Teach A Level
Maths
Resultant Velocity
Volume 4: Mechanics 1
Resultant Velocity
AQA
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This presentation uses the sections “1. Introducing
Vectors” and “2. Unit Vectors” from “Vectors for
Mechanics”.
Select from the options below.
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Vectors for Mechanics
Suppose a child sets out to swim to a sand bank in a
river.
She starts at A and heads
path
current
directly across in a direction
x
perpendicular to the bank.
A
Decide with your partner why she might reach the
opposite bank but miss the sandbank.
Ans: The current could push her downstream ( to
the left in the diagram ).
In which direction should
she try to swim in order
to reach the sandbank ?
current
path
x
A
Ans: She should head partly upstream ( towards the
right ).
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.
We say the velocity of the child in still water is the
velocity of the child relative to the water.
Since velocities are vectors, we can add the
velocities by drawing the vectors head-to-tail.
1. Child heads directly across, perpendicular to the
bank.
path
current
heading
x
path
A
2. Child heads upstream
current
path
x heading
A
path
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. )
We know the magnitude and
direction of the current.
path
current
heading
x
A
Velocity Triangle
1
v
1·5
(relative to
q the water)
We use Pythagoras’ Theorem
to find 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 q.
Tip: Avoid using the side
you calculated in case you
made a slip.
path
current
heading
x
A
Velocity
Triangle
1
1·5
v
q

tanq = 1
1·5
q = 33·7 ( 3 s.f. )
It’s convenient to give the
angle between the velocity
and the bank of the river.
So, the angle made with
the downstream bank . . .
= 90 - 33·7 = 56·3
path
current
heading
x
A
Velocity
Triangle
1
1·5
1·80
56·3
q
The resultant (actual) velocity of the child is 1·80 ms-1
making an angle of 56·3 to the bank.
2. The child heads upstream
at an unknown angle.
1
resultant
x
Working with your partner,
find v and q this time.
This line must end so that
the 3rd side of the triangle is
in the right direction for the
resultant.
current
Velocity
Triangle
1·5
A
1
v
1·5
q
Solution:
Pythagoras’ Theorem:

v 2 = 1·52 - 12
v = 1·12 ( 3 s.f. )
The resultant speed of
the child is 1·12 ms-1
sinq = 1

1·5
q = 41·8 ( 3 s.f. )
The angle made with the
upstream bank . . .
= 90 - 41·8 = 48·2
1
current
resultant
x
Velocity
Triangle
1·5
A
1
v
1·5
q
48·2
Velocities are also added when we have an aeroplane
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 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 ?
Solution: We start by drawing and labelling the
vectors separately.
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 ?
Solution: We start by drawing and labelling the
vectors separately.
Resultant velocity
v
Double headed
arrow.
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 ?
Solution: We start by drawing and labelling the
vectors separately.
Resultant velocity
v
Just write the magnitude of
thisspeed
vector= as
Air
350it’s quite
tricky to spot its direction.
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 ?
Solution: We start by drawing and labelling the
vectors separately.
Resultant velocity
v
Wind
Air speed = 350
100
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 ?
Solution: We start by drawing and labelling the
vectors separately.
Resultant velocity
v
Wind
Air speed = 350
We can now draw the triangle.
100
Resultant velocity
v
Wind
Air speed = 350
Which of the following diagrams is correct ?
(i)
(ii)
350
100
v Not head-to-tail.
(iii)
350
100
v
100
350
100
v
Not the sum of the
other vectors.
Ans: (iii) is correct.
The two vectors which are equivalent to the
resultant must be head-to-tail.
What is the resultant velocity of the plane and
what bearing must the pilot choose ?
350
100
v 2 = 3502 - 1002
v = 335 ( 3 s.f. )
v
The resultant velocity is 335 km h-1 due east.
What is the resultant velocity of the plane and
what bearing must the pilot choose ?
q
350
We can use q to find
the bearing.
cosq = 100
100
335
350
q = 73·4
180 - 73·4 = 106·6
The pilot needs to set a bearing of 107
( nearest degree ).
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 (0·5i - 0·3 j )m s -1.
A current is flowing with a velocity of ( i - 0·2 j ) m s -1.
What is the resultant speed of the boat ?
Solution:
The resultant velocity v is given by the sum of the
velocities of the boat and current
v = (0·5 i - 0·3 j ) + ( i - 0·2 j )
v = 1·5 i - 0·5 j
The speed is v = 1·5 2 + 0·5 2
 v = 1·58 m s -1 ( 3 s.f. )


