a river problem a crocodile is lurking beside the platte river. it spots an unsuspecting juggler on...

A River Problem

A crocodile is lurking beside the Platte River. It spots an unsuspecting juggler

on the bank of the river exactly opposite.

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.

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.

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

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

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

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

. .. .

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

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.

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

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.

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.

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.

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.

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:

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.

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:

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)

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


Solution:

Resultant velocity

Just write the magnitude of this vector as it’s

quite tricky to spot its direction.

Air speed = 350

100


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

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


Solution:

We can now draw the triangle.

Air speed = 350v

Resultant velocity

Wind

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

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.

100

350

335


100350

335


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


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

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,

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.

1.2

ms

-1

0.6 ms-1

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.

1

1.2

ms

-1

0.6 ms-1

0.6

1.2Sin

30

Mary should aim to swim 30° to the right of Q.

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

HOMEWORK

• Page 322 (3 – 5)

a river problem a crocodile is lurking beside the platte river. it spots an unsuspecting juggler on...

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