1. College Physics, 7th Edition
Lecture Outline
Chapter 3
College Physics, 7th Edition
Wilson / Buffa / Lou
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2. Chapter 3 Motion in Two Dimensions
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3. Units of Chapter 3 Components of Motion
Vector Addition and Subtraction
Projectile Motion
Relative Velocity
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4. 3.1 Components of Motion
An object in motion on a plane can be located using two numbers—the x and y coordinates of its position. Similarly, its velocity can be described using components along the x- and y-axes.
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5. 3.1 Components of Motion The velocity components are:
The magnitude of the velocity vector is:
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6. 3.1 Components of Motion
The components of the displacement are then given by:
Note that the x- and y-components are calculated separately.
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7. 3.1 Components of Motion The equations of motion are:
When solving two-dimensional kinematics problems, each component is treated separately. The time is common to both.
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8. 3.1 Components of Motion
If the acceleration is not parallel to the velocity, the object will move in a curve:
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9. 3.2 Vector Addition and Subtraction
Geometric methods of vector addition
Triangle method:
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10. 3.2 Vector Addition and Subtraction
The negative of a vector has the same magnitude but is opposite in direction to the original vector. Adding a negative vector is the same as subtracting a vector.
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11. 3.2 Vector Addition and Subtraction
Vector Components and the Analytical Component Method
If you know A and B, here is how to find C:
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12. 3.2 Vector Addition and Subtraction
The components of C are given by:
Equivalently,
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13. 3.2 Vector Addition and Subtraction
Vectors can also be written using unit vectors (i, j, k):
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14. 3.2 Vector Addition and Subtraction
Vectors can be resolved into components and the components added separately; then recombine to find the resultant.
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15. Non-Collinear Vectors
When 2 vectors are perpendicular, you must use the Pythagorean theorem.
A man walks 95 km, East then 55 km, north. Calculate his RESULTANT DISPLACEMENT.
Finish
The hypotenuse in Physics is called the RESULTANT.
55 km, N
Vertical Component
Horizontal Component
Start
95 km,E
The LEGS of the triangle are called the COMPONENTS

16. What about the direction?
In the previous example, DISPLACEMENT was asked for and since it is a VECTOR we should include a DIRECTION on our final answer. N
W of N
E of N
N of E
N of W W E
N of E
S of W
S of E
NOTE: When drawing a right triangle that conveys some type of motion, you MUST draw your components HEAD TO TOE.
W of S
E of S S

17. What about the VALUE of the angle?
Just putting North of East on the answer is NOT specific enough for the direction. We MUST find the VALUE of the angle.
To find the value of the angle we use a Trig function called TANGENT.
109.8 km
55 km, N q
N of E
95 km,E
So the COMPLETE final answer is : km, 30 degrees North of East

18. What if you are missing a component?
Suppose a person walked 65 m, 25 degrees East of North. What were his horizontal and vertical components?
The goal: ALWAYS MAKE A RIGHT TRIANGLE!
To solve for components, we often use the trig functions since and cosine.
H.C. = ?
V.C = ? 25
65 m

19. Example 23 m, E - = 12 m, W - = 14 m, N 6 m, S 20 m, N 35 m, E R
A bear, searching for food wanders 35 meters east then 20 meters north. Frustrated, he wanders another 12 meters west then 6 meters south. Calculate the bear's displacement.
23 m, E - =
12 m, W - =
14 m, N
6 m, S
20 m, N
35 m, E R
14 m, N q
23 m, E
The Final Answer: m, 31.3 degrees NORTH of EAST

20. Example
A boat moves with a velocity of 15 m/s, N in a river which flows with a velocity of 8.0 m/s, west. Calculate the boat's resultant velocity with respect to due north.
8.0 m/s, W
15 m/s, N Rv q
The Final Answer : degrees West of North

21. Example
A plane moves with a velocity of 63.5 m/s at 32 degrees South of East. Calculate the plane's horizontal and vertical velocity components.
H.C. =? 32
V.C. = ?
63.5 m/s

22. Example
A storm system moves 5000 km due east, then shifts course at 40 degrees North of East for 1500 km. Calculate the storm's resultant displacement.
1500 km
V.C. 40
5000 km, E
H.C.
5000 km km = km R
964.2 km q
The Final Answer: degrees, North of East km

23. 3.3 Projectile Motion
A projectile launched in an arbitrary direction may have initial velocity components in both the horizontal and vertical directions, but its acceleration is still downward.
Projectile - Any object which projected by some means and continues to move due to its own inertia (mass). Its parabolic in path of motion.

