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

# Analysing motion in a medium

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

Summary: Resultant motion follows the principle of independence of motion in mutually perpendicular directions.

The motion of an object body in a medium is composed of two motions : (i) motion of the object and (ii) motion of the medium. The resulting motion as observed by an observer in the reference frame is the resultant of two motions. The basic equation that governs the context of study is the equation of relative motion in two dimensions :

v A B = v A v B v A B = v A v B
(1)

As discussed in the previous module, the motion of the body can alternatively be considered as the resultant of two motions. This concept of resultant motion is essentially an equivalent way of stating the concept of relative motion. Rearranging the equation, we have :

v A = v A B + v B v A = v A B + v B
(2)

Both these equations are a vector equations, which can be broadly dealt in two ways (i) analytically (using graphics) and (ii) algebraically (using unit vectors). The use of a particular method depends on the inputs available and the context of motion.

## Concept of independence of motion

Analysis of the motion of a body in a medium, specifically, makes use of independence of motions in two perpendicular directions. On a rectilinear coordinate system, the same principle can be stated in terms of component velocities. For example, let us consider the motion of a boat in a river, which tries to reach a point on the opposite bank of the river. The boat sails in the direction perpendicular to the direction of stream. Had the water been still, the boat would have reached the point exactly across the river with a velocity ( v y v y ). But, water body is not still. It has a velocity in x-direction. The boat, therefore, drifts in the direction of the motion of the water stream ( v x v x ).

The important aspect of this motion is that the drift (x) depends on the component of velocity in x-direction. This drift (x) is independent of the component of velocity in y-direction. Why? Simply because, it is an experimental fact, which is fundamental to natural phenomena. We shall expand on this aspect while studying projectile motion also, where motions in vertical and horizontal directions are independent of each other.

In the case of boat, the displacements in the mutually perpendicular coordinate directions are :

x = v x t x = v x t
(3)

y = v y t y = v y t
(4)

## Motion of boat in a stream

In this section, we shall consider a general situation of sailing of a boat in a moving stream of water. In order to keep our context simplified, we consider that stream is unidirectional in x-direction and the width of stream, “d”, is constant.

Let the velocities of boat (A) and stream (B) be “ v A v A “ and “ v B v B “ respectively with respect to ground. The velocity of boat (A) with respect to stream (B), therefore, is :

v A B = v A v B v A B = v A v B

v A = v A B + v B v A = v A B + v B

These velocities are drawn as shown in the figure. This is clear from the figure that boat sails in the direction, making an angle “θ” with y-direction, but reaches destination in different direction. The boat obviously is carried along in the stream in x-direction. The displacement in x-direction (x = QR) from the directly opposite position to actual position on the other side of the stream is called the drift of the boat.

### Resultant velocity

The magnitude of resultant velocity is obtained, using parallelogram theorem,

v A = v A B 2 + v B 2 + 2 v A B v B cos α v A = v A B 2 + v B 2 + 2 v A B v B cos α
(5)

where “α” is the angle between v B v B and v AB v AB vectors. The angle “β” formed by the resultant velocity with x-direction is given as :

tan β = v A B sin α v B + v A B cos α tan β = v A B sin α v B + v A B cos α
(6)

### Time to cross the stream

The boat covers a linear distance equal to the width of stream “d” in time “t” in y-direction Applying the concept of independence of motions in perpendicular direction, we can say that boat covers a linear distance “OQ = d” with a speed equal to the component of resultant velocity in y-direction.

Now, resultant velocity is composed of (i) velocity of boat with respect to stream and (ii) velocity of stream. We observe here that velocity of stream is perpendicular to y – direction. As such, it does not have any component in y – direction. We, therefore, conclude that the component of resultant velocity is equal to the component of the velocity of boat with respect to stream in y -direction,. Note the two equal components shown in the figure. They are geometrically equal as they are altitudes of same parallelogram. Hence,

v A y = v A B y = v A B cos θ v A y = v A B y = v A B cos θ

where “θ” is the angle that relative velocity of boat w.r.t stream makes with the vertical.

t = d v A y = d v A B cos θ t = d v A y = d v A B cos θ
(7)

We can use either of the two expressions to calculate time to cross the river, depending on the inputs available.

### Drift of the boat

The displacement of the boat in x-direction is independent of motion in perpendicular direction. Hence, displacement in x-direction is achieved with the component of resultant velocity in x-direction,

x = v A x t = v B v AB x t = v B v A B sin θ t x = v A x t = v B v AB x t = v B v A B sin θ t

Substituting for time “t”, we have :

x = v B v A B sin θ d v A B cos θ x = v B v A B sin θ d v A B cos θ
(8)

## Special cases

### Shortest interval of time to cross the stream

The time to cross the river is given by :

t = d v A y = d v A B cos θ t = d v A y = d v A B cos θ

Clearly, time is minimum for greatest value of denominator. The denominator is maximum for θ = 0°. For this value,

t min = d v A B t min = d v A B
(9)

This means that the boat needs to sail in the perpendicular direction to the stream to reach the opposite side in minimum time. The drift of the boat for this condition is :

x = v B d v A B x = v B d v A B
(10)

#### Example 1

Problem : A boat, which has a speed of 10 m/s in still water, points directly across the river of width 100 m. If the stream flows with the velocity 7.5 m/s in a linear direction, then how far downstream does the boat touch on the opposite bank.

