Chapter 3: Motion in a Plane
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Vector Addition
Velocity
Acceleration
Projectile motion
Relative Velocity
CQ: 1, 2.
P: 3, 5, 7, 13, 21, 31, 39, 49, 51.
Two Dimensional Vectors
• Displacement, velocity, and acceleration
each have (x, y) components
• Two methods used:
• geometrical (graphical) method
• algebraic (analytical) method
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2
Graphical, Tail-to-Head
3
Addition Example
• Giam (11)
Order Independent (Commutative)
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Subtraction, tail-to-tail
6
Subtraction Example
• Giam (19)
Algebraic Component Addition
• trigonometry & geometry
• “R” denotes “resultant” sum
• Rx = sum of x-parts of each vector
• Ry = sum of y-parts of each vector
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Vector Components
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Examples
• Magnitude || (g4-5) Notation, Example
• Component Example Animated
• Phet Vectors
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Trigonometry
o
sin  
h
a
cos  
h
h
o

a
o
tan  
a
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Using your Calculator: Degrees and
Radians
sin 30  0.5
Check this to verify your
calculator is working with
degrees
sin 60  0.866
cos 30  0.866
cos 60  0.5
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Example:
h
o
Given:
 = 10°, h = 3

a
Find o and a.
o
sin 10 
3
a
cos10 
3
3 sin 10  o
o  0.52
3 cos10  a
a  2.95
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Inverse Trig
• Determine angle from length ratios.
• Ex. o/h = 0.5:
1
sin (0.5)  30
• Ex. o/a = 1.0:
1
tan (1.0)  45
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Pythagorean Theorem
h
o a h
2
2
o
2

a
Example: Given,
2 3  h
o = 2 and a = 3
49  h
Find h
h  13
2
2
2
2
2
h  13  3.605
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Azimuth:
Angle measured counterclockwise from +x direction.
Examples:
East 0°, North 90°, West 180°, South 270°.
Northeast = NE = 45°
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Check your understanding:
What are the Azimuth angles?
A: 180°
B: 60°
70°
C: 110°
Note: All angles measured from east.
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Components:
Given A = 2.0m @ 25°, its x, y components are:
Ax  A cos   2.0 cos 25  1.81m
Ay  A sin   2.0 sin 25  0.85m
Check using Pythagorean Theorem:
A  A  A  1.81  0.85  1.9996  2.0
2
x
2
y
2
2
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Vector Addition by Components:
Rx  A cos  A  B cos  B  
Ry  A sin  A  B sin  B  
R R R
2
x
2
y
 Ry 
  tan    180(quads.II , III )
 Rx 
1
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Example Vector Addition
R = (10cm, 0°) + (10cm, 45°):
Rx  Ax  Bx  10 cos 0  10 cos 45
Ry  Ay  By  10 sin 0  10 sin 45
Rx  10  7.07  17.07
Ry  0  7.07  7.07
R  R  R  17.07  7.07  18.5cm
2
x
2
y
2
2
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(cont) Magnitude, Angle:
 Ry 
  tan    180(quads.II , III )
 Rx 
1  7.07 
  tan 
  0  22.5
 17.07 
1

