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This technique is a very useful tool for consideration of relative motion in two dimensions.
For a pair of two moving objects moving uniformly, there are two values of relative velocity corresponding to two reference frames. The values differ only in sign – not in magnitude. This is clear from the example here.
Problem : Two cars start moving away from each other with speeds 1 m/s and 2 m/s along a straight road. What are relative velocities ? Discuss the significance of their sign.
Solution : Let the cars be denoted by subscripts “1” and “2”. Let us also consider that the direction ${v}_{2}$ is the positive reference direction, then relative velocities are :
$$\begin{array}{l}\Rightarrow {v}_{12}={v}_{1}-{v}_{2}=-1-2=-3\phantom{\rule{2pt}{0ex}}m/s\\ \Rightarrow {v}_{21}={v}_{2}-{v}_{1}=2-(-1)=3\phantom{\rule{2pt}{0ex}}m/s\end{array}$$
The sign attached to relative velocity indicates the direction of relative velocity with respect to reference direction. The directions of relative velocity are different, depending on the reference object.
However, two relative velocities with different directions mean same physical situation. Let us read the negative value first. It means that car 1 moves away from car 2 at a speed of 3 m/s in the direction opposite to that of car 2. This is exactly the physical situation. Now for positive value of relative velocity, the value reads as car 2 moves from car 1 in the direction of its own velocity. This also is exactly the physical situation. There is no contradiction as far as physical interpretation is concerned. Importantly, the magnitude of approach – whatever be the sign of relative velocity – is same.
It is very important to understand that relative velocity refers to two moving bodies – not a single body. Also that relative velocity is a different concept than the concept of "difference of two velocities", which may pertain to the same or different objects. The difference in velocities represents difference of “final” velocity and “initial” velocity and is independent of any order of subscript. In the case of relative velocity, the order of subscripts are important. The expression for two concepts viz relative velocity and difference in velocities may look similar, but they are different concepts.
We had restricted out discussion up to this point for objects, which moved with constant velocity. The question, now, is whether we can extend the concept of relative velocity to acceleration as well. The answer is yes. We can attach similar meaning to most of the quantities - scalar and vector both. It all depends on attaching physical meaning to the relative concept with respect to a particular quantity. For example, we measure potential energy (a scalar quantity) with respect to an assumed datum.
Extending concept of relative velocity to acceleration is done with the restriction that measurements of individual accelerations are made from the same reference.
If two objects are moving with different accelerations in one dimension, then the relative acceleration is equal to the net acceleration following the same working relation as that for relative velocity. For example, let us consider than an object designated as "1" moves with acceleration " ${a}_{1}$ " and the other object designated as "2" moves with acceleration " ${a}_{2}$ " along a straight line. Then, relative acceleration of "1" with respect to "2" is given by :
$$\begin{array}{l}\Rightarrow {a}_{12}={a}_{1}-{a}_{2}\end{array}$$
Similarly,relative acceleration of "2" with respect to "1" is given by :
$$\begin{array}{l}\Rightarrow {a}_{21}={a}_{2}-{a}_{1}\end{array}$$
Problem : Two trains are running on parallel straight tracks in the same direction. The train, moving with the speed of 30 m/s overtakes the train ahead, which is moving with the speed of 20 m/s. If the train lengths are 200 m each, then find the time elapsed and the ground distance covered by the trains during overtake.
Solution : First train, moving with the speed of 30 m/s overtakes the second train, moving with the speed of 20 m/s. The relative speed with which first train overtakes the second train,
$$\begin{array}{l}{v}_{12}={v}_{1}-{v}_{2}=30-20=\mathrm{10\; m/s.}\end{array}$$
The figure here shows the initial situation, when faster train begins to overtake and the final situation, when faster train goes past the slower train. The total distance to be covered is equal to the sum of each length of the trains (L1 + L2) i.e. 200 + 200 = 400 m. Thus, time taken to overtake is :
$$\begin{array}{l}t=\frac{\mathrm{400}}{\mathrm{10}}=\mathrm{40\; s.}\end{array}$$
In this time interval, the two trains cover the ground distance given by:
$$\begin{array}{l}s=30x40+20x40=1200+800=\mathrm{2000\; m.}\end{array}$$
In the question given in the example, if the trains travel in the opposite direction, then find the time elapsed and the ground distance covered by the trains during the period in which they cross each other.
$$\begin{array}{l}{v}_{12}={v}_{1}-{v}_{2}=30-(-20)=\mathrm{50\; m/s.}\end{array}$$
The total distance to be covered is equal to the sum of each length of the trains i.e. 200 + 200 = 400 m. Thus, time taken to overtake is :
$$\begin{array}{l}t=\frac{\mathrm{400}}{\mathrm{50}}=\mathrm{8\; s.}\end{array}$$
Now, in this time interval, the two trains cover the ground distance given by:
$$\begin{array}{l}s=30x8+20x8=240+160=\mathrm{400\; m.}\end{array}$$
In this case, we find that the sum of the lengths of the trains is equal to the ground distance covered by the trains, while crossing each other.
Check the module titled Relative velocity in one dimension (Check your understanding) to test your understanding of the topics covered in this module.
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