a river problem a crocodile is lurking beside the platte river. it spots an unsuspecting juggler on...
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A River Problem
A crocodile is lurking beside the Platte River. It spots an unsuspecting juggler
on the bank of the river exactly opposite.
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The river flows parallel to its banks at a a velocity of 2.25 m s-1.
The maximum velocity of the crocodile in still water is 3.5 m s-1.
and the river is 750 m wide.
Assume that the crocodile travels at a constant velocity.
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1. Draw a sketch to show how the crocodile must swim in order to reach its prey on the opposite bank.
2. Find the resultant velocity of the crocodile
3. Calculate the time taken between the crocodile sliding into the water and the juggler being eaten.
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1. Draw a sketch to show how the crocodile must swim in order to reach its prey on the
opposite bank.
The river
Current flows this way
Crocodile starts here
If the croc swims straight across
It will be washed downstream by the
currentIt will travel straight
across the river.If the croc swims at the correct angle up
stream
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1. Draw a sketch to show how the crocodile must swim in order to reach its prey on the
opposite bank.
Crocodile starts here
Velocity of croc in
water
Velocity of river
Resultant velocity of croc
Croc ends up here
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2. Find the resultant velocity vector of the croc.
3.5 m.s -1
2.25 m.s-1
3.52 - 2.252
= 2.69 m s-1
=
Pythagorus:Resultant velocity
RESULTANT VELOCITY = 2.69 m s-1
Perpendicular to the bank of the river
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3. Calculate the time taken between the crocodile sliding into the water and the juggler being eaten.
Since croc. moves at a constant velocity
v = 2.69 m s-1
t = d/v
∴ t = 750/2.69
= 278.8 s
. .. .
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current
Suppose some kid sets out to swim across the Platte. The kid sees a croc and swims to the nearest sand bank.
She starts at A and heads directly across in a direction perpendicular
to the bank.Why might she reach the opposite bank but miss the sandbank.
Ans: The current could push her downstream
(to the left in the diagram).
path
Ax
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A
currentIn which direction
should she try to swim in order to reach the
sandbank ?
Ans: She should head partly upstream (towards the right ).
path
We say the velocity of the child in still water is the velocity of the child relative to the water.
x
To find the actual, or resultant, velocity of the child we must add the velocity of the current to the velocity that the child herself would have in
still water.
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Ax
current
headingpath
Ax
currentpathheading
Since velocities are vectors, we can add the velocities by drawing the vectors head-to-
tail.
path
path
1. Child heads directly across, perpendicular to the bank.
2. Child heads upstream
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Ax
currentpathheading
Velocity Triangle
Suppose the child can swim at 1.5 ms-1 in still water and the current flows at 1 ms-1.
1. Child heads straight across river. ( So, we
know the direction and magnitude of the relative velocity. )
1.5
1
v
(relative
to the water)
We know the magnitude and direction of the
current.
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We use Pythagoras’ Theorem to find v:
Ax
currentpathheading
Velocity
Triangle
1·5
1
v
v 2 12 1.52
v 1.80 ( 3 s.f. )
We have a right angled triangle and we can use any trig ratio to find Tip: Avoid using the side you calculated in case you made a slip.
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Ax
currentpathheading
Velocity
Triangle
1.5
1
1.80
tan
11.5
33.7 ( 3 s.f. )
The resultant (actual) velocity of the child is 1.80 ms-1 making an angle of 56.3 to the bank.
So, the angle made with the downstream
bank . . . 90 33.7 56.356·3
It’s convenient to give the angle between the velocity and the bank
of the river.
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1
1.5v
2. The child heads upstream at an unknown angle.
A
current
1.5resultant
x
1
Velocity
Triangle
Find v and this time.
This line must end so that the 3rd side of the triangle is in the right
direction for the resultant.
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A
current
1.5resultant
x
1
48·2
Pythagoras’ Theorem:v
2 1.52 12
v 1.12 ( 3 s.f. )
sin
11.5
41.8 ( 3 s.f. )
The resultant speed of the child is 1.12 ms-1
The angle made with the upstream bank . . .
90 41.8 48.2
Velocity
Triangle
1
1.5v
Solution:
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Velocities are also added when we have an airplane travelling in a wind.
The air speed of a plane is the speed relative to the air. This is the speed the plane would
have in still air.The ground speed of a plane is the speed at
which it covers the ground, so it is the actual or resultant speed.
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e.g. A plane flying due North needs to fly in a straight line due east. It’s air speed is 350
km h-1. The wind is blowing at 100 km h-1 from the south. What is the resultant velocity of the plane and what bearing must the pilot
choose ?We start by drawing and labelling the vectors separately.
