Time, Speed & Distance Primer  Ravi Handa

There are some topics in Quantitative Aptitude, like permutation & combination, where you can easily find out the answer, (it is in the options) but it turns out to be wrong. And then there are some topics in which you read the question, understand it but cannot even begin solving it. You get stuck at the first step and you have no idea about how to even approach the question.
The irritating fact is that you understood the question properly. It happens very frequently with questions on Time Speed & Distance (TSD). I have always been a big advocate of skipping questions which you cannot solve. More often than not, TSD questions should be skipped if you cannot figure out how to start within the first minute. Typically the questions on TSD are based upon certain ideas, which if you are not aware of can make solving the question extremely difficult and time consuming. I do not think that I am even aware of all ideas / types of questions in TSD but there are a few popular ones which have been doing the rounds in the past few years.
I am going to cover some of them in this post and probably revisit some more in the months to come. I am also going to discuss the reasoning behind those ideas. It is very important that you understand the logic behind the formulae before you actually start using them. If you don't, there is a very high probability that you will make a mistake.
To begin with, some of the very basic ideas that you should be aware of are:
Speed = Distance / Time
If Distance is constant, then Speed and Time are inversely proportional to each other
Two bodies moving in the same direction would have the relative speed of S(1)  S(2) and two bodies moving in the opposite direction would have the relative speed of S(1) + S(2)In this particular post, I am going to talk about the motion of two bodies in a straight line starting from opposite ends.
Case 1:
Two bodies start from opposite ends P & Q at the same time and move towards each other with speeds S(1) & S(2).
They take times of T(1) & T(2) to reach their destinations.In such a case, the relationship between the times taken and the speeds will be S(1) / S(2) = T(2) / T(1)
The distances that both bodies have travelled are PQ and QP respectively.
PQ = S(1) T(1)
QP = S(2) T(2)Also, PQ = QP so the above equations can be equated to get the desired result.
Let us say that they meet at a point R in between, then
R would divide the distance PQ in the ratio of S(1):S(2)
They started at the same time and they are meeting at point R, so the time taken by both the bodies will be the same.
If we assume that time as T
PR = S(1) T
QR = S(2) TDividing the above two equations would give us the desired result.
Case 2:
Two bodies start from opposite ends P & Q at the same time and move towards each other with speeds S(1) & S(2). After meeting each other, they take times of T(1) & T(2) to reach their destinations.
In such a case, the relationship between the times taken and the speeds will be S(1)/S(2) = √(T(2)/T(1))
Also, the time taken for the two bodies to to meet will be √(T(1) x T(2))
Let us assume that the two bodies meet at a point R, after time T.
For the first body, PR = S(1)T and RQ = S(1)T(1)
For the second body, QR = S(2)T and RP = S(2)T(2)
We know that PR = RP
=> S(1)T = S(2)T(2)
We know the QR = RQ
=> S(2)T = S(1)T(1)
Dividing the above two equations, we will get
Using the above result, we can obtain the value of T
Case 3:
Two bodies start from opposite ends P & Q to reach their destinations at the same time and move towards each other with speeds S(1) & S(2). Before meeting, they take times of T(1) & T(2) to reach the meeting point.
In such a case, the relationship between the times taken and the speeds will be S(1) / S(2) = √ (T(2) / T(1))
Also, the time taken for the two bodies to reach their destinations after meeting will be √ (T(1) x T(2))
The logic for this would be exactly the same as Case 2. Try working this out on your own.
Case 4:
Two bodies start from opposite ends P & Q at the same time and move towards each other with speeds S(1) & S(2). They reach the opposite ends and reverse directions. They continue this to and fro motion.
If the distance between the two bodies in the beginning is D, then
Time taken by them to meet for the first time, T = D / S(1) + S(2)
If R is the first meeting point, PR / QR = S(1) / S(2)
The idea that I am going to discuss now is only valid in the case when the greater speed is less than twice of the lesser speed. {If S(1) > S(2), then S(1) < 2S(2)}
After the first meeting, if the bodies continue to move they will reach their respective ends and start the return journey. They will meet again on the return journey and then proceed further. After the first meeting, they would have covered 2D distance. Since the distance has doubled, time taken will also double.
So, the total distance covered by the bodies for their first, second, third ... nth meetings will be:
D, 3D, 5D ... and (2n1)D
So, the total times taken by the bodies for their first, second, third ... nth meetings will be:
D / S(1) + S(2), 3 D / S(1) + S(2), 5 D / S(1) + S(2) ... and (2n1) D / S(1) + S(2)
In case you need to figure out the point at which the bodies meet for the nth time, consider only one of the bodies, say 1.
Distance covered by 1 till the nth meeting = [S(1)/(S(1)+S(2))] * (2n1)D
The remainder of the above when divided by 2D will give you the exact location of the nth meeting point.
For example, if the distance covered by 1 till the nth meeting comes out as 700 meters and D = 150 meters, the nth meeting point can be obtained by the remainder of 700/300 which is 100. So, the nth meeting point will be 100 m from P.
I know you would hate me for saying this, but this is not the end of TSD. This is not even the end of motion of two bodies in a straight line. The destination is still far away and hopefully we will reach there in time.