1.1 Vectors

Scalars
A scalar quantity requires only size (magnitude) to completely describe it.
Vectors
A vector quantity requires size (magnitude) and a direction to completely describe it.

Here are some vector and scalar quantities:
Scalar time temperature volume distance speed energy mass frequency power
Vector force weight acceleration displacement velocity momentum impulse
** Familiarise yourself with these scalar and vector quantities **

N
A helicopter takes off from Edinburgh and drops a package over Inverness before landing at Glasgow as shown.
W
S
E
Inverness
To calculate how much fuel is needed for the journey, the total distance is required.
200 km
75 km
300 km
If the pilot wanted to know his final position relative to his starting position , the displacement is required.
Glasgow Edinburgh

Distance
Distance travelled by the helicopter: distance  500 km distance - total distance travelled along a route
Displacement
Helicopters final position relative to starting position : displaceme nt  75 km due West (270  ) displacement - final position relative to starting position

Summary
Distance has only size , whereas displacement has both size and direction .

Speed is the rate of change of distance : speed  distance time
Say the helicopter journey lasted 2 hours, the speed would be: speed 
500
2
 250 km h  1

Velocity however, is the rate of change of displacement : velocity  displaceme time nt
So for the 2 hour journey, the velocity is: velocity 
75
2
 37.5
km h  1 due west (270  )
Speed has only size , whereas velocity has both size and direction .

Q1 – Q9

Vectors are represented by a line with an arrow.
The length of the line represents the size of the vector.
The arrow represents the direction of the vector.
The sum of two or more vectors is called the resultant .

Vectors can be added using a vector diagram .
Vector diagrams are drawn so that vectors are joined “tip-to-tail”
Vector 2
Vector 2
Vector 1
RESULTANT
VECTOR
Vector 1
RESULTANT
VECTOR
Resultant of a Vector
The resultant of a number of forces is that single force which has the same effect , in both magnitude and direction, as the sum of the individual forces.

Example 1
A man walks 40 m east then 50 m south in one minute.
W
(a)
(b)
(c)
(d)
(e)
N
Draw a diagram showing the journey.
Calculate the total distance travelled.
Calculate the total displacement of the man.
Calculate his average speed.
Calculate his velocity.
S
(a) Draw a diagram showing the journey.
40 m
50 m
Vectors are joined
“ tip-to-tail ”
E

(b) Calculate the total distance travelled.
distance  40  50
 90 m
(c) Calculate the total displacement of the person.
40 m displacement
50 m
The displacement is the size and direction of the line from start to finish.
Size
By Pythagoras:
 displaceme nt a 2

2
 b 2
 40
 c 2
2  50 2
 displaceme nt 

4100
4100
64 m

Direction
40 m
displacement
(d)
50 m tan θ  opp adj tan θ 
θ
50
40
 tan  1

1.25

θ  51.3

Calculate the speed of the man.
speed  distance time speed 
90
60 speed  1.5
ms  1
So the total displacement of the man is: s  64m on a bearing of 141.3

90 + 51.3 = 141.3°
(bearing)

(e) Calculate the velocity of the man.
velocity  displaceme time nt velocity 
64
60 velocity  1.07 ms  1 on a bearing of 141.3

Speed has only size , whereas velocity has both size and direction .

Example 2
A plane is flying with a velocity of 20 ms -1 due east. A crosswind is blowing with a velocity of 5 ms -1 due north.
Calculate the resultant velocity of the plane.
N
Size
By Pythagoras velocity
θ
20 ms -1
5 ms -1
W
S
E
Direction a 2  b 2  c 2 v 2 
 v  v 
20 2 
425
425
5 2
20.6
ms -1 tan θ  tan θ 
θ  opp adj
5
20 tan  1
 0.25

