Transcript Giancoli, PHYSICS,6/E

Chapter 3
Kinematics in Two Dimensions
Vectors
AP Physics
Motion in two dimensions
Giancoli, PHYSICS,6/E © 2004. Electronically
reproduced by permission of Pearson Education, Inc.,
Upper Saddle River, New Jersey
Module 5
Vectors
Giancoli, Sec 3-1, 2, 3, 4
The following is an excellent lecture on this material.
Giancoli, PHYSICS,6/E © 2004. Electronically
reproduced by permission of Pearson Education, Inc.,
Upper Saddle River, New Jersey
Vectors
•A vector has magnitude as well as direction.
•Examples: displacement, velocity, acceleration,
force, momentum
•A scalar has only magnitude
•Examples: time, mass, temperature
Module 5 - 1
Vector Addition – One Dimension
A person walks 8 km
East and then 6 km East.
Displacement = 14 km
East
A person walks 8 km East
and then 6 km West.
Displacement = 2 km
Module 5 - 2
Vector Addition
Example 1: A person walks 10 km
East and 5.0
North
 km 
D
R D
1 D
2
2
2
D

D

D
R
1
2
2
2
D

(
10
.
0
km
)

(
5
.
0
km
)

11
.
2
k
R
D2
sin
DR
D
5
.
0
km

1
1
0


sin
(2
)

sin
(
)

26
.
5
D
11
.
2
km
R
Order doesn’t matter
Module 5 - 3
Graphical Method of Vector Addition
Tail to Tip Method

V1

VR
Module 5 - 4

V2

V3
Graphical Method of Vector Addition
Tail to Tip Method

V1

V2

V1

V3

VR
Module 5 - 5

V2

V3
Parallelogram Method
Module 5 - 6
Subtraction of Vectors
Negative of vector has same
magnitude but points in the
opposite direction.
For subtraction, we add the negative vector.
Module 5 - 7
Multiplication by a Scalar
A vector V can be multiplied by a scalar c; the result is a
vector cV that has the same direction but a magnitude
cV. If c is negative, the resultant vector points in the
opposite direction.
Module 5 - 8
Adding Vectors by Components
Any vector can be expressed as the sum
of two other vectors, which are called its
components. Usually the other vectors are
chosen so that they are perpendicular to
each other.
Module 5 - 9
Trigonometry Review
Hypotenuse
Opposite
Adjacent
sin
 Opposite
Hypotenuse
cos
 Adjacent
Hypotenuse
Module 5 - 10
Opposite
sin

tan



Adjacent
cos

Adding Vectors by Components
If the
components are
perpendicular,
they can be
found using
trigonometric
functions.
Vy
sin
 Opposite
V
Vsin


y
Hypotenuse
V
Vx
cos
 Adjacent


V

V
cos

x
Hypotenuse
V
sin

tan
Opp

Adj
cos

Module 5 - 11
Adding Vectors by Components
The components are effectively one-dimensional,
so they can be added arithmetically:
Module 5 - 12
Signs of Components
y
Rx  
Rx  
Ry  
Ry  
x
Rx  
Ry  
Rx  
Ry  
Module 5 - 13
3-4 Adding Vectors by Components
Adding vectors:
1. Draw a diagram; add the vectors graphically.
2. Choose x and y axes.
3. Resolve each vector into x and y components.
4. Calculate each component using sines and cosines.
5. Add the components in each direction.
6. To find the length and direction of the vector, use:
Module 5 - 14
sin 
Vy
V
Module 6
Vector Problems and Relative Velocity
Giancoli, Sec 3- 4, 8
The following is an excellent lecture on this material.
Giancoli, PHYSICS,6/E © 2004. Electronically
reproduced by permission of Pearson Education, Inc.,
Upper Saddle River, New Jersey
Example 2
A man pushing a mop across a floor causes it to undergo two
displacements. The first has a magnitude of 150 cm and makes a angle of 1200
with the positive x-axis. The resultant displacement has a magnitude of 140 cm
and is directed at an angle of 35.00 to the positive x axis. Find the magnitude
and direction of the second displacement.

