Next Week
Experiment 1:
Measurement of Length and Mass
WileyPLUS Assignment 1 now available
Due Monday, October 5 at 11:00 pm
Chapters 2 & 3
Friday, September 25, 2009
44
Relative Velocity
If:
A moves at velocity �vA (relative to the ground)
and:
B moves at velocity �vB (relative to the ground)
then, the velocity of B relative to A is:
�vBA =�vB −�vA
This is the velocity of B as seen by A.
Friday, September 25, 2009
45
Focus on Concepts, Question 16
The drawing shows two cars travelling in different directions with different
speeds. Their velocities are:
vAG = velocity of car A relative to the Ground = 31.3 m/s, due east
vBG = velocity of car B relative to the Ground = 21.0 m/s, due north
The driver of car B looks out the window and sees car A. What is the velocity
(magnitude and direction) of car A as observed by the driver of car B? In
other words, what is the velocity vAB of car A relative to car B?
21.0 m/s
31.3 m/s
Friday, September 25, 2009
46
3.59/53: A hot air balloon is moving relative to the ground at 6 m/s due
east. A hawk flies at 2 m/s due north relative to the balloon. What is
the velocity of the hawk relative to the ground?
Velocity of hawk relative to balloon is vH - vB
and is 2 m/s to the north
900
vH =
�
62 + 22 = 6.3 m/s
�vH = �vB + (�vH − �vB )
�vH
(relative to the ground)
2
tan θ = , θ = 18.4◦ N of E
6
Friday, September 25, 2009
θ
900
47
Clickers!
Strategies for swimming across the river in the shortest time.
Which is fastest? The swimmers swim at the same speed vs in the
water. The water flows at speed vw to the right.
(A)
(B)
�vw
(C)
�vtot =�vs +�vw
�vw
�vw
�vs
�vtot
(A) Head
straight across,
get carried
downstream
�vs
�vs
�vtot
(B) Head
upstream to end
at closest point
on far bank
�vtot
(C) Head
downstream to take
advantage of the
current
Friday, September 25, 2009
48
Relative Velocity
�vw 0.91 m/s
1.4 m/s
�vs
�vtot
�vtot =�vs +�vw
3.53/47: A swimmer swims directly across a river that is 2.8 km wide.
He can swim at 1.4 m/s in still water (vs), i.e. at 1.4 m/s relative to the
water. The river flows at 0.91 m/s (vw), i.e. at 0.91 m/s relative to the
riverbank.
How long to cross the river?
Where does he end up on the other bank?
Friday, September 25, 2009
49
Relative Velocity contd
�vw 0.91 m/s
1.4 m/s
�vs
As he’s swimming directly across the
river, his speed toward the other bank
is 1.4 m/s.
�vtot
Therefore, time to cross the river is
2800 m
= 2000 s
1.4 m/s
In this time, the current will carry him downstream by:
(0.91 m/s) × (2000 s) = 1820 m
Friday, September 25, 2009
50
Relative Velocity contd
Alternative strategy: he
heads upstream a little so as
to swim directly across the
river – the most direct route.
�vw 0.91 m/s
1.4 m/s
�vtot
�vs
θ
�vtot =�vs +�vw
Pythagoras:
2
v2s = v2w + vtot
�
So, vtot = v2s − v2w
Takes
=
�
1.42 − 0.912 = 1.064 m/s
2800 m
= 2630 s to cross the river
1.064 m/s
Friday, September 25, 2009
sin θ =
0.91
→ θ = 41◦
1.4
51
3.-/69: An aircraft is headed due south with a speed of 57.8 m/s relative
to still air. Then, for 900 s a wind blows the plane so that it moves in a
direction 450 west of south, even though the plane continues to point due
south. The plane travels 81 km with respect to the ground in this time.
Determine the velocity of the wind with respect to the ground.
To south:
(v p)S + (vw)S = 90 cos 45◦
57.8 + (vw)S = 63.64 → (vw)S = 5.84 m/s
90 m/s
To west:
S
(v p)W + (vw)W = 90 sin 45◦
0 + (vw)W = 63.64 → (vw)W = 63.64 m/s
�
63.84
vw = 5.842 + 63.642 = 63.9 m/s
tan θw =
→ θw = 84.8◦
5.84
west of south
Friday, September 25, 2009
52
3.62/56: Relative to the ground, a car has velocity 18 m/s, directed
due north. Relative to this car, a truck has a velocity of 22.8 m/s,
directed at 52.10 south of east. Find the magnitude and direction of
the truck’s velocity relative to the ground.
