For example, if an object has a certain velocity, it must be described with respect to a given reference frame. In most examples, the reference frame is Earth.

Suppose a train X moving at 30 m/s along the east, then it is considered that the train X is moving relative to the surface of Earth at 30 m/s. Suppose another train Y overtakes the train X from behind. What actually happens is that any passenger in train X from the behind sees the train X coming in the backward direction and eventually goes back. But a person standing on the ground doesn’t observe that the train X is moving backward. This is what relative velocity is. To understand such observations, the concept of relative velocity is introduced.

Suppose, two objects A and B are moving with uniform velocities $V_{A}$ and $V_{B}$ along the same direction respectively.

If $X_{OA}$and $X_{OB}$be their distances from origin at time t = 0, and $X_{A}$and $X_{B}$be their distances at time t, then,

For object A, $X_{A}$= $X_{OA}$+ $V_{A}$t

For object B, $X_{B}$= $X_{OB}$+ $V_{B}$t

Now, $X_{B}$- $X_{A}$= $(X_{OB}$- $X_{OA}$) + ($V_{B}-V_{A}$) t ………(1)

From equation (1), object B seems to have velocity ($V_{B}-V_{A}$) with respect to A.

Then, ($V_{B}-V_{A}$) is the velocity of B relative to A.

Thus, the relative velocity of B with respect to A is $V_{BA}=V_{B}-V_{A}$ ………….. (2)

Similarly, the relative velocity of A with respect to B is $V_{AB}=V_{A}-V_{B}$ ……………. (3)

The relative velocity of A with respect to B is $V_{AB}=V_{A}-V_{B}$

The relative velocity of B with respect to A is $V_{BA}=V_{B}-V_{A}$

If $V_{A}$ = $V_{B}$, then from equation (1), $X_{B}$- $X_{A}$= $(X_{OB}$- $X_{OA}$). It signifies that objects A and B stay apart at a constant distance. It is shown below.

*Figure:4.a*

If $V_{A}$ {$>$ }$V_{B}$, then $(V_{B}-V_{A})$ is negative and the distance between them will go on decreasing by an amount $(V_{B}-V_{A}$) in every unit of time and after some time they will meet and object A will overtake object B. It is shown below.

*Figure:4.b*

If $V_{A}$ and $V_{B}$ are opposite signs:

Then the magnitude of $V_{BA}$ or $V_{AB}$ will be greater than the magnitude of the velocity of object A or that of object B.

Therefore, considering the example of the motion of two trains X and Y as stated above, a person sitting on either of the two trains, the other train seems to go very fast.

## Conceptual problems:

**1. A train travels at 30 m/s to the east with respect to the earth’s reference frame. A passenger on the train runs at 2 m/s to the opposite with respect to the train. Find the velocity of the man with respect to the ground.**

Given:

Velocity of the train with respect to the ground (Earth’s reference frame) = 30 m/s

Velocity of the passenger with respect to the train = – 2 m/s

Velocity of the passenger with respect to ground = – 2 + 30 = 28 m/s

**2. A sail travels a distance of 500 m along east from point A to B in 120 s and turns back to A in 70 s. If the velocity of the river is zero, find the total time taken.**

The direction of the velocity of the sail is along the east and that of the river is west.

Velocity of sail with respect to river = $V_{SR}$

Velocity of sail with respect to ground = $ V_{SG}$

Velocity of river with respect to ground = $V_{RG}$

Therefore, velocity of sail with respect to ground, when it travels from A to B is

$ V_{SG}$= $V_{SR}$- $V_{RG}$ …………….(1)

And, velocity of sail with respect to ground, when it travels from B to A is

$ V_{SG}$= $V_{SR}$+ $V_{RG}$ ……..(2)

From equation (1), $V_{SR}$- $V_{RG}$ = 500/120 = 4.16 m/s

From equation (2), $V_{SR}$+ $V_{RG}$= 500/70 = 7.14 m/s

Solving the equations, we get

$V_{SR}$= 5.65 m/s and $V_{RG}$= 1.49 m/s

Given, velocity of river is zero. Therefore, the total time taken to cover (500 m + 500 m) = 1000 m is 1000/5.65 = 177 s.