In this article, we are going to discuss the concept, definition and formula for relative velocity with the help of solved problems and examples. Before going through this article, kindly learn the concept of speed, velocity, distance and displacement.

## Concept of relative velocity

Suppose we are travelling in the car. Now say, another car overtakes us. We will not feel the actual speed of the overtaking car, as felt by a person who looks it, standing by the side of the road.

If both the cars are moving with the same speed in the same direction, a person in one car observes a person in the other car at rest, even though both are in motion.

If both cars are in the opposite direction, then the passengers observe a greater speed, greater than their individual speed.

In all these situations, we measure the relative velocities. Thus, the relative velocity is the velocity of one body with respect to another body. It is measured as the rate of change of position of a body with respect to another body.

## Relative velocity – Definition

The velocity of a body with respect to the velocity of another body is called the relative velocity of the first body with respect to the second.

If VA and VB represent the velocities of two bodies A and B respectively at any instant, then, the relative velocity of A with respect to B is represented by VAB

Then, VAB   = VA – VB                    ——————  (1)

Similarly, the relative velocity of B with respect to A is given by,
VBA   = VB – VA                    ——————–(2)

From equation (1) and (2), we can write
VBA   =  – VAB

### Case 1: Two bodies moving in the same direction with equal velocities.

Let two buses A and B be moving in the same direction with equal velocities (i.e., VA = VB ). To a person seated in A, the bus B would appear to be at rest, if he forgets for a moment the fact that he himself is in motion. Hence the velocity of B relative to A is zero. (figure 1).

i.e.,  VBA = VB – V= 0

Same is the case with a person in bus B and observing the bus A. The relative velocity of A with respect to B is also zero.

i.e., VAB  = VA – VB   = 0

Therefore, we can conclude that, when both the objects are moving in the same direction along parallel straight lines with equal velocities, their position-time graphs are parallel straight lines as shown in figure 1. The distance between these objects will be the same at all instants. Therefore, the velocity of each one with respect to the other is zero.

### Case 2: Two bodies starting from the same point in the same direction with unequal velocities.

Let A be moving with a velocity VA and B,(starting from two points close to each other at the same time), be moving with a greater velocity VB   in the same direction. Then, the person in A feels that bus B is moving away from him with a velocity,
VBA = VB – VA

For an observer in B, the bus A appears to go back with a velocity, VAB   =  – (VB – VA )

That is, the velocity of A relative to B is the negative of the velocity of B relative to A.

### Case 3: Two bodies moving in the same direction starting from different positions with unequal velocities.

As a general case, let the position coordinates of the two buses A and B be different, say X1(0)   and X2(0)   at time t = 0, moving with a velocity VA   and Vrespectively. After an interval of time t, let X1(t)   and X2(t)  be the position coordinates of the two buses, then

X1(t)   = X1(0) + VA   t            ———————–(3)

X2(t)   = X2(0) + VB t             ————————(4)

eqn (4) – (3),
then,  X2(t)  –  X1(t)  =  X2(0)  –  X1(0) + (VB – VA )t               —————(5)

At the instant t, the distance between A and B, or their relative displacement is given by, X2(t) – X1(t)

This quantity may be positive, negative or zero. A positive value means that B is ahead of A and a negative value means that A is ahead of B. From equation (5), it is clear that the distance between A and B changes with time.

Also, equation (5) shows that as observed from the bus A, the bus B has a velocity (VB – VA ) because the displacement between A and B changes steadily by the amount (VB – VA ) in every unit of time.

Therefore, the velocity of B relative to A is given by
VBA = VB – VA

If VB – V= 0, means the two cars are moving in the same direction and are keeping the same distance between them always as in case 1. So their position-time graphs are two parallel straight lines as shown in figure 1.

If VB – VA   ≠ 0, means the two buses are moving with unequal velocities. Their position-time graphs are straight lines inclined to the time axis and one of them will be steeper than the other. Hence they will intersect at a point and the object along the steeper line overtakes the other. Here, if the bus A which is having the higher velocity moving behind the bus B, then the bus A will overtake the bus B at some instant and the position-time graphs of the two buses will intersect at a point as shown in figure 3.   Also, from the very beginning, if the bus with greater velocity is moving ahead, there is no chance of overtaking the other and hence the position-time graphs will not intersect as shown in figure 4.

