# An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 12 KJ to  36 KJ over t in [0, 12 s]. What is the average speed of the object?

Dec 1, 2017

The average speed is $= 108.3 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

The mass is $m = 4 k g$

The initial velocity is $= {u}_{1} m {s}^{-} 1$

The final velocity is $= {u}_{2} m {s}^{-} 1$

The initial kinetic energy is $\frac{1}{2} m {u}_{1}^{2} = 12000 J$

The final kinetic energy is $\frac{1}{2} m {u}_{2}^{2} = 36000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{4} \cdot 12000 = 6000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{4} \cdot 36000 = 18000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 6000\right)$ and $\left(12 , 18000\right)$

The equation of the line is

${v}^{2} - 6000 = \frac{18000 - 6000}{12} t$

${v}^{2} = 1000 t + 6000$

So,

$v = \sqrt{1000 t + 6000}$

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

$\left(12 - 0\right) \overline{v} = {\int}_{0}^{12} \left(\sqrt{1000 t + 6000}\right) \mathrm{dt}$

12 barv=[1000t+6000)^(3/2)/(3/2*1000))] _( 0) ^ (12)

$= \left({\left(1000 \cdot 12 + 6000\right)}^{\frac{3}{2}} / \left(1500\right)\right) - \left({\left(1000 \cdot 0 + 6000\right)}^{\frac{3}{2}} / \left(1500\right)\right)$

$= {18000}^{\frac{3}{2}} / 1500 - {6000}^{\frac{3}{2}} / 1500$

$= 1300.1$

So,

$\overline{v} = \frac{1300.1}{12} = 108.3 m {s}^{-} 1$

The average speed is $= 108.3 m {s}^{-} 1$

Dec 1, 2017

Kinetic energy $K E$ of an object of mass $m$ moving with velocity $v$ is given by the expression

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

Given $m = 4 k g$
Let initial velocity be $= u \text{ } m {s}^{-} 1$
The final velocity be $= v \text{ } m {s}^{-} 1$

Initial kinetic energy

$\frac{1}{2} m {u}^{2} = 12000 J$

Final kinetic energy is

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

As kinetic energy changes uniformly, the graph of $t \text{ vs } K E$ is a straight line.

The points on the graph are $\left(0 , 12000\right)$ and $\left(12 , 36000\right)$

The equation of the line is of the type $y = m x + c$

$\frac{1}{2} m {\left(v \left(t\right)\right)}^{2} = \frac{36000 - 12000}{12} t + 12000$
or $\frac{1}{2} \times 4 {\left(v \left(t\right)\right)}^{2} = 2000 t + 12000$
$\implies {\left(v \left(t\right)\right)}^{2} = 1000 t + 6000$
$\implies v \left(t\right) = \sqrt{1000 t + 6000}$

We know that velocity can be written as $\frac{\mathrm{ds}}{\mathrm{dt}}$ where $s$ is distance moved

$\implies \frac{\mathrm{ds} \left(t\right)}{\mathrm{dt}} = \sqrt{1000 t + 6000}$

We need to calculate total distance traveled in time $t$ from $0$ to$12 s$ by integrating over the interval

$\therefore s = {\int}_{0}^{12} \sqrt{1000 t + 6000} \cdot \mathrm{dt}$
$\implies s = {\left[{\left(1000 t + 6000\right)}^{\frac{3}{2}} / \left(\frac{3}{2} \times 1000\right)\right]}_{0}^{12}$
$\implies s = \frac{1}{1500} \left[{\left(1000 \times 12 + 6000\right)}^{\frac{3}{2}} - {\left(1000 \times 0 + 6000\right)}^{\frac{3}{2}}\right]$
$\implies s = \frac{1}{1500} \left({18000}^{\frac{3}{2}} - {6000}^{\frac{3}{2}}\right)$
$\implies s = 1300.13 m$

We know that
$\text{Average speed"="Total distance traveled"/"Time of travel}$
$\implies \overline{v} = \frac{1300.13}{12} = 108.3 m {s}^{-} 1$, rounded to one decimal place