This is a continuation of tutorial on solving problems related to time and distance. In this chapter, we'll look at important concepts in time and distance related problems like relative speed, linear races, circular races and meeting points in circular race. This would help you to solve majority of time and distance related problems quickly using simple shortcuts and tricks.
Objectives
- Relative Speed
- Linear Races
- Circular Races
- Meeting Points
Relative speed
Normally the speed of a moving body is calculating as per a stationary object, or the calculation of speed is as per a stationary base. When speed of one moving body in relation with another moving body then the effective speed of both the movements is called the relative speed of these two moving bodies
Bodies moving in opposite direction
Consider two cars travelling from a particular point and first car is travelling towards East at 40 kmph and the second car is travelling towards west at 60 kmph
Initially the distance between the cars at the starting point is zero or both are in the same line.
After 1 hour; Car A covered = `40 kmph` and Car B covered = `60 kmph`
Distance between the cars after 1 hr = `40 + 60 =100 km`
In one hour both the cars together covered a distance of `100 km`
Speed of the entire activity `= "distance"/"time" = "100 km"/"1 hr" = 100 kmph`
This speed is called the relative speed of both the cars.
`:.` Relative speed of two moving bodies moving in the opposite direction is the sum of their individual speeds.
Bodies moving in the same direction.
Speeds of two cars are 40 kmph and 60 kmph. Suppose they start simultaneously and travelling in the same direction; after one hour the faster car will cover 60 km and the slower car will cover 40 km.
At the starting time , distance between the cars is '0' km and at the finishing time distance between them is 20 km.ie, both the cars made a distance of 20 km in between them in one hour
`:.`20 kmph is the relative speed of both the cars, i.e. two moving bodies are travelling in the same direction and start simultaneously then their relative speed is the difference of their individual speed.
Examples on relative speed
Speed of thief `= 5 "m/s"`
They run in the same direction, hence their relative speed `= 7 - 5 = 2 "m/s"`
Distance to be covered by cop to catch thief `= 100 m`
Time taken to catch the thief `= 100 / 2 = 50 s`
And the speed of the slower train `= S2`
Relative speed, when they are in opposite direction `= S_1 + S_2`
Relative speed, when they are in same direction `= S_1 - S_2`
Distance covered in both directions = Sum of the length of train `= 200 + 200 = 400m`
Time taken for crossing (opposite direction) `= 400/(S_1 + S_2) = 10s -> S_1 + S_2 = 40s`
Time taken for crossing (same direction) `= 400/(S_1 - S_2) = 40s -> S_1 - S_2 = 10s`
So, `S_1 = 25 "m/s"` and `S_2 = 15 "m/s"`
`-> S_1 : S_2 = 25:15 = 5:3`
When the cat covered 3 unit distance, the train reach at the entrance.
Consider if the cat moves towards the exit, after covering 3 unit distance train will reach at entrance A. ie; from the diagram when the cat at the point Q, train at the entrance A.
Cat covered QB and the train covered AB in a constant time. ie; within a constant time cat covered 2 unit lengths then train covered 8 unit lengths. Ratio of the distance covered by the train and the cat in a constant time is `8:2` or `4:1`.
When time is constant the ratio of distance covered = the ratio of the respective speeds. `:.` Ratio of speeds of train to cat = `4:1`
Races and meeting points
It is a frequently testing area in all types of B school entrance exams. Especially in CAT at least one question from races is a usual condition. A proper understanding of the basic terminologies and the effective interpretation of the given situation with a visualization skill will help the student to crack the question. Difficulty level of the question varies exam to exam.
Basic terminologies on Races
There are basically two types of races
- Linear race: In this case participants in the race are moving in a row or in a linear track.
- Circular Race: Here the shape of the race track is circular such as a circular stadium or circular running track.
In the most of the races related questions containing some common terminologies, which are given below.
- Head Start: When a runner allows to start the race 'x' meters ahead from the starting point , then we can say that the runner got a head start of (start up) x meters.
If the runner allows starting the race by 't' seconds earlier than the other runners, then the runner got a head start of t seconds. - Beats by 'x' meters or 't' seconds: Runner A beats B by 'x' meters means, when A finishes the race then B is at 'x' meters behind the finishing point. Runner A beats B by 't' seconds means, after A finishes the race B will take 't' seconds more to finish the race.
