1. Chapter 4 – 2D and 3D Motion Definitions Projectile motion
Uniform circular motion
Relative motion

2. Motion in Two Dimensions
Using + or – signs is not always sufficient to fully describe motion in more than one dimension
Vectors can be used to more fully describe motion
Still interested in displacement, velocity, and acceleration
Will serve as the basis of multiple types of motion in future chapters

3. Position and Displacement
The position of an object is described by its position vector, r
The displacement of the object is defined as the change in its position
Δr = rf - ri

4. General Motion Ideas
In two- or three-dimensional kinematics, everything is the same as in one-dimensional motion except that we must now use full vector notation
Positive and negative signs are no longer sufficient to determine the direction

5. Definitions
Position vector: extends from the origin of a coordinate system to the particle.
Displacement vector: represents a particle’s position change during a certain time interval.

6. Average Velocity
The average velocity is the ratio of the displacement to the time interval for the displacement
The direction of the average velocity is the direction of the displacement vector, Δr

7. Average Velocity
The average velocity between points is independent of the path taken
This is because it is dependent on the displacement, also independent of the path

8. Definitions
Average velocity:

9. Instantaneous Velocity
The instantaneous velocity is the limit of the average velocity as Δt approaches zero
The direction of the instantaneous velocity is along a line that is tangent to the path of the particle’s direction of motion

10. Instantaneous velocity:
The direction of the instantaneous velocity of a particle is always tangent to the particle’s path at the particle’s position

11. Average Acceleration
The average acceleration of a particle as it moves is defined as the change in the instantaneous velocity vector divided by the time interval during which that change occurs.

12. Average Acceleration
As a particle moves, Δv can be found in different ways
The average acceleration is a vector quantity directed along Δv

13. Instantaneous Acceleration
The instantaneous acceleration is the limit of the average acceleration as Δv/Δt approaches zero

14. Instantaneous acceleration:

15. Producing An Acceleration
Various changes in a particle’s motion may produce an acceleration
The magnitude of the velocity vector may change
The direction of the velocity vector may change
Even if the magnitude remains constant
Both may change simultaneously

16. Kinematic Equations for Two-Dimensional Motion
When the two-dimensional motion has a constant acceleration, a series of equations can be developed that describe the motion
These equations will be similar to those of one-dimensional kinematics

17. Kinematic Equations Position vector Velocity vf = vi + at
Since acceleration is constant, we can also find an expression for the velocity as a function of time:
vf = vi + at

18. Kinematic Equations
The velocity vector can be represented by its components
vf is generally not along the direction of either vi or at

19. Kinematic Equations
The position vector can also be expressed as a function of time:
rf = ri + vit + ½ at2
This indicates that the position vector is the sum of three other vectors:
The initial position vector
The displacement resulting from vi t
The displacement resulting from ½ at2

20. Kinematic Equations The vector representation of the position vector
rf is generally not in the same direction as vi or as ai
rf and vf are generally not in the same direction

21. Kinematic Equations, Components
The equations for final velocity and final position are vector equations, therefore they may also be written in component form
This shows that two-dimensional motion at constant acceleration is equivalent to two independent motions
One motion in the x-direction and the other in the y-direction

22. Kinematic Equations, Component Equations
vf = vi + at becomes
vxf = vxi + axt and
vyf = vyi + ayt
rf = ri + vi t + ½ at2 becomes
xf = xi + vxi t + ½ axt2 and
yf = yi + vyi t + ½ ayt2

23. Projectile Motion
An object may move in both the x and y directions simultaneously
The form of two-dimensional motion we will deal with is called projectile motion

24. Assumptions of Projectile Motion
The free-fall acceleration g is constant over the range of motion
And is directed downward
The effect of air friction is negligible
With these assumptions, an object in projectile motion will follow a parabolic path
This path is called the trajectory

25. Verifying the Parabolic Trajectory
Reference frame chosen
y is vertical with upward positive
Acceleration components
ay = -g and ax = 0
Initial velocity components
vxi = vi cos q and vyi = vi sin q

26. II. Projectile motion Horizontal motion: ax=0  vx=v0x
Motion of a particle launched with initial velocity, v0 and free fall acceleration g.
The horizontal and vertical motions are independent from each other.
Horizontal motion:
ax=0  vx=v0x

27. II. Projectile motion
Range (R): horizontal distance traveled by a projectile before returning to launch height.
Vertical motion: ay= -g

28. Trajectory: projectile’s path.
We can find y as a function of x by eliminating time

29. (Maximum for a launch angle of 45º )
Horizontal range: R = x-x0; y-y0=0.
(Maximum for a launch angle of 45º )

