OpenStax-CNX module: m14034
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∗
Resultant motion (application)
Sunil Kumar Singh
This work is produced by OpenStax-CNX and licensed under the
†
Creative Commons Attribution License 2.0
Abstract
Solving problems is an essential part of the understanding process.
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.
1 Representative problems and their solutions
We discuss problems, which highlight certain aspects of the study leading to the resultant velocity.
questions are categorized in terms of the characterizing features of the subject matter :
•
•
•
•
Velocity of the object
Time to cross the stream
Multiple references
Minimum time, distance and speed
2 Velocity of the ob ject
Example 1
Problem :
A person can swim at 1 m/s in still water. He swims to cross a river of width 200 m
to a point exactly opposite to his/her initial position. If the water stream in river ows at 2 m/s
in a linear direction, then nd the time taken (in seconds) to reach the opposite point.
Solution :
Let the direction of stream be x-direction and the direction across stream be
y-direction. Let us also denote person with A and water stream with B.
To reach the point across, the person has to swim upstream at an angle such that the velocity
of the person with respect to ground(
shown in the gure.
∗ Version
1.6: Oct 10, 2008 11:09 pm -0500
† http://creativecommons.org/licenses/by/2.0/
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vA
) is across the direction of water stream. The situation is
The
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Relative velocity
Figure 1
Here,
Speed of the person (A) with respect to stream (B) : vAB
Speed of water stream (B) with respect to ground : vB
Speed of the person (A) with respect to ground :vA
= 1 m/s
= 2 m/s
=?
d = 200 m
From the
∆OAB,
OQ2 = OP2 + PQ2
√
⇒ OP =
{OQ2 − PQ2 }
⇒ t =
√
d
{OB2 − AB2 }
It is clear from the denominator of the expression that for nite time, OB
values as given in the question, OB
<
>
AB. From the
AB and the denominator becomes square root of negative
number. The result is interpreted to mean that the physical event associated with the expression
is not possible.The swimmer, therefore, can not reach the point, which is exactly opposite to his
position. The speed of the swimmer should be greater than that of the stream to reach the point
lying exactly opposite.
Note that we had explained the same situation in the module on the subject with the help of
the value of "sinθ ", which can not be greater than 1. We have taken a dierent approach here to
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illustrate the same limitation of the swimmer's ability to cross the water stream, showing that the
interpretation is consistent and correct.
Example 2
Problem :
The direction of water stream in a river is along x direction of the coordinate
system attached to the ground. A swimmer swims across the river with a velocity (
0.8i + 1.4j
) m/s, as seen from the ground. If the river is 70 m wide, how long (in seconds) does he take to
reach the river bank on the other side ?
Solution :
We recognize here that the given velocity represents the resultant velocity (
vA
)
of the swimmer (A). The time to reach the river bank on the other side is a function of component
velocity in y-direction.
Relative velocity
Figure 2
Here,
vx = 0.8 m/s
vy = 1.4 m/s
t =
t =
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Width of the river
vy
70
=
50 s
1.4
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Example 3
Problem :
4
A person can swim at a speed u in still water. He points across the direction of
water stream to cross a river. The water stream ows with a speed v in a linear direction. Find
the direction in which he actually swims with respect to the direction of stream.
Solution :
Let the direction of stream be x-direction and the direction across stream be
y-direction. Let us also denote person with "A" and water stream with B.
Here,
Speed of the person (A) with respect to stream (B) : vAB
Speed of stream (B) with respect to ground : vB
= v
Speed of the person (A) with respect to ground : vC
Using equation,
vAB = vA − vB
= u
=?
,
⇒ vA = vB + vAB
From the gure,
Relative velocity
Figure 3
tanθ' =
vAB
vB
=
u
v
The direction in which he actually swims with respect to the direction of stream is
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θ‘ = tan − 1
u
v
3 Time to cross the river
Example 4
Problem :
A person can swim at a speed 1 m/s in still water. He swims perpendicular to the
direction of water stream, owing at the speed 2 m/s.
If the linear distance covered during the
motion is 300 m, then nd the time taken to cross the river.
Solution :
Let the direction of stream be x-direction and the direction across stream be
y-direction. Let us also denote ground person with "A" and water stream with B. This is clearly
the situation corresponding to the least time for crossing the river.
Here,
Speed of the person (A) with respect to stream (B) : vAB
Speed of water stream (B) with respect to ground : vB
Speed of the person (A) with respect to ground : vA
= u = 1m / s
= v = 2m / s
=?
We note here that the perpendicular linear distance i.e. the width of river is not given. Instead,
the linear distance covered during the motion is given. Hence, we need to nd the resultant speed
in the direction of motion to nd time. Using equation for the resultant velocity,
⇒ vA = vB + vAB
From the gure, we have :
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Relative velocity
Figure 4
vA =
√
{vAB 2 + vB 2 } =
√
√
{12 + 22 } =
√
√
⇒ t = 500
5s
5 = 100
vA =
Example 5
Problem :
A person can swim at a speed of
√
√
{u2 + v 2 }
5m / s
3 m/s in still water. He swims at an angle of 120
◦
from the stream direction while crossing a river. The water stream ows with a speed of 1 m/s. If
the river width is 300 m, how long (in seconds) does he take to reach the river bank on the other
side ?
Solution :
Let the direction of stream be x-direction and the direction across stream be
y-direction. Here, we need to know the component of the resultant velocity in the direction perpendicular to the stream.