θ  14  90 – 14 = 076°
(bearing) velocity  20.6
ms -1 on bearing of 076 

Q1.
A person walks 65 m due south then 85 m due west.
(a) draw a diagram of the journey
(b)
(c) calculate the total distance travelled calculate the total displacement.
[ 150 m ]
[ 107 m at bearing of 232.6°]
Q2.
A person walks 80 m due north, then 20 m south.
(a) draw a diagram of the journey
(b)
(c) calculate the total distance travelled calculate the total displacement.
[ 100 m ]
[ 60 m due north]
Q3.
A yacht is sailing at 48 ms -1 due south while the wind is blowing at 36 ms -1 west.
Calculate the resultant velocity.
[ 60 ms -1 on bearing of 216.9°]

Q1 – Q12

Vectors are not always at right angles with each other.
To add such vectors together, it is easiest to use a scale diagram .
Example 1
An aircraft travels due north for 100 km. The aircraft changes its course to 25° west of north and travels for a further 250 km.
Find the displacement of the aircraft.

Step 1
Choose a suitable scale.
25 km : 1 cm
Step 2
Draw diagram using a pencil and a protractor.
13.7 cm
Step 3
Measure the length of the resultant vector and convert using your scale.
13.7 x 25 km = 342.5 km
Step 4
Measure the size of the angle using a protractor.
W
10 cm
N
S
θ
4 cm
E

Example 2
A ship sailing due west passes buoy X and continues to sail west for 30 minutes at a speed of 10 km h -1 .
It changes its course to 20° west of north and continues on this course for 1½ hours at a speed of 8 km h -1 until it reaches buoy Y.
(a) Show that the ship sails a total distance of 17 km between marker buoys X and Y.
(b) By scale drawing or otherwise, find the displacement from marker buoy X to marker buoy Y.
(a) Stage 1 d  v 
 d  t
10 
5 km
0.5
Stage 2 d  v 
 d  t
8  1.5
12 km
Total d  17 km

(b)
1 km : 1 cm
N
W E
S
14.4 cm
Length of Vector
14.4 x 1 km = 14.4 km
12 cm
Direction of Vector
θ = 52°
5 cm
θ displaceme nt  14.4
km with angle θ of 52 
Answer Range
14.5 km ± 0.4 km
52° ± 2°

Q1 – Q3

Horizontal and Vertical Components
To analyse a vector, it is essential to ‘ break-up ’ or resolve a vector into its rectangular components .
The rectangular components of a vector are the horizontal and vertical components .
V
=
V
V
OR
V
V
V
H
V
H

The horizontal and vertical component of the vector can be calculated as shown.
θ
V
V
H
V
V sin θ
V
V

V
V
V
 V sin θ cos θ
V
H

V
V
H
 V cos θ

Example 1
A ship is sailing with a velocity of 50 ms -1 on a bearing of 320°.
Calculate its component velocity
(a) north
V
W
N
50 ms -1
360° - 320° = 40°
40°
V
N cos θ 
V
V
N cos 40 
V
N
50
V
N
 50  cos 40
V
N
 38.3
ms -1
W
S
E

(b) west sin θ 
V
W
V sin 40 
V
W
50
V
W
V
W
 50  sin 40
 32.1
ms  1
V
W
50 ms -1
40°
V
N

Example 2
A ball is kicked with a velocity of 16 ms -1 at an angle of 30° above the ground.
Calculate the horizontal and vertical components of the balls velocity.
Horizontal
16 ms -1
30°
V
H
V
V cos θ 
V
V
H cos 30 
V
H
16
V
H
 16  cos 30
V
H
 13.9
ms  1

Vertical sin θ 
V
V
V sin 30 
V
V
16
V
V
V
V
 16  sin 30
 8 ms  1
16 ms -1
30°
V
H
V
V

Slopes – Parallel and Perpendicular Components
On a slope , the components of a vector are parallel and perpendicular to the slope.
Vectors are joined
“ tip-to-tail ”
θ x resultant
θ W  m g y
Perpendicular Component cos θ  adj hyp cos θ  x mg x  m g cos θ
Parallel Component sin θ  opp hyp sin θ  y mg y  m g sin θ

Example 1
A 10 kg mass sits on a 30° slope.
Calculate the component of weight acting down (parallel) the slope.
30
W  m g mg
(resultant)
30 x y sin θ  opp hyp sin 30  sin 30  y mg
10  y
9.8
y  98  sin 30 y  49 N

Q1 – Q8