Relative Motion

Relative motion is the concept of how the position or velocity of one object appears to change when observed from another moving object. In physics, motion is always measured relative to a chosen frame of reference. For example, a person walking inside a moving train may seem still to another person on the train but appears to be moving to someone standing outside. Motion is not absolute; it's always relative to a reference point.It's especially useful in problems involving multiple moving objects, like cars, boats, or airplanes.

1.0Definition Relative Motion

• It refers to the movement of an object as seen from a specific frame of reference, which might also be in motion.In simpler terms, it describes how one object appears to move in relation to another moving object.
• Motion is a combined property of the object under study as well as the observer. It is always relative; there is no such thing like absolute motion or absolute rest. Motion is always defined with respect to an observer or reference frame.

2.0Reference Frame

• A reference frame is an axis system from which motion is observed along with a clock attached to the axis, to measure time. The reference frame can be stationary or moving.
• Suppose there are two persons A and B sitting in a car moving at constant speed. Two stationary persons C and D observe them from the ground.

• Here B appears to be moving fr C and D, but at rest for A. Similarly, C appears to be at rest for D but moving backward for A and B.

3.0Relative Motion Analysis

• In a two dimensional situation, Consider three observers at Rest O, A and B. The position of B from the reference frame of A is ${\vec{r}}_{BA}$. We must understand that if A wants to go and meet B he needs to look only at ${\vec{r}}_{BA}$ and decide the direction of his velocity which will take him to B. For A, the origin of the reference frame is not O it is A itself. From our knowledge of vectors we can deduce that

$\begin{array}{rl}& {\vec{r}}_B={\vec{r}}_A+{\vec{r}}_{BA}\\ & {\vec{r}}_{BA}={\vec{r}}_B-{\vec{r}}_A\dots \dots \dots .\end{array}$

Position vector of B w.r.t A is defined as ${\vec{r}}_{BA}$

Differentiating this equation W.R.T. time we get

${\vec{v}}_{BA}={\vec{v}}_B-{\vec{v}}_A$ ……….(2)

On further differentiating we get

${\vec{a}}_{BA}={\vec{a}}_B-{\vec{a}}_A$ ……….(3)

Illustration-1:Suppose that a car A travelling on a straight road at 80 km/h passes a car B going in the same direction at 60 km/h, (Fig. a). Then, velocity of A relative to B is given by

${\vec{v}}_A-{\vec{v}}_B=80\mathrm{km}/\mathrm{h}-60\mathrm{km}/\mathrm{h}=20\mathrm{km}/\mathrm{h}$ (to the right in fig. b)

We know that if somebody passes us by in the same direction we don't find them moving very fast.

If A and B are travelling in opposite directions, Fig. c, we show this by giving one velocity + Sign says $V_A$and the other– sign . Hence, we can write

${\vec{v}}_A-{\vec{v}}_B=+80\mathrm{km}/\mathrm{h}-(-60\mathrm{km}/\mathrm{h})=140\mathrm{km}/\mathrm{h}$ (to the right in fig. d)

We know that if somebody passes us in the opposite direction we find them moving really fast.

In effect, in both cases, to find the velocity of A relative to B, we have applied B’s velocity reversed to both cars. It is then just as if B is at rest and A has two velocities $V_A$ and $V_B$, which are subtracted when $V_A$ and $V_B$

are in the same direction and added when they are in opposite directions.

4.0Equations of Motion When Relative Acceleration is Constant

$\begin{array}{rl}& v_{rel}=u_{rel}+a_{rel}t\\ & s_{rel}=u_{rel}+\frac{1}{2}{a}_{rel}{t}^2\\ & v_{rel}^2=u_{rel}^2+2a_{rel}{s}_{rel}\end{array}$

Illustration:Two particles P and Q, initially separated by 75 m, are moving towards each other along a straight line as shown in figure with$a_P=2\mathrm{m}/{\mathrm{s}}^2,u_p=15\mathrm{m}/\mathrm{s},and{a}_P=4\mathrm{m}/{\mathrm{s}}^2,u_Q=25\mathrm{m}/\mathrm{s}$. Calculate the time when they meet.            

