Transcript Climate Change - Nuffield Foundation

Nuffield Free-Standing Mathematics Activity
Vectors
© Nuffield Foundation 2011
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
force
momentum
velocity
acceleration
Unit vectors
Suppose 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
Column vector notation
v=





12

5 
Magnitude and direction of a vector
v1
v=
v2 





v
v2

v1
Magnitude
Direction
v=
v12  v22
v2
tan  =
v1


–1



v2 
 = tan

v1 
Think about
How can you use the triangle
to find the magnitude and
direction of v?
N
Example
v=
Speed
Direction





12

5 
bearing
Think
about
v
How can you find the
5 speed

and direction of the yacht?
12
v = 122  52 = 13
tan  =
5

= 0.416
12
 = 22.6
The yacht is sailing at 13 ms–1 on bearing 067 (nearest )
To add or subtract vectors
Add or subtract the components
Example
Forces acting on an object
(in newtons)










7
- 4
3



5



where i is a horizontal unit vector to the right
and j is a vertical unit vector upwards
Think about
how to find the total force
Total force acting on the object
7 - 4 3
    
5  3  8





To multiply a vector by a scalar
Multiply each component by the scalar
Example
 
1
Displacement s =  
- 2
 
(in metres)
s
3s
Think about
What do you get if you multiply both
components of the vector by 3?
3s =





3 

- 6
Multiplying by 3 gives a displacement 3 times as big
in the same direction
Constant acceleration equations
Equation 1
v  u  at
where
u = initial velocity
Equation 2
Equation 3
Momentum
s  ut  1at 2
2
s  1(u  v) t
2
mv is a vector
v = final velocity
a = acceleration
t = time taken
s = displacement
Forces and acceleration
F1
Resultant force is the sum of the forces acting on a
body, in this case F1 + F2 + F3
F3
Newton’s First Law
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 Law
Resultant force causes acceleration F = ma
Newton’s Third Law
Action and reaction are equal and opposite.
This means if a body A exerts a force on a body B, then B exerts
an equal and opposite force on A.
F2
Swimmer
Find the magnitude and direction of the
swimmer’s resultant velocity.
v=
S
Resultant velocity
Speed
2.4  0.5   2.9 

 





1.5 - 2.1 - 0.6
vR =
2.92  0.62
= 2.96
Direction
v=
C
i = unit vector to the east
j = unit vector to the north
ms–1
tan  =
0.6
= 0.2068 …
2.9
 = 11.7
2.4 

1.5 





0.5 

- 2.1
(ms–1)
N
bearing
2.9

vR=










vR
The swimmer will travel at 2.96 ms–1 on bearing 102 (nearest )
0.6





25

u=
16
Golf ball
Find
a the velocity at time t
b the velocity when t = 2
c the ball’s displacement from O,
when t = 2
a
c
i = horizontal unit vector
j = vertical unit vector

25 
25  0 

v  u  at    

t  
16 - 9.8t 



16 - 9.8







25 
v 

16 - 9.82



0 

a=
- 9.8
O





b When t = 2






(ms–1)





25 
–1

 (ms )
- 3.6


25
0  2  25t 
s  ut  1at 2   t  1 

t

2
2
16




2 - 9.8


16t - 4.9t 
When t = 2



252
 50 

s 
 (m)
 

2

162 - 4.92  12.4 





(m)





- 8

u=
6 
Skier
a Find the skier’s acceleration.
b Find the speed and direction
of the skier 20 seconds later.





24 

F=
-15
60 kg
i = unit vector to the east
j = unit vector to the north
a) F = ma
b)





24 
 = 60a
-15





0.4 
a=
 (ms–2)
- 0.25
v  u  at
v
v






-8






6
0

1










0.4
-0.25
(ms–1)
The skier is travelling at 1 ms–1 to the north.






20
Ship
Ship travels at a constant velocity u ms–1
a What is the force, F, from the tug?
b





300
Ship’s initial position vector r 

500





- 5600


2.5



R=


u
=
1200 
 -1 


i Find the position vector of the boat at time t.




x




ii The ship is aiming for a buoy which has position vector
100
Assuming the ship reaches the buoy, find x.
Ship
a Ship travels at a constant velocity u ms–1
This means there is no acceleration





- 5600


2.5 

R=
1200  u =  -1 







5600 

F=
-1200


 


1 0 2 2.5t 
2.5 
1
2

t

 t

b i Displacement s  ut  at  



2
2
1
0




 
 -t 


300
Ship’s initial position vector r   
500


At time t,
b ii












2.5t  300  2.5t 
r

 



500
- t   500 - t 
300





When ship reaches
500 – t = 100




x




300

500
t = 400
= 1300
2.5t 

- t 
r
O
100
x = 300 + 2.5t = 300 + 2.5 400













x




100
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?