Solution : Let the direction of stream be in x-direction and the direction across stream is y-direction. We further denote boat with “A” and stream with “B”. Now, from the question, we have :

v A B = 10 m / s v B = 7.5 m / s v A B = 10 m / s v B = 7.5 m / s

The motions in two mutually perpendicular directions are independent of each other. In order to determine time (t), we consider motion in y – direction,

t = OP v A B = 100 10 = 10 s t = OP v A B = 100 10 = 10 s

The displacement in x-direction is :

PQ = v B x t PQ = v B x t

Putting this value, we have :

x = PQ = v B x t = 7.5 x 10 = 75 m x = PQ = v B x t = 7.5 x 10 = 75 m

The velocity of the boat w.r.t stream and the stream velocity are perpendicular to each other in this situation of shortest time as shown here in the figure. Magnitude of resultant velocity in this condition, therefore, is given as :

v A = v A B 2 + v B 2 v A = v A B 2 + v B 2
(11)

The angle that the resultant makes with y-direction (perpendicular to stream direction) is :

tan θ = v B v A B tan θ = v B v A B

Time to cross the river, in terms of linear distance covered during the motion, is :

t = O Q v A B 2 + v B 2 t = O Q v A B 2 + v B 2
(12)

### Direction to reach opposite point of the stream

If the boat is required to reach a point directly opposite, then it should sail upstream. In this case, the resultant velocity of the boat should be directed in y –direction. The drift of the boat is zero here. Hence,

x = v B v A B sin θ d v A B cos θ = 0 x = v B v A B sin θ d v A B cos θ = 0

v B v A B sin θ = 0 v B v A B sin θ = 0

sin θ = v B v A B sin θ = v B v A B
(13)

Thus. the boat should sail upstream at an angle given by above expression to reach a point exactly opposite to the point of sailing.

### The velocity to reach opposite point of the stream

The angle at which boat sails to reach the opposite point is :

sin θ = v B v A B sin θ = v B v A B

This expression points to a certain limitation with respect to velocities of boat and stream. If velocity of boat in still water is equal to the velocity of stream, then

sin θ = v B v A B = 1 = sin 90 0 sin θ = v B v A B = 1 = sin 90 0

θ = 90 0 θ = 90 0

It means that boat has to sail in the direction opposite to the stream to reach opposite point. This is an impossibility from the point of physical reality. Hence, we can say that velocity of boat in still water should be greater than the velocity of stream ( v A B > v B v A B > v B ) in order to reach a point opposite to the point of sailing.

In any case, if v A B < v B v A B < v B , then the boat can not reach the opposite point as sine function can not be greater than 1.

### Shortest path

The magnitude of linear distance covered by the boat is given by :

s = d 2 + x 2 s = d 2 + x 2
(14)

It is evident from the equation that linear distance depends on the drift of the boat, “x”. Thus, shortest path corresponds to shortest drift. Now, there are two situations depending on the relative magnitudes of velocities of boat and stream.

#### Note:

We should be aware that though the perpendicular distance to stream (width of the river) is the shortest path, but boat may not be capable to follow this shortest path in the first place.

1: v A B > v B v A B > v B

We have seen that when stream velocity ( v B v B ) is less than the velocity of boat in still water, the boat is capable to reach the opposite point across the stream. For this condition, drift (x) is zero and represents the minimum value. Accordingly, the shortest path is :

s min = d s min = d
(15)

The boat needs to sail upstream at the specified angle. In this case, the resultant velocity is directed across the river in perpendicular direction and its magnitude is given by :

v A = v A B 2 v B 2 v A = v A B 2 v B 2
(16)

The time taken to cross the river is :

t = d v A B 2 v B 2 t = d v A B 2 v B 2
(17)

2: v A B < v B v A B < v B

In this case, the boat is carried away from the opposite point in the direction of stream. Now, the drift “x” is given as :

x = v B v A B sin θ d v A B cos θ x = v B v A B sin θ d v A B cos θ

For minimum value of “x”, first time derivative of “x” is equal to zero,

đ x đ t = 0 đ x đ t = 0
(18)

We need to find minimum drift and corresponding minimum length of path, subject to this condition.

## Motion of an object in a medium

We have discussed the motion in the specific reference of boat in water stream. However, the consideration is general and is applicable to the motion of a body in a medium. For example, the discussion and analysis can be extended to the motion of an aircraft, whose velocity is modified by the motion of the wind.

### Example 2

Problem : An aircraft flies with a wind velocity of 200√2 km/hr blowing from south. If the relative velocity of aircraft with respect to wind is 1000 km/hr, then find the direction in which aircraft should fly such that it reaches a destination in north – east direction.

Solution : The figure here shows the velocities. OP denotes the velocity of the aircraft in the still air or equivalently it represents the relative velocity of aircraft with respect to air in motion; PQ denotes the velocity of the wind and OQ denotes the resultant velocity of the aircraft. It is clear that the aircraft should fly in the direction OP so that it is ultimately led to follow the north-east direction.

We should understand here that one of the velocities is resultant velocity of the remaining two velocities. It follows then that three velocity vectors are represented by the sides of a closed triangle.

We can get the direction of OP, if we can find the angle “θ”. The easiest technique to determine the angle between vectors composing a triangle is to apply sine law,

O P sin 45 0 = P Q sin θ O P sin 45 0 = P Q sin θ

Putting values, we have :

sin θ = P Q sin 45 0 O P = 200 2 1000 x 2 = 1 5 = 0.2 sin θ = P Q sin 45 0 O P = 200 2 1000 x 2 = 1 5 = 0.2

θ = sin - 1 0.2 θ = sin - 1 0.2

Hence the aircraft should steer in the direction, making an angle with east as given by :

θ = 45 0 sin 1 0.2 θ = 45 0 sin 1 0.2

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