A
A  150
cm
  
R

A

B

 

B

R
1200
35
R  140
cm
BRA
B
R
A
X
0
X
X


B

(
140
cm
)
cos
35

(
150
cm
)
co
12
x
Bx 190
cm
BRA
y
y
y
B

(
140
cm
)
sin
35

(
150
cm
)
sin
12
y

Module 6 - 1
B
49
.6cm
y 

Example 2
Alternative Solution. In the solution below, the angles
for vector A are measured from the negative x axis. In this case, we have to
assign the signs for the components. The answer is the same.

A

B
60
A  150
cm
  
R

A

B

 

R
0
35
R  140
cm
BRA
B
R
A
X
0
X
X


B

(
140
cm
)
cos
35

(

150
cm
)
co
6
x
Bx 190
cm
BRA
y
y
y
B

(
140
cm
)
sin
35

(
150
cm
)
sin
60
y

Module 6 - 2
B
49
.6cm
y 

A man pushing a mop across a floor
Example 2 Continued
causes it to undergo two displacements. The first has a magnitude
of 150 cm and makes a angle of 1200 with the positive x-axis. The
resultant displacement has a magnitude of 140 cm and is directed
at an angle of 35.00 to the positive x axis. Find the magnitude and
direction of the second displacement.
B
 B
B
2
2
x
y
2
2
196cm
B

(
190
cm
)

(

49
.
6
cm
)

49
.6
cm
sin

196
cm
49
.6

 
sin


196



1
Module 6 - 3
14.6
Relative Velocity
•Will consider how observations made in different
reference frames are related to each other.
A person walks toward the front of a train at 5 km / h
(VPT). The train is moving 80 km / h with respect to the
ground (VTG). What is the person’s velocity with respect
to the ground (VPG)?



V
V
V
PG
PT
TG
V

5
km
/
h

80
km
/
h

85
km
/
h
PG
Module 6 - 4
Relative Velocity
•Boat is aimed upstream
so that it will move
directly across.
•Boat is aimed directly
across, so it will land at a
point downstream.
•Can expect similar
problems with airplanes.
Module 6 - 5
Example 6 An airplane is capable of flying at 400 mi/h in still air.
At what angle should the pilot point the plane in order for it to
travel due east, if there is a wind of speed 50.0 mi/h directed due
south? What is the speed relative to the ground?
VPA


VAG

VPG
VAG
sin

V
PA
mi


50
.
0


V



1
0

1
h
AG



7
.
18
sin  sin


mi
400
PA
V

h

North of East
0
mi
mi
V

(
400
)
cos
(
7
.
18
)

397
PG
h
h
Module 6 - 6
Vectors
Vector Addition Applet
Module 6 - 7
Module 7
Projectile Motion
Giancoli, Sec 3-5, 6, 7
Giancoli, PHYSICS,6/E © 2004. Electronically
reproduced by permission of Pearson Education, Inc.,
Upper Saddle River, New Jersey
3-5 Projectile Motion
A projectile is an object
moving in two
dimensions under the
influence of Earth's
gravity; its path is a
parabola.
Module 7 - 1
Projectile Motion
•Neglect air resistance
•Consider motion only after release and before it hits
•Analyze the vertical and horizontal components separately
(Galileo)
•No acceleration in the horizontal, so velocity is constant
•Acceleration in the vertical is – 9.8 m/s2 due to gravity and
thus velocity is not constant.
•Object projected horizontally will reach the ground at the
same time as one dropped vertically
Module 7 - 2
Equations for Projectile Motion
Horizontal
ax=0
Vertical
ay = - g
vx= constant
v0  vx0
x x0 vx0 t
vy vy0 gt
1 2
y
y
v
t
0
y
0t g
2
2
2
v

v
2
g
(y

y
)
y
y
0
0
Module 7 - 3
Initial Velocity
vy0 v0sin

vx0 v0cos

•If the ball returns to the y = 0 point, then the velocity at
that point will equal the initial velocity.
•At the highest point, v0 y = 0 and v = vx0
Module 7 - 4
Example 3A A football is kicked at an angle of 50.00 above the
horizontal with a velocity of 18.0 m / s. Calculate the maximum
height. Assume that the ball was kicked at ground level and lands
at ground level.
m
.6m
v