24. 3.3 Projectile Motion
An object projected horizontally has an initial velocity in the horizontal direction, and acceleration (due to gravity) in the vertical direction. The time it takes to reach the ground is the same as if it were simply dropped.
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25. 3.3 Projectile Motion Horizontally launched vs dropped.
See how the x component is different from the y?
This is ALWAYS true for projectile motion.
Solve the x separate from the y.
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26. 3.3 Projectile Motion
The vertical motion is the same as if the object were thrown straight up or down with the same initial y velocity, and the horizontal velocity is constant.
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27. 3.3 Projectile Motion
The range of a projectile is maximum (if there is no air resistance) for a launch angle of 45°.
What else do you notice?
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28. 3.3 Projectile Motion
With air resistance, the range is shortened, and the maximum range occurs at an angle a little less than 45°. Its still projectile motion but now the relationship is not parabolic. Complex algebra or calculus.
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29. Horizontally Launched Projectiles
Projectiles which have NO upward trajectory and NO initial VERTICAL velocity.

30. Horizontally Launched Projectiles
To analyze a projectile in 2 dimensions we need 2 equations. One for the “x” direction and one for the “y” direction. And for this we use kinematic #2.
Remember, the velocity is CONSTANT horizontally, so that means the acceleration is ZERO. Range is only solved for in the x direction. Constant velocity -> Essentially DIRT! d=rt.
Remember that since the projectile is launched horizontally, the INITIAL VERTICAL VELOCITY is equal to ZERO. Time is always solved for in the y direction ONLY.

31. Horizontally Launched Projectiles
Example: A plane traveling with a horizontal velocity of 100 m/s is 500 m above the ground. At some point the pilot decides to drop some supplies to designated target below. (a) How long is the drop in the air? (b) How far away from point where it was launched will it land?
What do I know?
What I want to know?
vox=100 m/s
t = ?
y = 500 m
x = ?
voy= 0 m/s
g = -9.8 m/s/s
1010 m
10.1 seconds

32. Vertically Launched Projectiles
NO Vertical Velocity at the top of the trajectory.
Vertical Velocity decreases on the way upward
Horizontal Velocity is constant
Vertical Velocity increases on the way down
Component
Magnitude
Direction
Horizontal
Constant
Vertical
Decreases up, top, Increases down
Changes

33. Vertically Launched Projectiles
Since the projectile was launched at a angle, the velocity MUST be broken into components!!! vo
voy q
vox

34. Vertically Launched Projectiles
There are several things you must consider when doing these types of projectiles besides using components. If it begins and ends at ground level, the “y” displacement is ZERO: y = 0

35. Vertically Launched Projectiles
You will still use kinematic #2, but YOU MUST use COMPONENTS in the equation. vo
voy q
vox

36. Example
A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees.
(a) How long is the ball in the air?
(b) How far away does it land?
(c) How high does it travel?
vo=20.0 m/s
q = 53

37. Example
What I know
What I want to know
vox=12.04 m/s
t = ?
voy=15.97 m/s
x = ?
y = 0
ymax=?
g = m/s/s
A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees.
(a) How long is the ball in the air?
3.26 s

38. Example
A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees.
(b) How far away does it land?
What I know
What I want to know
vox=12.04 m/s
t = 3.26 s
voy=15.97 m/s
x = ?
y = 0
ymax=?
g = m/s/s
39.24 m

39. Example What I know What I want to know t = 3.26 s x = 39.24 m y = 0
vox=12.04 m/s
t = 3.26 s
voy=15.97 m/s
x = m
y = 0
ymax=?
g = m/s/s
A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees.
(c) How high does it travel?
CUT YOUR TIME IN HALF!
13.01 m

40. 3.4 Relative Velocity
Velocity may be measured in any inertial reference frame. At top, the velocities are measured relative to the ground; at bottom they are measured relative to the white car.
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41. 3.4 Relative Velocity
In two dimensions, the components of the velocity, and therefore the angle it makes with a coordinate axis, will change depending on the point of view.
What if the runner’s velocity is vbr?
What if the walker’s velocity is vbs?
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42. Review of Chapter 3
Two-dimensional motion is analyzed by considering each component separately. Time is the only common factor and only solved for in y.
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43. Review of Chapter 3
Vector components:
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44. Review of Chapter 3 Range is the maximum horizontal distance traveled.
Relative velocity is expressed relative to a particular reference frame.
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