - \left({\left(3333.3 \cdot 0 + 16000\right)}^{\frac{3}{2}} / \left(5000\right)\right)$

$= {36000}^{\frac{3}{2}} / 5000 - {16000}^{\frac{3}{2}} / 5000$

$= 961.3$

So,

$\overline{v} = \frac{961.3}{6} = 160.2 m {s}^{-} 1$

Apr 3, 2017

Average velocity $\overline{v} \approx 160.2 \text{ m/s}$

So average speed is $\overline{\dot{s}} \approx 160.2 \text{ m/s}$.

#### Explanation:

During the period we can say that:

$\frac{\mathrm{dT}}{\mathrm{dt}} = \frac{36 , 000 - 16 , 000}{6} = \frac{10 , 000}{3} \text{ Joules / s}$

Integrating:

$T \left(t\right) = \frac{1}{2} m v {\left(t\right)}^{2} = \int \frac{10 , 000}{3} \setminus \mathrm{dt} = \frac{10 , 000}{3} t + \alpha$

$T \left(0\right) = 16 , 000 \implies \alpha = 16 , 000$

And as $m = 2$, we have:

$v \left(t\right) = \sqrt{\frac{10 , 000}{3} t + 16 , 000}$

For average velocity $\overline{v}$:

$\overline{v} = \frac{{\int}_{0}^{6} v \left(t\right) \setminus \mathrm{dt}}{6 - 0}$ (simply put, this calculates that constant velocity that would cover the same distance (sense displacement!) in the same time)

$= \frac{{\left[{\left(\frac{10 , 000 t}{3} + 16 , 000\right)}^{\frac{3}{2}}\right]}_{0}^{6}}{6 \cdot 5 , 000} \approx 160.2 \text{ m/s}$

Reality check:

$v = \sqrt{\frac{2 T}{m}}$

$v \left(0\right) \approx 126.5 \text{ m/s}$

$v \left(6\right) \approx 190 \text{ m/s}$