Let’s have a look at this topic. The questions which come from this section are lengthy. Many people just leave the question by seeing the length of it. They do not even try to read the question. If you read this article from the start to the end, you will have a different view on these kinds of questions. So without much ado, I shall take you right to the concepts.
In relationships, they say distance is inversely proportion to love. I am not sure how true this is but one thing I am sure of is that you have to understand what inversely proportionate means and what directly proportionate means (I am assuming all my readers are aware of inverse / direct proportions and hence not digging in).
The most important thing to remember in Time, Speed and Distance (TSD) problems is that Speed = Distance/Time (ahhh! Who does not know this?). Now, the shortcut or the understanding we can develop from the above basic concept is:
- T1*S1 = T2*S2
- D1/S1 = D2/S2
- D1/T1 = D2/T2 (Generally not used)
That’s it. You are done and with these concepts, I am sure all of you will be able to solve many questions which you would have left otherwise.
Should have read this article before…
Generally, you will encounter questions in TSD that ask you to calculate either of the three in two case types i.e.
- Two cars are travelling in any direction but in the end meeting at some place [Important thing to note is time is constant for both the vehicle i.e. they take same time to get to a location]. Assumption in this case is that both the cars started at the same time. In case, they start at different times, we still can solve it by considering the time from which the later vehicle started. Concept to be used is D1/S1 = D2/S2.
- Two cars are travelling from once location and stopping at another location or coming back to the same location [Important thing to note is that the distance travelled by both the cars is same]. Concept to be used is T1*S1 = T2*S2.
Mr. X uses his car to go to office which is 50 km from his place. It takes him 1 hour to reach office if he travels at usual speed of 50Km per hour. Suppose, Mr. X on a particular day managed to reach office in 30 min. Hence, we can very well say that he drove fast that day. Now the question is how fast?
Solution: We can very well say that this is case 2 problem. This is because, distance is constant in this case.
Applying the concept T1*S1 = T2*S2.
ð 1 hr * 50 = 1/2hr * S2
ð 50 = 1/2 * S2
ð S2 = 100 Km per hour
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