Solution:
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e.g. A plane needs to fly in a straight line due east. It’s air speed is 350 km h-1. The wind is blowing at 100 km h-1 from the south. What is the resultant velocity of the plane and what bearing must the pilot choose ? The
plane is flying due North
v
We start by drawing and labelling the vectors separately.
Solution:
Resultant velocity (This is what we want)
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e.g. A plane needs to fly in a straight line due east. It’s air speed is 350 km h-1. The wind
is blowing at 100 km h-1 from the south. What is the resultant velocity of the plane and what bearing must the pilot choose ?
v
We start by drawing and labelling the vectors separately.
Solution:
Resultant velocity
Just write the magnitude of this vector as it’s
quite tricky to spot its direction.
Air speed = 350
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100
We start by drawing and labelling the vectors separately.
Solution:
Wind
Air speed = 350v
Resultant velocity
e.g.A plane needs to fly in a straight line due east. It’s air speed is 350 km h-1. The wind is blowing at 100 km h-1 from the south. What is the resultant velocity of the plane and what bearing must the pilot choose ? Plane flying due North
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e.g. A plane needs to fly in a straight line due east. It’s air speed is 350 km h-1. The wind
is blowing at 100 km h-1 from the south. The plane is fling due North. What is the
resultant velocity of the plane and what bearing must the pilot choose ?
100
We start by drawing and labelling the vectors separately.
Solution:
We can now draw the triangle.
Air speed = 350v
Resultant velocity
Wind
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v
100
v
100
v
100
(i) (ii)
(iii)
350 350
350
100Air speed = 350
v
Resultant velocity
Wind
Which of the following diagrams is correct ?
Ans: (iii) is correct.
Not head-to-tail. Not the sum of
the other vectors.
The two vectors which are equivalent to the resultant must be head-to-tail.
Wrong
direction
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v
100350
What is the resultant velocity of the plane and what bearing must the pilot choose ?
v 2 3502 1002
v 335 ( 3 s.f. )
The resultant velocity is 335 km h-1 due east.
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100
350
335
What is the resultant velocity of the plane and what bearing must the pilot choose ?
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100350
335
What is the resultant velocity of the plane and what bearing must the pilot choose ?
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100350
335
We can use to find the bearing.cos
100350
73.4
The pilot needs to set a bearing of 107 ( nearest degree ).
180 73.4 06.6
What is the resultant velocity of the plane and what bearing must the pilot choose ?
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Instead of giving the vectors as magnitudes and directions, the velocities can be given
using i and j
e.g. A boat is rowed across a river. In still water the velocity of the boat is m s
-1. 0.5 0.3 )i j
A current is flowing with a velocity of m s -1. 0.2
)i j
What is the resultant speed of the boat ?Solutio
n:The resultant velocity v is given by the sum of the velocities of the boat and current
jv0.5 0.3 ) + ( 0.2 ) ji ii= 1.5 0.5 j
The speed is v 1.58 m s
-1 ( 3 s.f. )
v 1.5
2 0.5
2
v
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SUMMARY
To find the resultant velocity of a boat or swimmer in water or an aeroplane in the
wind we
• the resultant velocity is given by closing a triangle,
• we always use a double headed arrow for the resultant and check that this vector
is the sum of the other two.
The relative velocity of a swimmer, boat or plane is the velocity in still water or still
air.
or by drawing the vectors head-to-tail.
add the relative velocity and the velocity of the wind or water
jeither by adding the i and components,
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If the athlete is assisted by a wind of 1 ms-1, his
speed will be 7 ms-1.
If the athlete runs into a head wind of1 ms-1, his
speed will be 5 ms-1.
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Page 321
1.2
ms
-1
0.6 ms-1
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Page 321
1.2
ms
-1
0.6 ms-1
2 2
2 Swimming Current
Speed
Actual
Speed
2
2 20.6 1.2
Actual
Speed
1.8
1.3
4 m
s-1
0.6
1.2Tan 26.6
11.8 1.34ms
Mary’s actual velocity is about 1.34 ms-1 in the direction of 26.6° to the left of the
intended line.
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Page 321
1.2
ms
-1
0.6 ms-1
0.6
1.2Sin
30
Mary should aim to swim 30° to the right of Q.
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Page 321
1.2
ms
-1
0.6 ms -1
2 2
2Swimming Current
Speed
Actual
Speed
2
2 21.2 0.6
Actual
Speed
1.08
11.08 1.04ms
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HOMEWORK
• Page 322 (3 – 5)