Ch.3 Topics
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x and y parts of motion
adding vectors
properties of vectors
projectile and circular motion
relative motion
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Motion in Two Dimensions
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displacements: x and y parts
thus: x and y velocities
Ex:
30m/s North + 40m/s East =  50m/s
vx + vy = v
component set = vector
2
Two Dimensional Motion
(constant acceleration)
x  xo  voxt  a x t
1
2
2
y  yo  voyt  a yt
1
2
2
vx  vox  axt
v y  voy  a y t
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Vector Math
• Two Methods:
• geometrical (graphical) method
• algebraic (analytical) method
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Graphical, Tail-to-Head
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Order Independent (Commutative)
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Subtraction, head-to-head
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Example Subtraction: Dv.
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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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Addition by Parts (Components)
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Vector Components
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Quadrants of x,y-Plane
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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:
A: 180°
B: 60°
C: > 90°
Note: All angles measured from east.
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Unit Vectors, i, j, k
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
A  Axiˆ  Ay ˆj  Ax kˆ
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Point-Style Vector Notation
iˆ  (1,0,0)
Example:
ˆj  (0,1,0)
kˆ  (0,0,1)
A  3iˆ  4 ˆj
 3(1,0,0)  4(0,1,0)
 (3,0,0)  (0,4,0)
 (3,4,0)
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Components Example:
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  Ax  Bx  C x  
Ry  Ay  By  C y  
R R R
2
x
2
y
 Ry 
  tan    180(quads.II , III )
 Rx 
1
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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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General Properties of Vectors
• size and direction define a vector
• location independent
• change size and/or direction when
multiplied by a constant
• 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
• begins when projecting force ends
• ends when object hits something
• gravity alone acts on object
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Projectile Motion
ax = 0 and ay = -9.8 m/s/s
vo
Dy
Dx = “Range”
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Horizontal V Constant
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Range vs. Angle
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Circular Motion
• centripetal, tangential components
• general acceleration vector
• case of uniform circular motion
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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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Example:
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Ex. A Plane has an air speed vpa = 75m/s. The wind has a
velocity with respect to the ground of vag = 8 m/s @ 330°.
The plane’s path is due North relative to ground. a) Draw
a vector diagram showing the relationship between the air
speed and the ground speed. b) Find the ground speed
and the compass heading of the plane.
(similar
situation)
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


v pg  v pa  vag
 (75 cos  ,75 sin  )  (8 cos 330,8 sin 330)  (0, v pg )
75 cos  8 cos 330  0
   95.3
75 sin 95.3  8 sin 330  70.7 m/s  v pg
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Summary
• Vector Components & Addition using trig
• Graphical Vector Addition & Azimuths
• Example planar motions: Projectile
Motion, Circular Motion
• Relative Motion
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Example 1: Calculate Range (R)
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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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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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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Ex. If v1(0.00s) = 12m/s, +60° and v2(0.65s) = 7.223 @
+33.83°, find aave.
v1  (12.0 cos(60),12.0 sin( 60))  (6.00,10.39)
v 2  (7.223 cos(33.83),7.223 sin( 33.83))  (6.00,4.02)
Dv  v 2  v1  (6.00,4.02)  (6.00,10.39)  (0,6.37)m / s
Dv (0,6.37)m / s
a

 (0,9.8)m / s / s
Dt
0.65  0.00s
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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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Q2. If v3(1.15s) = 6.06m/s, -8.32° and v4(1.60s) = 7.997, -41.389°, write the
coordinate-forms of these vectors and calculate the average acceleration.
v3  (6.06 cos( 8.32),6.06 sin( 8.32))  (6.00,0.8777)
v 4  (7.997 cos( 41.39),7.997 sin( 41.39))  (6.00,5.2877)
Dv  v 2  v1  (6.00,5.2877)  (6.00,0.8777)  (0,4.41)m / s
Dv (0,4.41)m / s
a

 (0,9.8)m / s / s
Dt
1.60  1.15s
v3
v4
Dv
a
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