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Course by: Sunil Kumar Singh. E-mail the author

# Relative velocity in one dimension(application)

Module by: Sunil Kumar Singh. E-mail the author

Summary: Solving problems is an essential part of the understanding

Questions and their answers are presented here in the module text format as if it were an extension of the treatment of the topic. The idea is to provide a verbose explanation, detailing the application of theory. Solution presented is, therefore, treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.

## Hints on solving problems

1. Foremost thing in solving problems of relative motion is about visualizing measurement. If we say a body "A" has relative velocity "v" with respect to another moving body "B", then we simply mean that we are making measurement from the moving frame (reference) of "B".
2. The concept of relative velocity applies to two objects. It is always intuitive to designate one of the objects as moving and other as reference object.
3. It is helpful in solving problem to make reference object stationary by applying negative of its velocity to both objects. The resultant velocity of the moving object is equal to the relative velocity of the moving object with respect to reference object. If we interpret relative velocity in this manner, it gives easy visualization as we are accustomed to observing motion from stationary state.

## Representative problems and their solutions

We discuss problems, which highlight certain aspects of the study leading to the relative velocity in one dimension.

### Example 1

Problem : A jet cruising at a speed of 1000 km/hr ejects hot air in the opposite direction. If the speed of hot air with respect to Jet is 800 km/hr, then find its speed with respect to ground.

Solution : Let the direction of Jet be x – direction. Also, let us denote jet with “A” and hot air with “B”. Here,

v A = 1000 km / hr v B = ? v BA = -800 km / hr v A = 1000 km / hr v B = ? v BA = -800 km / hr

Now,

v B A = v B - v A v B = v A + v A B = 1000 + ( - 800 ) = 200 km / hr v B A = v B - v A v B = v A + v A B = 1000 + ( - 800 ) = 200 km / hr

The speed of the hot air with respect to ground is 200 km/hr.

### Example 2

Problem : If two bodies, at constant speeds, move towards each other, then the linear distance between them decreases at 6 km/hr. If the bodies move in the same direction with same speeds, then the linear distance between them increases at 2 km/hr. Find the speeds of two bodies (in km/hr).

Solution : Let the speeds of the bodies are “u” and “v” respectively. When they move towards each other, the relative velocity between them is :

v A B = v A - v B = u - ( - v ) = u + v = 6 km / hr v A B = v A - v B = u - ( - v ) = u + v = 6 km / hr

When they are moving in the same direction, the relative velocity between them is :

v A B = v A - v B = u - v = 2 km / hr v A B = v A - v B = u - v = 2 km / hr

Solving two linear equations, we have :

u = 4 km/hr

v = 2 km/hr

### Example 3

Problem : Two cars A and B move along parallel paths from a common point in a given direction. If “u” and “v” be their speeds (u > v), then find the separation between them after time “t”.

Solution : The relative velocity of the cars is :

v 1 2 = v 1 - v 2 = u - v v 1 2 = v 1 - v 2 = u - v

The separation between the cars is :

x = v 1 2 t = ( u - v ) t x = v 1 2 t = ( u - v ) t

### Example 4

Problem : Two trains of length 100 m each, running on parallel track, take 20 seconds to overtake and 10 seconds to cross each other. Find their speeds (in m/s).

Solution : The distance traveled in the two cases is 100 + 100 = 200m.

Let their speeds be “u” and “v”. Now, relative velocity for the overtake is :

v 1 2 = v 1 - v 2 = u - v v 1 2 = v 1 - v 2 = u - v

200 = ( u - v ) x 20 u - v = 10 200 = ( u - v ) x 20 u - v = 10

The relative velocity to cross each other is :

v 1 2 = v 1 - v 2 = u - ( - v ) = u + v v 1 2 = v 1 - v 2 = u - ( - v ) = u + v

200 = ( u + v ) x 10 u + v = 20 200 = ( u + v ) x 10 u + v = 20

Solving two linear equations, we have :

u = 15 m/s and v = 5 m/s.