Races on circular track
When two or more runners participating in a race around a circular track, the following type of questions can expect.
- When the runners meet at first time anywhere on the track?
- When the runners will meet first time at the starting point itself?
Application of 'relative speed' concept is essential in race related questions.
Generalization of concept
Let A and B are two runners participating in a circular race around a circular track of length 'L' meters with respective speeds `'a' " m/s"` and `'b' " m/s"` .Assume A is faster than B ,then `a > b`. Below table will help you to solve the problems instantly.
| Time Taken By `->` | A and B meet first time along anywhere on the track | A and B meet first time at the starting point |
|---|---|---|
| A and B running in the same direction | `L/(a-b)` | LCM of `L/a` and `L/b` |
| A and B are in opposite direction | `L/(a+b)` | LCM of `L/a` and `L/b` |
Examples
In the given situation with in a constant time Q covered 800 meters when P covered only 600 meters.
`"Ratio of speeds" = "Ratio of distance covered"`
`:.` Speeds are in the ratio `600 : 800` or `3:4`
Method I:
| Distance covered by | |||
|---|---|---|---|
| Ram | Rahul | ||
| `1^"st"` meeting | 100m | `(d - 100) m` | |
| `2^"st"` meeting< | `(d + 30) `' | `(2d - 100) m` | |
`100 (2d-30) = (d+30) (d-100)`
`200d - 3000 = d^2- 70d - 3000`
`d^2 - 70d - 200d=0`
`d^2 - 270d = 0 -> d = 0 "or" d = 270`
`d = 270` is a valid result, therefore the distance between A and B is 270 meters.
Method II:
Let the distance between A and B is d meters. Distance covered by Ram and Rahul together in the first meeting `= d " meters"`
Distance covered by Ram and Rahul together in the second meeting `= 3d " meters"`
In an each round of d meters covered by Ram and Rahul together, Ram alone can cover a distance of 100 meters. Therefore when they together covered '3d' distance Ram alone covered `3xx100=300` meters
`-> d+30 = 300 " meters"`
`d = 270 " meters"`
Generalization of the concept
When the speed of B expressed in terms of the speed of A such as twice, thrice etc, then the following results will generate.
| B's speed in terms of the speed of A and they are running on opposite directions. | Number of meeting points |
|---|---|
| Equal to A | 2 |
| Twice of A | 3 |
| Thrice of A | 4 |
| N times of A | N + 1 |
Concept review questions
`83.33% = 5/6`
`Speed of A = 5/6 B -> A/B =5/6`
`-> ` Ist meeting at 5th point from starting point
`2^(nd) -> 10^(th)` point
`3^(rd) -> 4^(th)` point.. and so on
ie; When A covered 5 units, B will cover 6 units. Hence they will meet a total of 11 points.
Given that the circumference of the track = `20 \pi` meters, `:.` Radius = 10 meters.
Hence the straight line distance between the first and second meeting point is `10 sqrt(2)`m
Let the distance between A and B `= d km`
Distance covered by Athul `= d + 3`
Distance covered by Vijay `= d - 3`
Time taken for meeting `= (d + 3)/30 = (d - 3)/20`
`20d + 60 = 30d - 90`
`10d = 150`
`-> d = 15 km`
Total distance (AB) `= "Distance between B and meeting point" xx "sum of speeds"/"difference between speeds"`
`-> AB = 3 xx (30 + 20)/(30 - 20) = 15km`
Distance covered by Athul `= (25 + x) km`
Distance covered by Vijay `= (25 - x) km`
Time taken for meeting `= (25 + x)/30 = (25 - x)/20`
`500 + 20x = 750 - 30x`
`50x = 250`
`-> x = 5km`
`-> x = 25 xx 10/50 = 5 km`
Ratios of the time required for Syam and Ram to cover a constant distance is; `t/9 = 36/t`
`-> t = 18 sec`
- `"Speed of S"/"Speed of R" = sqrt(b)/sqrt(a)`
- `T = sqrt(ab)`
