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.