### Case 4: Two bodies moving in the opposite direction.

In this case, the position-time graph is as shown in figure 5.

If VA  and  VB are moving in opposite directions, the relative velocity VBA   or VAB   will be
VB – ( – VA ) = VB + VA

This velocity is greater than both VA  and  VB   in magnitude. That is, when two buses A and B are moving in opposite directions each seems to go very fast relative to the other.

Also, if A and B are moving with same velocity V  in the opposite direction ( i.e., VA = VB = V), then the relative velocity of A with respect to B is given by,
v – (-V) = 2V = – relative velocity of B with respect to A.

## Relative Velocity Formula – Derivation

We already discussed the definition and formula to calculate relative velocity in the previous section of this article. The derivation of relative velocity formula is written in this section.

Consider two bodies, A and B moving along the X axis with uniform velocities VA and VB    respectively. Let XA(0) and XB(0) be their positions at t = 0. At any instant t, let XA(t) and XB(t) be their positions. Then

XA(t) = XA(0) + VA t         ———————- eqn (1)
XB(t) = XB(0) + VB t         ———————- eqn (2)

Substacting  eqn (1) from eqn (2)

XB(t) – XA(t) = XB(0) –  XA(0) + (V– VA )   t

or, [XB(t) – XA(t)] – [XB(0) –  XA(0) ] = (VB    – VA )   t large begin{aligned} therefore; V_{B}-V_{A}=frac{[X_{B}(t)-X_{A}(t)] - [X_{B}(0)-X_{A}(0)]}{t} ;;;;;-------eqn(3) end{aligned}

Here [XB(0) –  XA(0)] = distance of the object B from object A at time t = 0.
[XB(t) – XA(t)] = distance of the object B from object A at time t.

Hence, large begin{aligned} frac{[X_{B}(t)-X_{A}(t)] - [X_{B}(0)-X_{A}(0)]}{t} &=Relative;displacement;of;object;B;from;object;A;in;unit;time \&=Velcoity;of;object;B;relative;to;object;A \&=V_{BA} end{aligned}

Therefore, from equation (3) we get,
VBA   = VB –  VA
This equation gives the velocity of object B relative to object A.

Similarly, the velocity of object A relative to object b is given by,
VAB   = VA –  VB

From the above equations, we can write
VBA   =  -VAB

## Relative velocity problems & examples

The concept of relative velocity can be understood more clearly with the help of the following examples.

Problem 1:
Two parallel tracks run north-south. Train A moves north with a speed of 68 km/hr and train B moves south with a speed of 94 km/hr. Find the
a) relative velocity of B with respect to A.
b) relative velocity of ground with respect to B.
c) relative velocity of a monkey running on the roof of the train A against its motion with a velocity of 16 km/hr ( with respect to the train A) as observed by a man standing on the ground.

Solution:
Given the velocity of train A, VA = 68 km/h
= 68 × 1000/3600
= 18.89 m/s (assume south to north direction is taken as positive)

Velocity of train B, VB = -94 km/h
= – 94 × 1000/3600
= -26.11 m/s

Relative velocity of monkey with respect to train A, VMA = -16 km/h (negative beacuse the monkey is moving against the motion of A)
= -16 × 1000/3600
= -4.44 m/s

a) Relative velocity of B with respect to A = VB – VA
= – 26.11 – 18.89
= – 45 m/s

b) Relative velocity of ground with respect to B = 0 – VB
= 0 – –26.11
= 26.11 m/s

c) Here, we need to find out the relative velocity of the monkey with respect to ground, VM.
Given, the relative velocity of the monkey with respect to the train A,   VMA  = –4.44 m/s
i.e.,          VM – V= –4.44 m/s
i.e.,    VM – 18.89  = –4.44 m/s
∴    VM =  –4.44 + 18.89
= 14.45 m/s
i.e., the relative velocity of a monkey with respect to an observer at the ground = 14.45 m/s

Problem 2:
An aeroplane travelling at the speed of 400 km/h ejects its combustion gases at the speed of 1300 km/h relative to the aeroplane. What is the speed of the later with respect to an observer on the ground?

