You must be familiar with the experience of travelling in a train and being overtaken by another train moving in the same direction as you are. While that train must be travelling faster than you to be able to pass you, it does seem slower to you than it would be to someone standing on the ground and watching both the trains. In case both the trains have the same velocity with respect to the ground, then to you the other train would seem to be not moving at all. To understand such observations, we now introduce the concept of relative velocity.
Consider two objects A and B moving uniformly with average velocities vA and vB in one dimension, say along x-axis. (Unless otherwise specified, the velocities mentioned in this chapter are measured with reference to the ground). If $x_A$(0) and $x_B$(0) are positions of objects A and B, respectively at time t = 0, their positions $x_A$(t) and $x_B$(t) at time t are given by:
$x_A(t)=x_A(0)+v_At$ (1a)
$x_B(t)=x_B(0)+v_Bt$ (1b)
Then, the displacement from object A to object B is given by
$x_{BA}(t)=x_B(t)+x_A(t)$
$x_{BA}(t)=[x_B(0)+v_B(t)]-[x_A(0)+v_A(t)]$
$x_{BA}(t)=[x_B(0)-x_A(0)]-(v_B-v_A)t$ (2)
Equation (3.13) is easily interpreted. It tells us that as seen from object A, object B has a velocity $v_B$ – $v_A$ because the displacement from A to B changes steadily by the amount $v_B$ – $v_A$ in each unit of time. We say that the velocity of object B relative to object A is $v_B$ – $v_A$:
$v_{BA}=v_B-v_A$ (3a)
Similarly, velocity of object A relative to object B is:
$v_{AB}=v_A-v_B$ (3b)
 |
| Fig 1: Position-time graphs of two objects with equal velocities. |
This shows: $v_{BA}=-v_{AB}$
Now we consider some special cases:
(a) If $v_B$ = $v_A$, $v_B$ –$v_A$ = 0. Then, from Eq. (3.13), $x_B$(t) – $x_A$(t) = $x_B$(0) – $x_A$(0). Therefore, the two objects stay at a constant distance ($x_B$(0) – $x_A$(0)) apart, and their position–time graphs are straight lines parallel to each other as shown in Fig. 1. The relative velocity $v_{AB}$ or $v_{BA}$ is zero in this case.
(b) If $v_A$ > $v_B$, $v_B$ – $v_A$ is negative. One graph is steeper than the other and they meet at a common point. For example, suppose $v_A$ = 20 m$s^{-1}$ and $x_A$(0) = 10 m; and $v_B$ = 10 m$s^{-1}$, $x_B$(0) = 40 m; then the time at which they meet is t = 3 s (Fig. 2). At this instant they are both at a position $x_A$(t) = $x_B$(t) = 70 m. Thus, object A overtakes object B at this time. In this case, $v_{BA}$ = 10 m.s$^{–1}$ – 20 ms$^{–1}$ = – 10 ms$^{–1}$ = – $v_{AB}$.
 |
| Fig 2: Position-time graphs of two objects with unequal velocities, showing the time of meeting. |
(c) Suppose $v_A$ and $v_B$ are of opposite signs. For example, if in the above example object A is moving with 20 ms$^{–1}$ starting at $x_A$(0) = 10 m and object B is moving with – 10 ms$^{–1}$ starting at $x_B$(0) = 40 m, the two objects meet at t = 1 s (Fig. 3). The velocity of B relative to A, $v_{BA}$ = [–10 – (20)] ms$^{–1}$ = –30 ms$^{–1}$ = – $v_{AB}$. In this case, the magnitude of $v_{BA}$ or $v_{AB}$ (= 30 ms$^{–1}$) is greater than the magnitude of velocity of A or that of B. If the objects under consideration are two trains, then for a person sitting on either of the two, the other train seems to go very fast.
 |
| Fig. 3: Position-time graphs of two objects with velocities in opposite directions, showing the time of meeting. |
Note that Eq. (3) are valid even if $v_A$ and $v_B$ represent instantaneous velocities.
Example 1
Two parallel rail tracks run north-south. Train A moves north with a speed of 54 km h$^{–1}$, and train B moves south with a speed of 90 km h$^{–1}$. What is the (a) velocity of B with respect to A?, (b) velocity of ground with respect to B?, and (c) velocity of a monkey running on the roof of the train A against its motion (with a velocity of 18 km h$^{–1}$ with respect to the train A) as observed by a man standing on the ground?
Answer
Choose the positive direction of x-axis to be from south to north. Then,
$v_A$ = + 54 km.h$^{–1}$= 15 m.s$^{–1}$
$v_B$ = – 90 km.h$^{–1}$ = – 25 m$^{–1}$
Relative velocity of B with respect to A = $v_B$ – $v_A$= – 40 m.s$^{–1}$ , i.e. the train B appears to A to move with a speed of 40 m.s$^{–1}$ from north to south. Relative velocity of ground with respect to
B = 0 – $v_B$ = 25 m.s$^{–1}$.
In (c), let the velocity of the monkey with respect to ground be vM. Relative velocity of the monkey with respect to A,
$v_{MA}$ = $v_M$ – $v_A$ = –18 km.h$^{–1}$ =–5 m.s$^{–1}$. Therefore,
$v_M$ = (15 – 5) m.s$^{–1}$ = 10 m.s$^{–1}$.
Post a Comment for "Relative Velocity"