# An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 640 KJ to  160 KJ over t in [0, 12 s]. What is the average speed of the object?

Apr 11, 2017

The average speed is $= 440 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

mass is $= 4 k g$

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

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

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

$\frac{1}{2} m {u}_{2}^{2} = 160000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{4} \cdot 640000 = 320000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{4} \cdot 160000 = 80000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 320000\right)$ and $\left(12 , 80000\right)$

The equation of the line is

${v}^{2} - 320000 = \frac{80000 - 320000}{12} t$

${v}^{2} = - 20000 t + 320000$

So,

v=sqrt((-20000t+320000)

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

$\left(12 - 0\right) \overline{v} = {\int}_{0}^{12} \sqrt{\left(- 20000 t + 320000\right)} \mathrm{dt}$

12 barv=[((-20000t+320000)^(3/2)/(-3/2*20000)]_0^12

$= \left({\left(- 20000 \cdot 12 + 320000\right)}^{\frac{3}{2}} / \left(- 30000\right)\right) - \left({\left(- 20000 \cdot 0 + 320000\right)}^{\frac{3}{2}} / \left(- 30000\right)\right)$

$= {320000}^{\frac{3}{2}} / 30000 - {80000}^{\frac{3}{2}} / 30000$

$= 5279.7$

So,

$\overline{v} = \frac{5279.7}{12} = 440 m {s}^{-} 1$

Apr 11, 2017

average speed is defined as total distance travelled divided by total time taken to cover that distance .

#### Explanation:

the kinetic energy of a body of mass m moving with speed v is defined as : $\frac{1}{2} m {v}^{2}$
thus you can calculate velocity of body at the starting and the end of interval of time :
lets say initially the velocity was ${v}_{1}$ , $\frac{1}{2} \cdot 4 \cdot {v}_{1}^{2} = 640000$
${v}_{1} = 565.68 m {s}^{-} 1$
let at the end of interval of time the velocity was ${v}_{2}$ , $\frac{1}{2} \cdot 4 \cdot {v}_{2}^{2} = 160000$
${v}_{2} = 282.84 m {s}^{-} 1$
now , total distance travelled by the body over given time interval can be calculated by using ${v}_{2}^{2} - {v}_{1}^{2} = 2 \cdot a \cdot s$ and ${v}_{2} = {v}_{1} + a \cdot t$
thus, $282.84 = 565.68 + 12 a$
$a = - 23.57 m {s}^{-} 2$
${282.84}^{2} - {565.68}^{2} = 2 \cdot - 23.57 \cdot s$
$s = 5091.12 m$
average speed is defined as ratio of total distance travelled and total time taken :
$v = \frac{s}{t}$
$v = \frac{5091.12}{12}$
$v = 424.26 m {s}^{-} 1$

Apr 12, 2017

Let rate of change of kinetic energy be
$\frac{\mathrm{dK} E \left(t\right)}{\mathrm{dt}} = C$
where $C$ is constant with time $t$.

Integrating both sides with $t$ we get
$\int \frac{\mathrm{dK} E \left(t\right)}{\mathrm{dt}} \cdot \mathrm{dt} = \int C \cdot \mathrm{dt}$

$\implies K E = C \cdot t + c$ ......(1)
where $c$ is a constant of integration.

To evaluate $c$, use initial condition at $t = 0$
$640 \times {10}^{3} = C \cdot 0 + c$
$\implies c = 640 \times {10}^{3}$
We have expression for kinetic energy as
$K E = C \cdot t + 640 \times {10}^{3}$

To evaluate $C$, use final condition at $t = 12$
$160 \times {10}^{3} = C \cdot 12 + 640 \times {10}^{3}$
$\implies C = - 40 \times {10}^{3}$

Equation (1) becomes
$K E = \left(- t + 16\right) \times 4 \times {10}^{4}$ ......(2)

If $m$ is the mass and $v \left(t\right)$ is the velocity of body, Kinetic energy is given as
$K E = \frac{1}{2} m {v}^{2}$
$\implies \frac{1}{2} m {v}^{2} \left(t\right) = \left(- t + 16\right) \times 4 \times {10}^{4}$
Given is $m = 4 k g$. Above expression becomes
$\implies \frac{1}{2} \times 4 {v}^{2} \left(t\right) = \left(- t + 16\right) \times 4 \times {10}^{4}$
$\implies v {\left(t\right)}^{2} = \left(- t + 16\right) \times 2 \times {10}^{4}$
$\implies v = \pm \sqrt{\left(- t + 16\right) \times 2 \times {10}^{4}}$
$\implies \text{speed } | v | = \sqrt{\left(- t + 16\right) \times 2 \times {10}^{4}}$

Average speed $= \text{Total Distance traveled"/"Time taken} = \frac{{\int}_{0}^{12} | v \left(t\right) | \cdot \mathrm{dt}}{12 - 0}$

$\therefore$ Average speed $= \frac{{\int}_{0}^{12} \sqrt{\left(- t + 16\right) \times 2 \times {10}^{4}} \cdot \mathrm{dt}}{12}$
$= 100 \frac{\sqrt{2}}{12} {\left[- {\left(16 - t\right)}^{\frac{3}{2}} / \left(\frac{3}{2}\right)\right]}_{0}^{12}$
$= - \frac{50}{9} \sqrt{2} \left({4}^{\frac{3}{2}} - {16}^{\frac{3}{2}}\right)$
$= 439.98 m {s}^{-} 1$, rounded to two decimal places.