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1800-102-2727You sit in a moving train and observe a person sitting on the platform. Let’s call him/her
Person 1. At the same time, you see another train moving with the same velocity. Now you being the observer of Person 1 and the person in the other train, what can you say about them? You would have heard that a body which does not move is said to be at “rest”. On the other hand, a body which moves is said to be in “motion”. Diving a bit more deeper, we must understand that rest and motion are “relative” - they depend upon the observer's perspective. For example, from your perspective, Person 1 is moving, and the person in the other train is said to be at rest. However, when observed from Person 1’s perspective, you and the person in the other train are said to be in motion, since the trains are moving. In this article, we will explore relative motion in detail.
Table of contents
Before we understand what relative motion is, we must first understand the concept of frame of reference– it is defined as a region or zone from which observations are made. It serves as the perspective with respect to which measurements are made.
Position of one particle with respect to another particle is called relative position.
Let us consider the following example where two particles A and B are located at positions xA and xB from the origin as shown.
Then the relative position of B wrt A can be written as,
${x}_{BA}={x}_{B}-{x}_{A}--\left(i\right)$
Here, B is the object that is being looked at and A is the reference point.
Relative velocity of an object B with respect to another object A is the velocity with which object B would appear to move with respect to an observer situated at A.
Differentiating equation (i) wrt time on both sides, we get
$\frac{d{x}_{BA}}{dt}=\frac{d{x}_{B}}{dt}-\frac{d{x}_{B}}{dt}$
${v}_{BA}={v}_{B}-{v}_{A}--\left(ii\right)$
${v}_{BA}-$ Relative velocity of B wrt A.
${v}_{B}-$ Velocity of B.
${v}_{A}-$ Velocity of A.
In vector form, the equation (ii) can be written as,
$\overrightarrow{{v}_{BA}}=\overrightarrow{{v}_{B}}-\overrightarrow{{v}_{A}}.$
Similarly, the relative velocity of A wrt B can be written as,
$\overrightarrow{{v}_{AB}}=\overrightarrow{{v}_{A}}-\overrightarrow{{v}_{B}}.$
Note:
(i)When two bodies A and B are approaching each other, their relative speed is the sum of the two speeds.
${v}_{BA}={v}_{A}+{v}_{B}.$
(ii)When two bodies A and B are moving in the same direction, their relative speed is the difference of the two speeds.
${v}_{AB}={|v}_{A}-{v}_{B}|.$
Relative acceleration of an object B with respect to another object A is the acceleration with which object B would appear to accelerate with respect to an observer situated at A.
Differentiating equation (ii) on both sides with respect to time, we get
$\frac{d{v}_{BA}}{dt}=\frac{d{v}_{B}}{dt}-\frac{d{v}_{A}}{dt}$
Since $\frac{dv}{dt}=a,$
we get
${a}_{BA}={a}_{B}-{a}_{A}$
Ground frame is the default frame of reference. Consider the following example where ground is chosen as the default frame of reference.
(i)Observer is standing on the ground.
(ii)Both objects A and B move for the observer.
When the place where observer is present is chosen as the reference point, then
observer is said to be present at the origin.
Note vAG=vA= velocity of A w.r.t. ground
Let us consider two projectiles A and B being projected with velocities uA and uB. The angles made with the horizontal are A and B. Then from the ground frame of reference,
$\overrightarrow{{u}_{A}}={u}_{A}cos\left({\theta}_{A}\right)\hat{i}+{u}_{B}sin\left({\theta}_{A}\right)\hat{j};\overrightarrow{{a}_{A}}=-g\hat{j}$
$\overrightarrow{{u}_{B}}={u}_{B}cos\left({\theta}_{B}\right)\hat{i}+{u}_{B}sin\left({\theta}_{B}\right)\hat{j};\overrightarrow{{a}_{B}}=-g\hat{j}$
From frame of reference of A, we get
$\overrightarrow{{u}_{AA}}=0,\overrightarrow{{a}_{AA}}=\overrightarrow{0.}$
$\overrightarrow{{u}_{BA}}=\overrightarrow{{u}_{B}}-\overrightarrow{{u}_{A}}$
$\overrightarrow{{u}_{BA}}=\overrightarrow{{u}_{B}}-\overrightarrow{{u}_{A}}={u}_{B}cos\left({\theta}_{B}\right)\hat{i}+{u}_{B}sin\left({\theta}_{B}\right)\hat{j}-{u}_{A}cos\left({\theta}_{A}\right)\hat{i}+{u}_{A}sin\left({\theta}_{A}\right)\hat{j};$
