# An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 24 KJ to  42KJ over t in [0, 9 s]. What is the average speed of the object?

Mar 31, 2017

$4 \frac{m}{s}$

#### Explanation:

Let the rate of increase of kinetic energy with time be some $m \frac{K J}{s}$ where $m$ is a constant as given in the question.

$\therefore \frac{\mathrm{dE} \left(t\right)}{\mathrm{dt}} = m$ where $E$ is Kinetic Energy and $t$ is time.

$\implies \mathrm{dE} \left(t\right) = m \cdot \mathrm{dt}$
$\implies$$\int \mathrm{dE} = \int m \cdot \mathrm{dt}$

$\implies$ $E \left(t\right) = m \cdot t + c$ where $c$ is a constant of integration.
We have two conditions:-
1. $E \left(0\right) = 24$ &
2. $E \left(9\right) = 42$

from 1. & 2.
$24 = c$ & $42 = 9 \cdot m + 24 \implies m = 2$
$\therefore E \left(t\right) = 2 \cdot t + 24$

$\implies \left(\frac{1}{2}\right) \cdot m \cdot {v}^{2} = 2 \cdot t + 24$ where $m$ is the mass & $v$ is the velocity (speed) of body.

$\implies {v}^{2} = t + 12$
$\implies v = \sqrt{t + 12}$

Average speed = $\frac{{\int}_{{t}_{1}}^{{t}_{2}} v \cdot \mathrm{dt}}{{\int}_{{t}_{1}}^{{t}_{2}} \mathrm{dt}}$
where
${t}_{1} = 0$ & ${t}_{2} = 9$

$\therefore$ Avg. speed $= \frac{{\int}_{0}^{9} \sqrt{t + 12} \cdot \mathrm{dt}}{{\int}_{0}^{9} \mathrm{dt}}$
= (2/3*(t+12)^(3/2))/(t)]_0^9 = 2/3*(21sqrt21 - 12sqrt12)/9 approx 4

Mar 31, 2017

$128.05 m {s}^{-} 1$, rounded to two decimal places.

#### Explanation:

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 $t = 0$
$24 \times {10}^{3} = C \cdot 0 + c$
$\implies c = 24 \times {10}^{3}$
We have expression for kinetic energy as
$K E = C \cdot t + 24 \times {10}^{3}$

To evaluate $C$, use final condition at $t = 9$
$42 \times {10}^{3} = C \cdot 9 + 24 \times {10}^{3}$
$\implies C = 2 \times {10}^{3}$

Equation (1) becomes
$K E = \left(2 \cdot t + 24\right) \times {10}^{3}$ ......(2)

If $m$ is the mass and $v \left(t\right)$ is the velocity of body, Kinetic energy is given as
$K E = \frac{1}{2} m {v}^{2}$
$\implies \frac{1}{2} m {v}^{2} \left(t\right) = \left(2 \cdot t + 24\right) \times {10}^{3}$
Given is $m = 4 k g$. Above expression becomes
$\implies \frac{1}{2} \times 4 {v}^{2} \left(t\right) = \left(2 \cdot t + 24\right) \times {10}^{3}$
$\implies v {\left(t\right)}^{2} = \left(t + 12\right) \times {10}^{3}$
$\implies v = \pm \sqrt{\left(t + 12\right) \times {10}^{3}}$
$\implies \text{speed } | v | = \sqrt{\left(t + 12\right) \times {10}^{3}}$

Average speed $= \text{Total Distance traveled"/"Time taken} = \frac{{\int}_{0}^{9} | v \left(t\right) | \cdot \mathrm{dt}}{9 - 0}$

$\therefore$ Average speed $= \frac{{\int}_{0}^{9} \sqrt{\left(t + 12\right) \times {10}^{3}} \cdot \mathrm{dt}}{9}$
$= \frac{\sqrt{{10}^{3}}}{9} {\left[{\left(t + 12\right)}^{\frac{3}{2}} / \left(\frac{3}{2}\right)\right]}_{0}^{9}$
$= \frac{2 \sqrt{{10}^{3}}}{27} \left({21}^{\frac{3}{2}} - {12}^{\frac{3}{2}}\right)$
$= 128.05 m {s}^{-} 1$, rounded to two decimal places.