Solution:

Ground Frame

Choosing the origin at P and right as positive

At convergence,

$\begin{array}{rl}& \text{ Position of }P=\text{ Position of }Q\\ & \begin{array}{rl}& \Rightarrow x_P=x_Q\\ & \Rightarrow {\left(x_o+ut+\frac{1}{2}a{t}^2\right)}_P={\left(x_o+ut+\frac{1}{2}a{t}^2\right)}_Q\\ & \Rightarrow {\left(0+15t+\frac{1}{2}\times 2\times t^2\right)}_P={\left(75-25t+\frac{1}{2}\times -4\times t^2\right)}_Q\\ & \Rightarrow 15t+t^2=75-25t-2t^2\\ & \Rightarrow 3t^2+40t-75=0\end{array}\\ & \Rightarrow t=-15\mathrm{s}\text{ or }t=\frac{5}{3}\mathrm{s}\end{array}$

Neglecting negative value of time, $t=\frac{5}{3}s$

Alternate Method

P Frame

• Let the frame be P as shown in figure. Here, P acts as the origin and it is at rest. Position, velocity, and acceleration of Q is a relative quantity with respect to P.

Now, P and Q will collide when Q covers a distance of 75 m and hits P, which is at rest in this reference frame.

$\begin{array}{rl}& \begin{array}{rl}& s=ut+\frac{1}{2}a{t}^2\\ & \Rightarrow =-75=-40t-\frac{1}{2}6t^2\end{array}\\ & \begin{array}{rl}& \Rightarrow 3t^2+40t-75=0\\ & t=\frac{5}{3}s\text{ or }t=-15s\end{array}\end{array}$

Neglecting negative value of time, $t=\frac{5}{3}s$

5.0Relative Motion in Two Dimension

$\begin{array}{rl}& {\vec{r}}_A=\text{ position of }A\text{ with respect to }O\\ & {\vec{r}}_B=\text{ position of }A\text{ with respect to }O\\ & {\vec{r}}_{AB}=\text{ position of }A\text{ with respect to }B\\ & {\vec{r}}_{AB}={\vec{r}}_A-{\vec{r}}_B\\ & \frac{d\left({\vec{r}}_{AB}\right)}{dt}=\frac{d\left({\vec{r}}_A\right)}{dt}-\frac{d\left({\vec{r}}_B\right)}{dt}\\ & {\vec{v}}_{AB}={\vec{v}}_A-{\vec{v}}_B\\ & \frac{d\left({\vec{v}}_{AB}\right)}{dt}=\frac{d\left({\vec{v}}_A\right)}{dt}-\frac{d\left({\vec{v}}_B\right)}{dt}\\ & {\vec{a}}_{AB}={\vec{a}}_A-{\vec{a}}_B\end{array}$

Collision of Two Bodies with Help of Relative Motion

Illustration-3.Two towers, labeled AB and CD, are located a horizontal distance d apart. Tower AB is 20 meters tall, while tower CD is 30 meters tall.An object of mass m is projected horizontally from the top of tower AB toward tower CD with a speed of 10 m/s. At the same instant, another object of mass 2m is projected from the top of tower CD toward tower AB at an angle of 60° above the horizontal, with the same initial speed of 10 m/s.Both objects travel in the same vertical plane, collide in mid-air, and stick together upon impact.Determine the horizontal distance d between the two towers.

Solution:Acceleration of A and C both is $9.8\mathrm{m}/{\mathrm{s}}^2$ downwards.

Therefore, relative acceleration between them is zero i.e., the relative motion between them will be straight line.

Now assuming A to be at rest, the condition of collision will be that ${\vec{v}}_{CA}={\vec{v}}_C-{\vec{v}}_A$ relative velocity of C w.r.t. A should be along CA.