$\overrightarrow{{a}_{BA}}=\left[\right(-g\hat{j})-(-g\hat{j}\left)\right]=0.$
Q. Two particles A and B are fired simultaneously. Find the minimum distance between them during their flight.
A. Let uA and uB indicate the velocities of A and B.
Then,
$\overrightarrow{{u}_{A}}=15cos\left({53}^{0}\right)\hat{i}+15sin\left({53}^{0}\right)\hat{j}=15\times \frac{3}{5}\hat{i}+15\times \frac{4}{5}\hat{j}=9\hat{i}+12\hat{j}$
$\overrightarrow{{u}_{B}}=15cos\left({37}^{0}\right)(-\hat{i})+15sin\left({37}^{0}\right)\hat{j}=-15\times \frac{4}{5}\hat{i}+15\times \frac{3}{5}\hat{j}=-12\hat{i}+9\hat{j.}$
Now relative velocity of B wrt A,
$\overrightarrow{{u}_{B}}-\overrightarrow{{u}_{A}}=(-12\hat{i}+9\hat{j})-(9\hat{i}+12\hat{j})=-21\hat{i}-3\hat{j}.$
Now tan $\theta =\frac{Verticalcomponentofvelocity}{Horizontalcomponentofvelocity}=\frac{-3}{-21}=\frac{1}{7}$
$\Rightarrow sin\theta =\frac{1}{\sqrt{50}}=\frac{1}{5\sqrt{2}}$
Now $sin\theta =\frac{{d}_{short}}{20\sqrt{2}};$
${d}_{short}=20\sqrt{2}\times \frac{1}{5\sqrt{2}}=4m.$
Q. Two particles P and Q, initially separated by 75 m, are moving towards each other along a straight line as shown in figure with aP= 2 m s-2, uP= 15 m s-1, and aQ= -4 m s-2,
uQ =-25 m s-1.Calculate the time when they meet.
A. Given,
${a}_{Q}=-4m{s}^{-2};{a}_{P}=2m{s}^{-2}$
${u}_{P}=+15m{s}^{-1};{u}_{Q}=-25m{s}^{-1}$
Position of P= Position of Q
${x}_{p}={x}_{o}+{u}_{P}t+\frac{1}{2}{a}_{P}{t}^{2}$
${x}_{Q}={x}_{o}+{u}_{Q}t+\frac{1}{2}{a}_{Q}{t}^{2}$
${x}_{p}=0+15t+\frac{1}{2}\times 2\times {t}^{2}$
${x}_{Q}=75-25t+\frac{1}{2}(-4){t}^{2}$
$15t+\frac{1}{2}\times 2\times {t}^{2}=75-25t+\frac{1}{2}(-4){t}^{2}$
$15t+{t}^{2}=75-25t-2{t}^{2}$
${3t}^{2}+40t-75=0$
$t=\frac{5}{3}sandt=-15s.$
We will neglect the negative value of t.
Q. Car P and car Q move along the same direction. Car P moves with a constant acceleration a=4 m s-2 while car Q moves with a constant velocity v= 1 m/s . Initially at t=0, car P is 10 m behind car Q. Find the time taken for car P to overtake Q .
A. Given, x=10 m.
${u}_{P}=0;{u}_{Q}=1m/s.$
${a}_{P}=4m{s}^{-2};{a}_{Q}=0.$
Relative velocity between cars P and Q, ${u}_{PQ}={u}_{P}-{u}_{Q}=-1m/s.$
${a}_{PQ}={a}_{P}-{a}_{Q}=4m{s}^{-2}.$
Now
$x={u}_{PQ}t+\frac{1}{2}{a}_{PQ}{t}^{2}$
$\Rightarrow 10=-t+\frac{1}{2}\left(4\right){t}^{2}$
$2{t}^{2}-t-10=0$
Solving, we get
t=2.5 s and t=-2 s.
Ignoring the negative value, the time at which the cars meet is 2.5 s.
Q. Two ships P and Q are located 10 km apart on a line running along the South - North direction. Ship P located at the farther North is speeding west at the rate of 20 km/hr. Ship Q is streaming North at 20 km/hr. Calculate their distance of closest approach and also how long do they take to reach it?
A. Given,
$\overrightarrow{{v}_{Q}}=20\hat{j};\overrightarrow{{v}_{p}}=-20\hat{i}$
Let vQP indicate the relative velocity of ship Q with respect to ship P. Then,
$\overrightarrow{{v}_{QP}}=\overrightarrow{{v}_{Q}}-\overrightarrow{{v}_{P}}=20\hat{j}+20\hat{i};$
$\left|\overrightarrow{{v}_{QP}}\right|=|20\hat{j}+20\hat{i}|=\sqrt{{20}^{2}+{20}^{2}}=20\sqrt{2}km/hr.$
Let dmin be the minimum distance between them.
${d}_{min}=AC=ABsin{45}^{0}=10\left(\frac{1}{\sqrt{2}}\right)=5\sqrt{2}km.$
The time taken
$t=\frac{BC}{\left|\overrightarrow{{v}_{QP}}\right|}=\frac{5\sqrt{2}}{20\sqrt{2}}=\frac{1}{4}h=15min.$
Q. Does relative velocity have dimension?
A. Yes, the dimension of relative velocity is the same as that of velocity.
Q. Who invented the concept of relative velocity?
A. Galileo was the first person to say that motion is relative and introduced the concept of relative velocity.
Q. Write one application of relative motion.
A. Relative velocity can be used to find the time at which two bodies meet, if we know the velocity of the bodies.
Q. In which condition is the relative velocity zero?
A. The relative velocity of two bodies is said to be zero if they are moving in the same direction at the same speed.