Components:
To east: (vt )E − (vc)E = 22.8 cos 52.1◦ = 14 m/s
(vt )E − 0 = 14 m/s → (vt )E = 14 m/s
To south: (vt )S − (vc)S = 22.8 sin 52.1◦ = 18 m/s
(vt )S + 18 = 18 → (vt )S = 0
vt = 14 m/s to the east
Friday, September 25, 2009
53
Prob. 3.78/58: Two boats are heading away from shore. Boat 1
heads due north at a speed of 3 m/s relative to the shore.
Relative to boat 1, boat 2 is moving 300 north of east at a speed
of 1.6 m/s.
A passenger on boat 2 walks due east
across the deck at a speed of 1.2 m/s
relative to boat 2.
What is the speed of the passenger
relative to the shore?
1.6 m/s
3 m/s
1.2 m/s
Friday, September 25, 2009
54
Chapter 3, Problem X3
A car drives horizontally off the edge of a cliff that is 35.4 m high. The
police at the scene of the accident note that the point of impact is 159
m from the base of the cliff. How fast was the car travelling when it
drove off the cliff?
Friday, September 25, 2009
55
3.65/59: A ship is travelling at a constant velocity of 4.2 m/s due east,
relative to the water. On his radar screen the navigator detects an object
that is moving at a constant velocity. The object is located at a distance of
2310 m with respect to the ship, in a direction 320 south of east. Six
minutes later, he notes that the object’s position relative to the ship has
changed to 1120 m, 570 south of west.
What are the magnitude and direction of the velocity of the object relative
to the water?
vsw 4.2 m/s
Velocity of ship relative to water
Ship
East
0
320
57
P1, P2: positions of object
2310 m
1120 m
relative to ship
Relative to ship:
P2
Displacement of object
relative to ship in 360 s
P1
= P2 – P1
Friday, September 25, 2009
56
vsw 4.2 m/s
Relative to ship:
P2
Velocity of ship relative to water
Ship
East
0
320
57
P1, P2: positions of object
2310
m
1120 m
relative to ship
P1
Displacement of object
= P2 – P1
relative to ship in 360 s
Displacement of object relative to ship in 360 s = P2 – P1
To the west: (P2 – P1)W = 1120 cos 570 + 2310 cos 320 = 2569 m to W
To the south: (P2 – P1)S = 1120 sin 570 – 2310 sin 320 = –285 m to S
Velocity of object relative to ship:
To the west: (vos)W = 2569/360 = 7.14 m/s to W 0.79 m/s
To the south: (vos)S = –285/360 = –0.79 m/s to S.
vos
7.14 m/s
Velocity of object relative to the water: vow = vos + vsw
Friday, September 25, 2009
57
Object relative to ship
vsw 4.2 m/s
0.79 m/s
(ship relative to water)
vos
7.14 m/s
Object relative to water
Velocity of object relative to the water: vow = vos + vsw
To the west: (vow)W = 7.14 – 4.2 = 2.94 m/s to W
To the north: (vow)N = 0.79 m/s to N
vow =
tan θ =
�
0.792 + 2.942
vow
0.79 m/s
θ
= 3.04 m/s
2.94 m/s
0.79
→ θ = 15◦ north of west
2.94
Friday, September 25, 2009
58
Cosine Rule
vsw 4.2 m/s
Relative to ship:
Ship
57 θ 320
East
0
b
P2
Velocity of ship relative to water
1120 m
c
a
Displacement of object
relative to ship in 360 s
a2 = b2 + c2 − 2ab cos θ
P1, P2: positions of object
relative to ship
2310 m
P1
θ = 1800 - 570 - 320 = 910
So, a2 = 11202 + 23102 − 2 × 1120 × 2310 cos 91◦ → a = 2585 m
Friday, September 25, 2009
59
Chapter 3 Summary
•
The laws of motion can be applied separately to motions in x
and y (negligible air resistance).
•
The time for a projectile to move up and down is the same
as the time for it to sideways.
•
Relative velocity is an application of the subtraction of
vectors covered in chapter 1.
•
If A travels at �vA and B travels at �vB the velocity of B relative
to A is �vB −�vA
Friday, September 25, 2009
60