SUMMARY
 The relative velocity of a swimmer, boat or plane
is the velocity in still water or still air.
 To find the resultant velocity of a boat or
swimmer in water or an aeroplane in the wind we
add the relative velocity and the velocity of the
wind or water
either by adding the i and j components,
or by drawing the vectors head-to-tail.
• 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.
EXERCISE
1. A river has a current of 1 m s -1 as shown in the
diagram.
A boat which can move at 3 m s -1
in still water crosses the river.
1 m s -1
The resultant velocity of the boat
is perpendicular to the bank.
Draw a velocity triangle and use
it to help you find the magnitude
of the resultant velocity.
EXERCISE
Solution:
1 m s -1
Either
v
Or
1
3
3
v
1
A boat can move at 3 m s -1 in still water.
This is the magnitude of the relative velocity and we
are not given the direction.
The resultant velocity is perpendicular to the bank.
(This is the line we draw with a double headed arrow.)
EXERCISE
Solution:
1 m s -1
Either
v
Or
1
3
3
v
1
(Pythagoras
Check that
both diagrams show the resultant as
theorem:
2 - 12
the sum of the other
)
v 2 = 3vectors.
 v = 2·83 m s -1 ( 3 s.f. )
EXERCISE
2. An aircraft can fly at a speed of 200 km h -1 in
still air.
Due to a wind blowing from the south, the actual
velocity of the aircraft is 180 km h -1 due west.
Find the direction the pilot is steering and the
magnitude of the velocity of the wind.
EXERCISE
2. An aircraft can fly at a speed of 200 km h -1 in
still air.
Due to a wind blowing from the south, the actual
velocity of the aircraft is 180 km h -1 due west.
Find the direction the pilot is steering and the
magnitude of the velocity of the wind.
Solution:
aircraft
relative speed = 200
( direction unknown )
EXERCISE
2. An aircraft can fly at a speed of 200 km h -1 in
still air.
Due to a wind blowing from the south, the actual
velocity of the aircraft is 180 km h -1 due west.
Find the direction the pilot is steering and the
magnitude of the velocity of the wind.
Solution:
wind
aircraft
relative speed = 200
( direction unknown )
v
( magnitude unknown )
EXERCISE
2. An aircraft can fly at a speed of 200 km h -1 in
still air.
Due to a wind blowing from the south, the actual
velocity of the aircraft is 180 km h -1 due west.
Find the direction the pilot is steering and the
magnitude of the velocity of the wind.
Solution:
wind
aircraft
aircraft
resultant velocity
relative speed = 200
( direction unknown )
v
( magnitude unknown )
180
EXERCISE
Solution:
wind
aircraft
relative speed = 200
( direction unknown )
aircraft
resultant velocity
v
180
( magnitude unknown )
Tip: I usually start with the
whose
Either
Orvector180
magnitude and direction are known.
200
v
180
v
200
EXERCISE
Tip: I usually start with the
whose
Either
Orvector180
magnitude and direction are known.
200
v
v
200
180
Solution:
Pythagoras’ theorem:
v 2 = 200 2 - 180 2
 v = 87·2 ( 3 s.f. )
The wind speed is 87·2 km h -1
EXERCISE
Tip: I usually start with the
whose
Either
Orvector180
magnitude and direction are known.
200
q
a
v
v
200
180
Solution:
We can use either angle to find the bearing the pilot
is steering. For example,
OR:
sina = 180
cosq = 180
200
a = 64
180 + 64 = 244
200
q = 26
270 - 26 = 244
The pilot needs to set a bearing of 244
( nearest degree ).
The following page contains the summary in a form
suitable for photocopying.
TEACH A LEVEL MATHS – MECHANICS 1
RESULTANT VELOCITY
Summary

The relative velocity of a swimmer, boat or plane is the velocity in still
water or still air.

To find the resultant velocity of a boat or swimmer in water or an aeroplane
in the wind we add the relative velocity and the velocity of the wind or
water
either by adding the
i
and
j
components,
or by drawing the vectors head-to-tail:
•
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.