$\begin{array}{rl}& {\vec{v}}_A=10\hat{i}\\ & \begin{array}{rl}& {\vec{v}}_C=-5\hat{i}-5\sqrt{3}\hat{j}\\ & {\vec{v}}_{CA}=-5\hat{i}-5\sqrt{3}\hat{j}-10\hat{i}\\ & {\vec{v}}_{CA}=-15\hat{i}-5\sqrt{3}\hat{j}\end{array}\end{array}$

Angle made by ${\vec{v}}_{CA}withx-axis,\tan \theta =\frac{-5\sqrt{3}}{-15}=\frac{1}{\sqrt{3}}$

$\Rightarrow \theta =30^{\circ }$

Now, by the geometry of figure

$\tan 30^{\circ }=\frac{10}{d}\Rightarrow d=10\sqrt{3}m$ Relative Motion in River Flow

• If a man can swim relative to water with velocity v_{m R} and water is flowing relative to ground with velocity ${\vec{v}}_R$, velocity of man relative to ground ${\vec{v}}_m$ will be given by :

${\vec{v}}_{mR}={\vec{v}}_m-{\vec{v}}_R$

• If  ${\vec{v}}_R=0then{\vec{v}}_m={\vec{v}}_{mR}$
• In words,$\text{ velocity of man in still water }=\text{ velocity of man w.r.t. river }$.

River Problem in One Dimension:

Case-1: Man swimming downstream (along the direction of river flow). In this case velocity of river v_R=+u velocity of man w.r.t. River $v_{mR}=+v$

Case-2: Man swimming upstream (opposite to the direction of river flow). In this case velocity of river v_R=-u velocity of man w.r.t. River v_{m R}=+v 

Motion of Man Swimming in a River

Consider a man swimming in a river with a velocity of \vec{v}_{M R} relative to river at an angle of with the river flow. The velocity of river is v_R .Let there be two observers I and II, observer I is on ground and observer II is on a raft floating along with the river and hence moving with the same velocity as that of river. Hence motion w.r.t. observer II is same as motion w.r.t. river. i.e., the man will appear to swim at an angle with the river flows for observer II.

For observer I the velocity of swimmer will be ${\vec{v}}_{mR}={\vec{v}}_m-{\vec{v}}_R$

${\vec{v}}_M={\vec{v}}_{MR}+{\vec{v}}_R$

Hence the swimmer will appear to move at an angle '' with the river flow.

River Problem in Two Dimension (Crossing River)

Consider a man swimming in a river with a velocity of ${\vec{v}}_{MR}$ relative to the river at an angle with the river flow. The velocity of river is $v_R$ and the width of the river is  d

$\begin{array}{rl}& {\vec{v}}_M={\vec{v}}_{MR}+{\vec{v}}_R\\ & {\vec{v}}_M=\left(v_{MR}\cos \theta \hat{i}+v_{MR}\mathrm{Sin}\theta \hat{j}\right)+v_R\hat{i}\\ & {\vec{v}}_M=\left(v_{MR}\cos \theta +v_R\right)\hat{i}+v_{MR}\mathrm{Sin}\theta \hat{j}\end{array}$

Here $v_{MR}\mathrm{Sin}\theta$ is the component of velocity of man in the direction perpendicular to the river flow. This component of velocity is responsible for the man crossing the river. Hence if the time to cross the river is t, 

$t=\frac{d}{v_y}=\frac{d}{v_{MR}\mathrm{Sin}\theta }$

Drift

• It is defined as the displacement of man in the direction of river flow. (See the figure). It is simply the displacement along the x-axis, during the period the man crosses the river $\left(v_{MR}\cos \theta +v_R\right)$ is the component of velocity of man in the direction of river flow and this component of velocity is responsible for drift along the river flow. If drift is x then,

$\begin{array}{rl}& \text{ Drift }=v_x\times t\\ & x=\left(v_{MR}\cos \theta +v_R\right)\times \frac{d}{v_{MR}\sin \theta }\end{array}$

Crossing the River in Shortest Time

• As we know that $t=\frac{d}{v_{MR}\mathrm{Sin}\theta }$, Clearly t will be minimum when $\theta =90^{\circ }$ i.e. time to cross the river will be minimum if man swims perpendicular to the river flow. Which is equal to $\frac{d}{v_{MR}}$