Let Vg = velocity of combustion gases
Given velocity of aeroplane, Vp = 400 km/h
Relative velocity of gases w.r.t aeroplane, Vgp = –1300 km/hr.

We know, Relative velocity of gases w.r.t aeroplane, Vgp = Vg – Vp
i.e.,   –1300 = Vg – 400
∴ Vg = –1300 + 400
= –900 km/h

∴ the speed of gases w.r.t the observer on the ground      = –900 km/h

Problem 3:
A police bus moving on a highway with a speed of 40 km/hr fires a bullet at a thief’s car speeding away in the same direction with a speed of 185 km/h. If the muzzle speed of the bullet is 160 m/s, with what speed does the bullet hit the thief’s car?

Given, Speed of the police bus, Vp = 40 km/h
= 40 × 1000/3600
= 11.11 m/s
Speed of thief’s car, Vt = 185 km/h
= 185 × 1000/3600
= 51.39 m/s

Muzzle speed of bullet = 160 m/s

The bullet will share the speed of the car, hence the speed of the bullet,
Vb = speed of police bus + muzzle speed of the bullet

= 11.11 + 160
= 171.11 m/s

The relative velocity of the bullet with respect to the thief’s car, Vbt = Vb- Vt
= 171.11 – 51.39
= 119.72 m/s
i.e., the bullet will hit the thief’s car at 119.72 m/s.

Problem 4:
How long will a boy sitting near the window of a train travelling at 68 km/hr see a train passing by in the opposite direction with a speed of 42 km/h? The length of the slow-moving train is 125 m.

Solution:
Given, distance = 125 m

The relative velocity of the slow-moving train w.r.t the boy = 68 + 42
= 110 km/h
= 110 × 1000/3600
= 30.6 m/s

∴ time, t = Distance / Relative velocity
= 125/30.6
= 4.08 seconds

Problem 5:
Two cities X and Y are connected by regular bus service with a bus leaving in either direction in every T minutes. A cyclist with a speed of 16 km/h in the direction X to Y observes that a bus crosses him in every 21 minutes in the direction of his motion and in every 8 minutes in the opposite direction. What is the period T of the bus service and what is the speed of the buses on the road?

Let v be the velocity of the bus.

Distance between the buses = vT

Velocity of the bus going from X to Y with respect to the cyclist = (v –16)km/h  large begin{aligned}We;know,;;frac{Displacment}{Velocity}&=time \thereforefrac{vT}{v - 16}&= frac{21}{60} \i.e., vT &= frac{21}{60}(v-16) end{aligned}———-eqn 1

Similarly velocity of the bus going from Y to X with respect to the cyclist = (v + 16) km/h large begin{aligned} thereforefrac{vT}{v + 16}&= frac{8}{60} \or; vT &= frac{8}{60}(v+16) end{aligned}————eqn 2

From equations (1) and (2), 21(v-16) = 8(v+16)
i.e.,   21v –336 = 8v + 128
ie.,  13 v = 464
∴ v = 464/13
= 35.69 km/h large begin{aligned} From;eqn;1,;Period;T &= frac{21(v-16)}{60v} \&=frac{21(35.6-16)}{60times35.6} \&=frac{21times19.6}{60times35.6} \&=frac{11.54}{60};h \&=11.54;minutes end{aligned}

Problem 6:
A cyclist is travelling at 10.8 km/h along a road parallel to a straight railway track. Along the track, a train of length 150 m is moving in the opposite direction. The cyclist takes 9 s to cross the length of the train. Find the speed of the train.

Solution:
Speed of the cyclist = 10.8 km/h
= 10.8 × 5/18 m/s
= 3 m/s

The direction of motion of the cyclist is taken as the positive X-direction.
Let the speed of the train be ( – v ).

The velocity of the cyclist relative to the train = 3 – (-v)
= 3 + v

Given, Length of the train = 150 m
Tme taken = 9 s

We know, large begin{aligned} Relative;velocity &= frac{Distance}{Time} \i.e.,; 3+v&=frac{150}{9} \i.e.,; 3+v&=16.66 \therefore v&=16.66-3 \&=13.66 end{aligned}

i.e., the speed of the train = 13.66 m/s

I hope the information in this article helps you to get a brief idea about the concept of relative velocity. If you think I have missed any points, or if you have any suggestions or feedback, do let me know in the comments section.

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