(
18
.
0
)(cos
50
.
0
)11
s
s
m
.8m
v

(
18
.
0
)(sin
50
.
0
)13
s
s

x
0

y
0
at top:
vy vy0 gt  0
 v  13.8ms
t   
1.41s
m
 g  9.80 s
1
y
y

v
tgt
2
y0
2
2
max 0
yo
1

m 2
m
y

0

(
13
.
8
)(
1
.
41
s
)

(
9
.
8
)(
1
.
41
s
)


2
max s
2

 s
ymax9.7m
Module 7 - 5
Level Horizontal Range
•Range is determined by time it takes for ball to return to
ground level or perhaps some other vertical value.
•If ball hits something a fixed distance away, then time is
determined by x motion
•If the motion is on a level field, when it hits: y = 0
1
1
2
2
y

y

v
t

g
t

0

0

v
t

g
t
0y
0
yo
2
2
2 vy0
Solving we find t  g

We can substitute this in the x equation to find the range R
2
2
v
2
v
v
2
v
sin
cos
y
0
x
0
yo
0
0
0
R

x

v
t

v
(
)


x
0 xo
g g
g
Module 7 - 6
Level Horizontal Range
sin
cos

sin
2

We can use a trig identity 2
v02sin
2
R
g
•Greatest range:  = 450
•  = 300 and 600 have
same range.
0
0
(45
15
)
Module 7 - 7
Caution– the range formula
has limited usefulness. It is
only valid when the
projectile returns to the same
vertical position.
Example 3B
A football is kicked at an angle of 50.00 above the horizontal with
a velocity of 18.0 m / s. Calculate the range. Assume that the ball
was kicked at ground level and lands at ground level.
Assume time down = time up
(2
)(
1
.41
s
) 2.82s
For Range: t
m
0

(
11
.
6
)(
2
.
82
s
)
R

x

x
vt
s
 33m
0
x
0
Could also use range formula



2
2
0
v
sin
2
(
18
m
/
s
)
sin
(
2
)
(
50
)
0
R



33
m
2
g
9
.
8
m
/
s
Module 7 - 8
Example 4A A football is kicked at an angle of 50.00 above the
horizontal with a velocity of 18.0 m / s. The football hits a window
in a house that is 25.0 m from where it was kicked. How high was
the window above the ground.
m
.6m
v

(
18
.
0
)(cos
50
.
0
)11
s
s
m
13
.8m
v

(
18
.
0
)(sin
50
.
0
)
s
s

x
0

y
0
x v t
Time to hit the window:
x0
25.0m

11.6m
s
x
t
v
x0
2.16s
1
y

y
vtgt
2
2
0
Module 7 - 9
y
0
1
2
m
m
y

0

(
13
.
8
)(
2
.
16
s
)

(
9
.
8
)(
2
.
16
s
)
2
s
s
2
y  6.9m
Example 4 B What is the final velocity and angle of the football
that hit the window in Example 4 A.
t 2.16
s
vv gt
y
y0
m
m
m
v

(
13
.
8
)

(
9
.
8
)
(
2
.
16
s
)
7
.
37
2
y
s
s
s
m
v

11
.6
s
x
m
m
m
v

(
11
.
6
)

(

7
.
37
)13
.
7
s
s
s
2
2
v
tan

v
y
x
.37
7


tan
   32
below x axis
.
4
11
.6

1
Module 7 - 10


Example 5. (35) A rescue plane wants to drop supplies to isolated mountain
climbers on a rocky ridge 235 m below. If the plane is traveling horizontally
with a speed of 250 km /h (69.4 m / s) how far in advance of the recipients
(horizontal distance) must the goods be dropped (Fig. 3–37a)? .
vy0 0
vx0 69
.4m
/s
Coordinate system is 235
m below plane
1 2
y

235
m

0
g
t
0
2
(
2
)
(
235
m
)
t

6
.
93
s
2
9
.
8
m
/s
Module 7 - 12
x

x

v
t

0

(
69
.
4
m
/
s
)
(
6
.
93
s
)
0
xo
x

481
m
Sec. 3-7 Projectile Motion Is Parabolic
In order to demonstrate that
projectile motion is parabolic,
the book derives y as a function
of x. When we do, we find that it
has the form:
This is the
equation for
a parabola.
Module 7 - 13