© Nuffield Foundation 2011

Nuffield Free-Standing Mathematics Activity

Vectors

Vectors

• Is the water skier moving in the same direction as the rope?

• What forces are acting on the water skier? • Which directions are the forces acting in?

Vectors

Scalar quantities have magnitude but no direction

Examples mass distance speed temperature

Vectors have magnitude and direction

Examples displacement velocity acceleration

force momentum

Unit vectorsSuppose the velocity of a yacht has an easterly component of 12 ms–1 and a northerly component of 5 ms–1

The velocity is v ms–1

where v = 12i + 5j

i represents a unit vector to the east

and j represents a unit vector to the north

v =

512Column vector notation

Magnitude and direction of a vector

v

v1

v2v =

2

1vv

Magnitude v = 22

21 vv

Direction tan = 1

2vv

=

1

2vv

tan–1

Think about How can you use the triangle to find the magnitude and direction of v?

v =

512

Example

Speed v = 22 512

Direction tan = 125

v

12

5

= 13

= 60.41

= 22.6

bearing

The yacht is sailing at 13 ms–1 on bearing 067 (nearest )

N

Think about How can you find the speed and direction of the yacht?

To add or subtract vectorsAdd or subtract the components

ExampleForces acting on an object

(in newtons)

34-

57

where i is a horizontal unit vector to the right

and j is a vertical unit vector upwards

Total force acting on the object

34-

57

83

Think about how to find the total force

To multiply a vector by a scalarMultiply each component by the scalar

ExampleDisplacement s = (in metres)

2-1

s

3s =

6-3

3s

Multiplying by 3 gives a displacement 3 times as big in the same direction

Think about What do you get if you multiply both components of the vector by 3?

Constant acceleration equationstauv Equation 1

t) (21 vus Equation 3

221 tt aus Equation 2

where

u = initial velocity

v = final velocity

a = acceleration

t = time taken

s = displacement

Momentum mv is a vector

Forces and acceleration

Newton’s First Law

Resultant force causes acceleration

Action and reaction are equal and opposite.

A particle will remain at rest or continue to move uniformly in a straight line unless acted upon by a non-zero resultant force.

Newton’s Second LawF = ma

Newton’s Third Law

This means if a body A exerts a force on a body B, then B exerts an equal and opposite force on A.

Resultant force is the sum of the forces acting on a body, in this case F1 + F2 + F3

F1

F2

F3

Swimmer

i = unit vector to the east

j = unit vector to the north

Find the magnitude and direction of the swimmer’s resultant velocity.

Resultant velocity

2.1-0.5

1.52.4

0.6-2.9

vR=

Speed vR = 22 0.62.9

Direction tan = 2.90.6

= 11.7

= 2.96 ms–1

= 0.2068 … vR

2.9

0.6

bearing

The swimmer will travel at 2.96 ms–1 on bearing 102 (nearest )

N

vS=

1.52.4

vC=

2.1-0.5

(ms–1)

Golf ball

O

u =

1625

i = horizontal unit vector

j = vertical unit vector

a =

9.8-0

Find a the velocity at time tb the velocity when t = 2c the ball’s displacement from O, when t = 2

tauv a t

9.8-0

1625

t9.8-16

25

b When t = 2

29.8-1625

v

c 221 s tt au 2

9.8-0

21

1625 tt

24.9-16

25tt

t

When t = 2 s

224.9-216225

(ms–1)

(m)

3.6-25

(ms–1)

12.450

(m)

Skiera Find the skier’s acceleration.

b Find the speed and directionof the skier 20 seconds later.

u =

68-

F =

15-24

a) F = ma

= 60a

15-24

a =

0.25-0.4

tauv b)

200.25-0.4

68-

v

10v

The skier is travelling at 1 ms–1 to the north.

(ms–2) (ms–1)

i = unit vector to the east

j = unit vector to the north

60 kg

Ship travels at a constant velocity u ms–1

a What is the force, F, from the tug?

b Ship’s initial position vector r i Find the position vector of the boat at time t.

ii The ship is aiming for a buoy which has position vector Assuming the ship reaches the buoy, find x.

Ship

u =

1-2.5R =

12005600-

500300

100x

a Ship travels at a constant velocity u ms–1

This means there is no acceleration

Ship

u =

1-2.5R =

12005600-

F =

1200-5600

200

21

1-2.5 tt

tt

-2.5

t

t - 500

2.5300

500300r

tt

-2.5

t = 400

x = 300 + 2.5t = 300 + 2.5 400 = 1300

500 – t = 100

O

500300

tt

-2.5

r

100x

Ship’s initial position vector r

500300

221 tt aus b i Displacement

At time t,

b ii When ship reaches

100x

Reflect on your work

• How have you used the fact that i and j are perpendicular unit vectors?

• Are there any similarities between the problems or the techniques you have used?

• Can you think of other scenarios which could be tackled using vectors in component form?