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Relative velocity
Figure 5
This approach, however, would be tedious. We shall use the fact that the component of "vA "
in any one of the two mutually perpendicular directions is equal to the sum of the components of
vAB
and
vB
in that direction.
⇒ vAy = vAB cos300
√
⇒ vAy =
3cos300 =
3
2
m/s
Thus, time taken to cross the river is :
t =
t =
Width of the river
vAy
300 x 2
= 200 s
3
4 Multiple references
Example 6
Problem :
A boat, capable of sailing at 2 m/s, moves upstream in a river. The water stream
ows at 1 m/s. A person walks from the front end to the rear end of the boat at a speed of 1 m/s
along the linear direction. What is the speed of the person (m/s) with respect to ground ?
Solution :
Let the direction of stream be x-direction and the direction across stream be
y-direction. We further denote boat with A, stream with B, and person with C.
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We shall work out this problem in two parts. In the rst part, we shall nd out the velocity of
boat (A) with respect to ground and then we shall nd out the velocity of person (C) with respect
to ground.
Here,
Velocity of boat (A) with respect to stream (B) : vAB
Velocity of the stream (A) with respect to ground : vA
= − 2m / s
= 1m / s
Velocity of the person (C) with respect to boat (A) : vCA
Velocity of the person (C) with respect to ground : vC
= 1m / s
= ?
Relative velocity
Figure 6
The velocity of boat with respect to ground is equal to the resultant velocity of the boat as
given by :
vA = vAB + vB
⇒ vA = − 2 + 1 = − 1 m / s
For the motion of person and boat, the velocity of the person with respect to ground is equal to
the resultant velocity of (i) velocity of the person (C) with respect to boat (A) and (ii) velocity of
the boat (A) with respect to ground. We note here that relative velocity of person with respect to
boat is given and that we have already determined the velocity of boat (A) with respect to ground
in the earlier step. Hence,
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Relative velocity
Figure 7
vC = vCA + vA
⇒ vC = 1 + ( − 1 ) = 0
5 Minimum time, distance and speed
Example 7
Problem :
A boy swims to reach a point Q on the opposite bank, such that line joining initial
and nal position makes an angle of 45 with the direction perpendicular to the stream of water. If
the velocity of water stream is u, then nd the minimum speed with which the boy should swim
to reach his target.
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Crossing a river
Figure 8:
The boy swims to reach point Q.
Solution :
Let A and B denote the boy and the stream respectively. Here, we are required
to know the minimum speed of boy,
vAB
(say v) such that he reaches point Q. Now, he can
adjust his speed with the direction he swims. Let the boy swims at an angle θ with a speed v.
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Crossing a river
Figure 9:
The boy swims to reach point Q.
Looking at the gure, it can be seen that we can make use of the given angle by taking trigonometric ratio such as tangent, which will involve speed of boy in still water (v) and the speed of
water stream (u). This expression may then be used to get an expression for the minimum speed
as required.
The slope of resultant velocity,
vA
, is :
tan450 =
vAx
=1
vAy
⇒ vAx = vAx
Now, the components of velocity in x and y directions are :
vAx = u − vsinθ
vAy = vcosθ
Putting in the equation we have :
u − vsinθ = vcosθ
Solving for v, we have :
⇒v=
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u
(sinθ + cosθ)
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The velocity is minimum for a maximum value of denominator. The denominator is maximum
for a particular value of the angle,
θ;
for which :
θ
(sinθ + cosθ) = 0
⇒ cosθ − sinθ = 0
⇒ tanθ = 1
⇒ θ = 450
It means that the boy swims with minimum speed if he swims in the direction making an angle
of 45 with y-direction. His speed with this angle is :
u
=
v=
sin450 + cos450
Example 8
Problem :
√
2u
u
=√
2
2
A boat crosses a river in minimum time, taking 10 minutes during which time the
it drifts by 120 m in the direction of stream. On the other hand, boat takes 12.5 minutes while
moving across the river. Find (i) width of the river (ii) velocity of boat in still water and (iii) speed
of the stream.
Solution :
There are three pieces of information about "minimum time", "drift" and "time
along shortest path".
Individually each of these values translate into three separate equations,
which can be solved to nd the required values.
The boat takes minimum time, when it sails in the direction perpendicular to the stream (current). The time to cross the river is given by dividing width with component of resultant velocity
(
vAy
).
The boat, in this case, sails in the perpendicular direction.
resultant velocity is equal to the velocity of boat in still water (
in this case is :
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vAB
Hence, the component of
). The time to cross the river
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Crossing a river
Figure 10:
The swims towards P.
tmin =
d
vAB
=
d
vAB
= 10
d = 10vAB
The drift in this time is given by :
x = vB tmin
Putting values,
120 = vB
vB = 12
x 10
meter/minute
Now we need to use the information on shortest path. It is given that the boat moves across
stream in 12.5 minutes. For this boat has to sail upstream at certain angle. The resultant speed is
given by :
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Crossing a river
Figure 11:
The swims towards P.
vA =
q
2
2
vAB
− vB
and the time taken is :
d
p
Substituting for d and vB
2
vAB
2
− vB
= 12.5
and squaring on both sides, we have :
2
2
vAB
− 122 12.52 = d2 = 102 vAB
2
vAB
12.52 − 102 = 122 x12.52
vAB =
12 x 12.5
= 20meter/minute
7.5
and
d = 10vAB = 10 x 20 = 200 m
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