30. Projectile Motion – Problem Solving Hints
Select a coordinate system
Resolve the initial velocity into x and y components
Analyze the horizontal motion using constant velocity techniques
Analyze the vertical motion using constant acceleration techniques
Remember that both directions share the same time

31. 122: A third baseman wishes to throw to first base, 127 feet distant
122: A third baseman wishes to throw to first base, 127 feet distant. His best throwing speed is 85 mi/h. (a) If he throws the ball horizontally 3 ft above the ground, how far from first base will it hit the ground? (b) From the same initial height, at what upward angle must the third baseman throw the ball if the first baseman is to catch it 3 ft above the ground? (c) What will be the time of flight in that case? y v0
h=3ft B1 B3 x
xmax
xB1=38.7m

32. 122: A third baseman wishes to throw to first base, 127 feet distant
122: A third baseman wishes to throw to first base, 127 feet distant. His best throwing speed is 85 mi/h. (a) If he throws the ball horizontally 3 ft above the ground, how far from first base will it hit the ground? (b) From the same initial height, at what upward angle must the third baseman throw the ball if the first baseman is to catch it 3 ft above the ground? (c) What will be the time of flight in that case? y v0
h=3ft B1 B3 x
xmax
xB1=38.7m

33. The ball will hit ground at 22.3 m from B3
(a) If he throws the ball horizontally 3 ft above the ground, how far from first base will it hit the ground?
Vertical movement y v0
h=3ft B1 B3 x
xmax
xB1=38.7m
Horizontal movement
The ball will hit ground at 22.3 m from B3

34. (b) From the same initial height, at what upward angle must the third baseman throw the ball if the first baseman is to catch it 3 ft above the ground? (c) What will be the time of flight in that case? y v0 θ
h=3ft x B3 B1
38.7m

35. (b) From the same initial height, at what upward angle must the third baseman throw the ball if the first baseman is to catch it 3 ft above the ground? (c) What will be the time of flight in that case? y v0 θ
h=3ft x B3 B1
38.7m

36. y v0
θ=45º x
x=R=R’?
N7: In Galileo’s Two New Sciences, the author states that “for elevations (angles of projection) which exceed or fall short of 45º by equal amounts, the ranges are equal…” Prove this statement.

37. N7: In Galileo’s Two New Sciences, the author states that “for elevations (angles of projection) which exceed or fall short of 45º by equal amounts, the ranges are equal…” Prove this statement. y v0
θ=45º x
x=R=R’?

38. y v0
θ=45º x
x=R=R’?

39. A ball is tossed from an upper-story window of a building
A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8.00 m/s at an angle of 20.0° below the horizontal. It strikes the ground 3.00 s later. (a) How far horizontally from the base of the building does the ball strike the ground? (b) Find the height from which the ball was thrown. (c) How long does it take the ball to reach a point 10.0 m below the level of launching?

40. A ball is tossed from an upper-story window of a building
A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8.00 m/s at an angle of 20.0° below the horizontal. It strikes the ground 3.00 s later. (a) How far horizontally from the base of the building does the ball strike the ground? (b) Find the height from which the ball was thrown. (c) How long does it take the ball to reach a point 10.0 m below the level of launching?
(a)
(b) Taking y positive downwards,

41. A ball is tossed from an upper-story window of a building
A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8.00 m/s at an angle of 20.0° below the horizontal. It strikes the ground 3.00 s later. (a) How far horizontally from the base of the building does the ball strike the ground? (b) Find the height from which the ball was thrown. (c) How long does it take the ball to reach a point 10.0 m below the level of launching?
(c)

42. Uniform Circular Motion
Uniform circular motion occurs when an object moves in a circular path with a constant speed
An acceleration exists since the direction of the motion is changing
This change in velocity is related to an acceleration
The velocity vector is always tangent to the path of the object

43. Changing Velocity in Uniform Circular Motion
The change in the velocity vector is due to the change in direction
The vector diagram shows Dv = vf - vi

44. Centripetal Acceleration
The acceleration is always perpendicular to the path of the motion
The acceleration always points toward the center of the circle of motion
This acceleration is called the centripetal acceleration

45. Centripetal Acceleration
The magnitude of the centripetal acceleration vector is given by
The direction of the centripetal acceleration vector is always changing, to stay directed toward the center of the circle of motion

46. Period The period, T, is the time required for one complete revolution
The speed of the particle would be the circumference of the circle of motion divided by the period
Therefore, the period is