Crossing the River in Shortest Path, Minimum Drift

• The minimum possible drift is zero. In this case the man swims in the direction perpendicular to the river flow as seen from the ground. This path is known as the shortest path.
• Here , $x_{\min }=0\Rightarrow \left(v_{MR}\cos \theta +v_R\right)=0\text{ or }\cos \theta =-\frac{v_R}{v_{MR}}$

Since $\cos \theta \text{ is }-ve$

$\theta >90^{\circ }$, i.e.for minimum drift the man must swim at some angle   with the perpendicular in backward direction. Where $\sin \varphi =\frac{v_R}{v_{MR}}$

$\theta ={\cos }^{-1}\left(-\frac{v_R}{v_{MR}}\right)$

$\left|\frac{v_R}{v_{MR}}\right|\le 1\text{ i.e. }{v}_R\le v_{MR}$

i.e. minimum drift is zero if and only if velocity of man in still water is greater than or equal to the velocity of river.

Time to cross the river along the shortest path

$t=\frac{d}{v_{MR}\mathrm{Sin}\theta }=\frac{d}{\sqrt{v_{MR}^2-v_R^2}}$

Note:

If $v_R>v_{MR}$ then it is not possible to have zero drift. In this case the minimum drift (corresponding to shortest possible path is non-zero and the condition for minimum drift can be proved to be $\cos \theta =-\left(\frac{v_{MR}}{v_R}\right)$ or $\sin \varphi =\left(\frac{v_{MR}}{v_R}\right)$ for minimum but non-zero drift.

for minimum but non-zero drift.

6.0Wind Aeroplane Problems

• This is very similar to boat river flow problems. The only difference is that boats are replaced by aeroplanes and rivers are replaced by wind. Thus, velocity of aeroplane with respect to wind

${\vec{v}}_{aw}={\vec{v}}_a-{\vec{v}}_w{\vec{v}}_a=\text{ velocity of aeroplane w.r.t. ground }{\vec{v}}_w=\text{ velocity of wind. }$

7.0Rain Problem

If rain is falling vertically with a velocity \vec{v}_R and an observer is moving horizontally with velocity ${\vec{v}}_m$, the velocity of rain relative to observer will be:

$\begin{array}{rl}& {\vec{v}}_{Rm}={\vec{v}}_R-{\vec{v}}_m\\ & v_{Rm}=\sqrt{v_R^2+v_m^2}\end{array}$

And direction $\theta ={\tan }^{-1}\left(\frac{v_m}{v_R}\right)$ with the vertical as shown in figure

8.0Velocity of Approach/Separation

• It is the component of relative velocity of one particle w.r.t. another, along the line joining them. If the separation is decreasing, we say it is velocity of approach and if separation is increasing, then we say it is velocity of separation. In one dimension, since relative velocity is along the line joining A and B, hence velocity of approach / separation is simply equal to magnitude of relative velocity of A w.r.t. B.

Illustration-4.A particle A is moving with a speed of 10 m/s towards right and another particle B is moving at speed of 12 m/s towards the left. Find their velocity of approach.

Solution:

$\begin{array}{rl}& V_A=+10,V_B=-12\\ & V_{AB}=V_A-V_B\Rightarrow 10-(-12)=22\mathrm{m}/\mathrm{s}\end{array}$

Since separation is decreasing, hence $V_{app}=\left|V_{AB}\right|=22\mathrm{m}/\mathrm{s}$

9.0Velocity of Approach / Separation in Two Dimension

• It is the component of relative velocity of one particle w.r.t. another, along the line joining them. If the separation is decreasing, we say it is velocity of approach and if separation is increasing, then we say it is velocity of separation.

Illustration-5.Two particles A and B are moving with constant velocities $v_1$ and  $v_2$.At t = 0, $v_1$ makes an angle ${\theta }_1$ with the line joining A and B and $v_2$ makes an angle ${\theta }_2$ with the line joining A and B. Find their velocity of approach.

Solution:Velocity of approach is relative velocity along line AB

$v_{app}=v_1\cos {\theta }_1+v_2\cos {\theta }_2$