R  18.5cm at 22.5 above  x axis
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General Properties of Vectors
• size and direction define a vector
• location independent
• change size and/or direction when
multiplied by a constant
• Vector multiplied by a negative
number changes to a direction
opposite of its original direction.
• written: Bold or Arrow
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these vectors are all the same
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Multiplication by Constants
A
0.5A
-A
-1.2A
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Projectile Motion
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time = 0: e.g. baseball leaves fingertips
time = t: e.g. baseball hits glove
Horizontal acceleration = 0
Vertical acceleration = -9.8m/s/s
Horizontal Displacement (Range) = Dx
Vertical Displacement = Dy
Vo = launch speed
o = launch angle
Range vs. Angle
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Example 1: 6m/s at 30
vo = 6.00m/s o = 30°
xo = 0, yo = 1.6m; x = R, y = 0
vox  vo cos  o  6.00 cos 30  5.20m / s
voy  vo sin  o  6.00 sin 30  3.00m / s
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Example 1 (cont.)
Step 1
Dy  voyt  a y t
1
2
2
 1.6  6 sin 30t  (9.8)t
1
2
 1.6  3t  4.9t
2
2
4.9t  3t  1.6  0
2
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Quadratic Equation
ax  bx  c  0
2
 b  b 2  4ac
x
2a
a  4.9
4.9t  3t  1.6  0
2
b  3
c  1.6
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Example 1 (cont.)
a  4.9
 b  b 2  4ac
x
2a
b  3
c  1.6
3  (3) 2  4(4.9)( 1.6)
t
2(4.9)
3  6.353
t
2(4.9)
t  0.342
t  0.954
End of Step 1
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Example 1 (cont.)
Step 2
Dx  voxt  a x t  voxt
1
2
2
(ax = 0)
Dx  vo cos  ot  6 cos 30(0.954)  4.96m
“Range” = 4.96m
End of Example
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Relative Motion
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Examples:
people-mover at airport
airplane flying in wind
passing velocity (difference in velocities)
notation used:
velocity “BA” = velocity of B – velocity of A
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Summary
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Vector Components & Addition using trig
Graphical Vector Addition & Azimuths
Projectile Motion
Relative Motion
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R = (2.0m, 25°) + (3.0m, 50°):
Rx  Ax  Bx  2.0 cos 25  3.0 cos 50
Ry  Ay  By  2.0 sin 25  3.0 sin 50
Rx  1.81  1.93  3.74
Ry  0.84  2.30  3.14
R  R  R  3.74  3.14  4.88m
2
x
2
y
2
2
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(cont) Magnitude, Angle:
R  R  R  3.74  3.14  4.88m
2
x
2
y
2
2
 Ry 
  tan    180(quads.II , III )
 Rx 
1
 3.14 
  tan 
  0  40.0
 3.74 
1
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PM Example 2:
vo = 6.00m/s o = 0°
xo = 0, yo = 1.6m; x = R, y = 0
vox  vo cos  o  6.00 cos 0  6.00m / s
voy  vo sin  o  6.00 sin 0  0m / s
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PM Example 2 (cont.)
Step 1
Dy  voyt  a y t
1
2
2
 1.6  6 sin 0t  (9.8)t
1
2
 1.6  0  4.9t
2
2
4.9t  1.6
2
1.6
t
 0.571
4.9
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PM Example 2 (cont.)
Step 2
Dx  voxt  a x t  voxt
1
2
2
(ax = 0)
Dx  vo cos  ot  6 cos 0(0.571)  3.43m
“Range” = 3.43m
End of Step 2
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PM Example 2: Speed at
Impact
t  0.571s
vx  vox  axt
v y  voy  a y t
vx  6  (0)t  6m / s
v y  (0)  (9.8)0.571
 5.59m / s
v  v  v  (6)  (5.59)  8.20m / s
2
x
2
y
2
2
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1. v1 and v2 are located on trajectory.
6
5
y(m)
4
3
2
v1
1
0
0
2
4
6
8
10
12
14
x(m )
v1
Dv
v2
a
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v1
Q1. Given
locate these on the
trajectory and form Dv.
v2
6
5
y(m)
4
3
2
1
0
0
2
4
6
8
10
12
14
x(m )
v1
Dv
v2
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Kinematic Equations in Two
Dimensions
v y  voy  a y t
vx  vox  axt
Dx  12 (vox  vx )t
2
1
Dx  voxt  2 a x t
Dy  12 (voy  v y )t
2
1
Dy  voyt  2 a yt
v  v  2a x Dx
v  v  2a y Dy
2
x
2
ox
2
y
2
oy
* many books assume that xo and yo are both zero.
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Velocity in Two Dimensions

vavg

Dr

Dt
• vavg // Dr
• instantaneous “v” is limit of “vavg” as Dt  0
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Acceleration in Two Dimensions


Dv
aavg 
Dt
• aavg // Dv
• instantaneous “a” is limit of “aavg” as Dt  0
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Conventions
• ro = “initial” position at t = 0
• r = “final” position at time t.
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Displacement in Two Dimensions
  
Dr  r  ro
Dr
ro
r
 

r  ro  Dr
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Acceleration ~ v change
• 1 dim. example: car starting, stopping
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Acceleration, Dv, in Two Dimensions
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Ex. Vector Addition
• Add A = 3@60degrees azimuth, plus B =
3@300degrees azimuth.
• Find length of A+B, and its azimuth.
Sketch the situation.
Ex.2:
• 10cm@10degrees + 10cm@30degrees
• Length and azimuth?
Calculate F3
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F1 = 8@60, F2 = 5.5@-45
F1 + F2 + F3 = 0
F3 = -(F1 + F2)
Rx = 8cos60+5.5cos(-45)=7.89
Ry = 8sin60+5.5sin(-45)=3.04
R = 8.46
Angle = tan-1(3.04/7.89) = 21 deg above +x axis
Answer book wants is 180 off this angle!
Addition by Parts (Components)
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