Placemant Preparation: Tips and Shortcuts on Relative Speed, Linear Races, Circular Races, Meeting Points

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 k m p h

and Car B covered =

60 k m p h

Distance between the cars after 1 hr =

40 + 60 = 100 k m

In one hour both the cars together covered a distance of

100 k m

Speed of the entire activity

= \frac{distance}{time} = \frac{100 km}{1 hr} = 100 k m p h

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.

If the speed of first moving object is

S_{1}

and the speed of the second moving object is

S_{2}

and they are travelling in the opposite directions, start simultaneously, then the

relative speed = S_{1} + S_{2}

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.

If the speed of first moving body is

S_{1}

and the speed of second moving body is

S_{2}

, they start simultaneously and moving in the same direction then their relative speed is

| S_{1} - S_{2} |

Examples on relative speed

A Cop and a thief start run simultaneously with the constant speed of

7 \frac{m}{s}

and

5 \frac{m}{s}

respectively. Initially they were 100 meters apart. After how many seconds the cop will catch the thief?

Speed of cop

= 7 m/s

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

= \frac{100}{2} = 50 s

Two trains of length 200 meters each take 10 seconds to cross each other when they are travelling towards each others. If the faster train will take 40 seconds to overtake the slower train. Find the ratio of the speed of the faster train to that of the slower train?

Let the speed of faster train

= S 1

And the speed of the slower train

= S 2

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 = 400 m

Time taken for crossing (opposite direction)

= \frac{400}{S_{1} + S_{2}} = 10 s \to S_{1} + S_{2} = 40 s

Time taken for crossing (same direction)

= \frac{400}{S_{1} - S_{2}} = 40 s \to S_{1} - S_{2} = 10 s

So,

S_{1} = 25 m/s

and

S_{2} = 15 m/s

\to S_{1} : S_{2} = 25 : 15 = 5 : 3

Trick

In the above situation, ratio of speeds =

\frac{t_{1} + t_{2}}{t_{1} - t_{2}} = \frac{40 + 10}{40 - 10} = \frac{50}{30} = 5 : 3

A train approaches a tunnel AB. Inside the tunnel a cat is located at a point that is 3/8 of the distance of A , from the entrance A. When the train whistles the cat starts to run. If the cat moves to the entrance A of the tunnel, the train catches the cat exactly at the entrance. If the cat moves to the exit B, then the train catches the cat exactly at the exit .Find the ratio of the speed of train to cat? (This is a CAT 2002 question.)

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

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 $\to$

	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	$\frac{L}{a - b}$

LCM of

\frac{L}{a}

and

\frac{L}{b}


A and B are in opposite direction	$\frac{L}{a + b}$

LCM of

\frac{L}{a}

and

\frac{L}{b}

Examples

In an 800 meters race P got a head start of 100 meters. If Q beats P by 100 meters find the ratio of the speeds of P and Q?

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

3 : 4

Ram and Rahul start running from two points A and B respectively and run towards each other. At the first time they meet at 100 meters away from A on the track. After completion of one full track AB they changed their directions and Ram run towards A and Rahul run towards B. In the second time they met at 30 meters away from B on the track. Find the distance between A and B?

This problem can be solved in two methods.
Method I:

		Distance covered by
	Ram	Rahul
$1^{st}$

meeting

100m

(d - 100) m

2^{st}

meeting<

(d + 30)

'	$(2 d - 100) m$

\frac{Distance covered by Ram}{Distance covered by Rahul} = \frac{100}{d - 100} = \frac{d + 30}{2 d - 30}

100 (2 d - 30) = (d + 30) (d - 100)

200 d - 3000 = d^{2} - 70 d - 3000

d^{2} - 70 d - 200 d = 0

d^{2} - 270 d = 0 \to 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

= 3 d 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

3 \times 100 = 300

meters

\to 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

In a circular race, A and B start running from a particular point, in opposite direction. If the speed of A is 83.33% of the speed of B, find the number of distinct meeting points? (CAT 2003 Model Question)

83.33 % = \frac{5}{6}

S p e e d o f A = \frac{5}{6} B \to \frac{A}{B} = \frac{5}{6}

\to

Ist meeting at 5th point from starting point

2^{n d} \to 10^{t h}

point

3^{r d} \to 4^{t h}

point.. and so on
ie; When A covered 5 units, B will cover 6 units. Hence they will meet a total of 11 points.

Generalization

A and B running in circular track and in opposite direction. If the speed of A is

\frac{x}{y}

of B , then the total number of meeting points

= x + y

Leena and Seena are running in opposite direction around a circular track of length

20 π

meters. Speed of Leena is 33.33% of the speed of Seena. Find straight line distance between their first meeting point and the second meeting point. It is given that they start simultaneously from a common starting point.

Speed of Leena

= \frac{1}{3}

speed of Seena.

∴

There are (1+3) = 4 distinct meeting points.

Given that the circumference of the track =

20 π

meters,

∴

Radius = 10 meters.

Hence the straight line distance between the first and second meeting point is

10 \sqrt{2}

Athul and Vijay start cycling simultaneously from town A to town B at their respective speeds of 30 kmph and 20 kmph. When Athul reaches town B, then he immediately return for town A. If he will meet Vijay at 3 km away from B. Find the distance between town A and town B?

Let the distance between A and B

= d k m

Distance covered by Athul

= d + 3

Distance covered by Vijay

= d - 3

Time taken for meeting

= \frac{d + 3}{30} = \frac{d - 3}{20}

20 d + 60 = 30 d - 90

10 d = 150

\to d = 15 k m

Shortcut

As per the above condition we can furnish the following result.
Total distance (AB)

= Distance between B and meeting point \times \frac{sum of speeds}{difference between speeds}

\to A B = 3 \times \frac{30 + 20}{30 - 20} = 15 k m

In the above question if the distance between A and B is 25 km then how far from B Athul will meet Vijay?

Distance covered by Athul

= (25 + x) k m

Distance covered by Vijay

= (25 - x) k m

Time taken for meeting

= \frac{25 + x}{30} = \frac{25 - x}{20}

500 + 20 x = 750 - 30 x

50 x = 250

\to x = 5 k m

Shortcut

As per the above condition, distance between meeting point and B

= A B \times \frac{difference between speeds}{sum of speeds}

\to x = 25 \times \frac{10}{50} = 5 k m

Syam and Ram start running simultaneously from A and B. Syam running from A to B and Ram running from B to A .They meet each other after a certain time 't'. There after Syam takes 36 seconds and Ram takes 9 seconds to reach their respective destinations. Find the value of t (in seconds)?

Ratios of the time required for Syam and Ram to cover a constant distance is;

\frac{t}{9} = \frac{36}{t}

\to t = 18 \sec

Shortcut

S is travelling from A to B at 's' kmph and R is travelling from B to A at 'r' kmph. They met at a point 't' seconds after their simultaneous start. S and R required 'a' seconds and 'b' seconds respectively for covering the further distances. Then;

$\frac{Speed of S}{Speed of R} = \frac{\sqrt{b}}{\sqrt{a}}$

$T = \sqrt{a b}$

Placemant Preparation

Pages

Tips and Shortcuts on Relative Speed, Linear Races, Circular Races, Meeting Points

Objectives

Relative speed

Bodies moving in opposite direction

Bodies moving in the same direction.

Examples on relative speed

Races and meeting points

Basic terminologies on Races

Races on circular track

Generalization of concept

Examples

Generalization of the concept

Concept review questions

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