### Example 5

Problem : Two cars are moving in the same direction at the same speed,”u”. The cars maintain a linear distance "x" between them. An another car (third car) coming from opposite direction meets the two cars at an interval of “t”, then find the speed of the third car.

Solution : Let “u” be the speed of either of the two cars and “v” be the speed of the third car. The relative velocity of third car with respect to either of the two cars is :

v rel = u + v v rel = u + v

The distance between the two cars remains same as they are moving at equal speeds in the same direction. The events of meeting two cars separated by a distance “x” is described in the context of relative velocity. We say equivalently that the third car is moving with relative velocity "u+v", whereas the first two cars are stationary. Thus, distance covered by the third car with relative velocity is given by :

x = ( u + v ) t x = ( u + v ) t

v = ( x - u t ) t = x t - u v = ( x - u t ) t = x t - u

### Example 6

Problem : Two cars, initially at a separation of 12 m, start simultaneously. First car “A”, starting from rest, moves with an acceleration 2 m / s 2 m / s 2 , whereas the car “B”, which is ahead, moves with a constant velocity 1 m/s along the same direction. Find the time when car “A” overtakes car “B”.

Solution : We shall attempt this question, using concept of relative velocity. First we need to understand what does the event “overtaking” mean? Simply said, it is the event when both cars are at the same position at a particular time instant. Subsequently, the car coming from behind moves ahead.

Now relative velocity has an appropriate interpretation suiting to this situation. The relative velocity of “A” with respect to “B” means that the car “B” is stationary, while car “A” moves with relative velocity. Thus, problem reduces to a simple motion of a single body moving with a certain velocity (relative velocity) and covering a certain linear distance (here sepeartion of 12 m) with certain acceleration. Note that car “B” has no acceleration.

Relative velocity of “A” with respect to “B”, i.e. v AB v AB , at the start of motion is :

v AB = v A - v B = 0 - 1 = 1 m / s v AB = v A - v B = 0 - 1 = 1 m / s

Relative acceleration of “A” with respect to “B”, i.e. aAB is :

a AB = a A - a B = 2 - 0 = 2 m / s 2 a AB = a A - a B = 2 - 0 = 2 m / s 2

Applying equation of motion for a single body (remember the second body is rendered stationary),

x = u t + 1 2 a t 2 12 = - 1 x t + 1 2 x 2 x t 2 t 2 - t - 12 = 0 t 2 + 3 t - 4 t - 12 = 0 ( t + 3 ) ( t - 4 ) = 0 x = u t + 1 2 a t 2 12 = - 1 x t + 1 2 x 2 x t 2 t 2 - t - 12 = 0 t 2 + 3 t - 4 t - 12 = 0 ( t + 3 ) ( t - 4 ) = 0

Neglecting negative value,

t = 4 s t = 4 s

Note : We can solve this problem without using the concept of relative velocity as well. The car “A” moves 12 m more than the distance covered by “B” at the time of overtake. If car “B” moves by “x” meters, then car “A” travels “x+12” meters. It means that :

Displacement of “A” = Displacement of “B” + 12 Displacement of “A” = Displacement of “B” + 12

1 2 a t 2 = v t + 12 1 2 a t 2 = v t + 12

Putting values,

1 2 x 2 x t 2 = 1 x t + 12 t 2 - t - 12 = 0 1 2 x 2 x t 2 = 1 x t + 12 t 2 - t - 12 = 0

This is the same equation as obtained earlier. Hence,

t = 4 s t = 4 s

Evidently, it appears to be much easier when we do not use the concept of relative velocity. This is the case with most situations in one dimensional motion of two bodies. It is easier not to use concept of relative velocity in one dimensional motion to the extent possible. We shall learn subsequently, however, that the concept of relative velocity has an edge in analyzing two dimensional motion, involving two bodies.

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