47. Tangential Acceleration
The magnitude of the velocity could also be changing
In this case, there would be a tangential acceleration

48. Total Acceleration
The tangential acceleration causes the change in the speed of the particle
The radial acceleration comes from a change in the direction of the velocity vector

49. Total Acceleration, equations
The tangential acceleration:
The radial acceleration:
The total acceleration:
Magnitude

50. Total Acceleration, In Terms of Unit Vectors
Define the following unit vectors
r lies along the radius vector
q is tangent to the circle
The total acceleration is

51. Uniform circular motion
Motion around a circle
at constant speed.
Magnitude of velocity and acceleration constant.
Direction varies continuously.
Velocity: tangent to circle in the direction of motion.
Acceleration: centripetal
Period of revolution:

52. 54. A cat rides a merry-go-round while turning with uniform circular motion. At time t1= 2s, the cat’s velocity is:
v1= (3m/s)i+(4m/s)j, measured on an horizontal xy coordinate system. At time t=5s its velocity is: v2= (-3m/s)i+(-4m/s)j. What is (a) the magnitude of the cat’s centripetal acceleration? v2 x v1 y

53. In 3s the velocity is reversed  the cat reaches the opposite
54. A cat rides a merry-go-round while turning with uniform circular motion. At time t1= 2s, the cat’s velocity is:
v1= (3m/s)i+(4m/s)j, measured on an horizontal xy coordinate system. At time t=5s its velocity is: v2= (-3m/s)i+(-4m/s)j. What is (a) the magnitude of the cat’s centripetal acceleration? v2 x v1 y
In 3s the velocity is reversed  the cat reaches the opposite
side of the circle

54. 54. A cat rides a merry-go-round while turning with uniform circular motion. At time t1= 2s, the cat’s velocity is:
v1= (3m/s)i+(4m/s)j, measured on an horizontal xy coordinate system. At time t2=5s its velocity is: v2=(-3m/s)i+(-4m/s)j. What is (b) the cat’s average acceleration during the time interval t2-t1? v2 x v1 y

55. Figure represents the total acceleration of a particle moving clockwise in a circle of radius 2.50 m at a certain of time. At this instant, find (a) the radial acceleration, (b) the speed of the particle, and (c) its tangential acceleration.

56. Figure represents the total acceleration of a particle moving clockwise in a circle of radius 2.50 m at a certain of time. At this instant, find (a) the radial acceleration, (b) the speed of the particle, and (c) its tangential acceleration.
(a)

57. Figure represents the total acceleration of a particle moving clockwise in a circle of radius 2.50 m at a certain of time. At this instant, find (a) the radial acceleration, (b) the speed of the particle, and (c) its tangential acceleration.
(b)

58. A ball swings in a vertical circle at the end of a rope 1. 50 m long
A ball swings in a vertical circle at the end of a rope 1.50 m long. When the ball is 36.9 past the lowest point on its way up, its total acceleration is At that instant, (a) sketch a vector diagram showing the components of its acceleration, (b) determine the magnitude of its radial acceleration, and (c) determine the speed and velocity of the ball.

59. A ball swings in a vertical circle at the end of a rope 1. 50 m long
A ball swings in a vertical circle at the end of a rope 1.50 m long. When the ball is 36.9 past the lowest point on its way up, its total acceleration is At that instant, (a) sketch a vector diagram showing the components of its acceleration, (b) determine the magnitude of its radial acceleration, and (c) determine the speed and velocity of the ball.
(a)

60. A ball swings in a vertical circle at the end of a rope 1. 50 m long
A ball swings in a vertical circle at the end of a rope 1.50 m long. When the ball is 36.9 past the lowest point on its way up, its total acceleration is At that instant, (a) sketch a vector diagram showing the components of its acceleration, (b) determine the magnitude of its radial acceleration, and (c) determine the speed and velocity of the ball.
(b)
(c)

61. Relative Velocity
Two observers moving relative to each other generally do not agree on the outcome of an experiment
For example, observers A and B below see different paths for the ball

62. Relative Velocity Reference frame S is stationary
Reference frame S’ is moving at vo
This also means that S moves at –vo relative to S’
Define time t = 0 as that time when the origins coincide

63. The coordinates of some event in frame S are (x,y,z,t).
Now what are the coordinates of the event (x,y,z,t) in S'?
It's easy to see t' = t - we synchronized the clocks when O‘
passed O. Also, evidently, y' = y and z' = z, from the figure.
We can also see that x = x' +vt. Thus (x,y,z,t) in S
corresponds to (x',y',z', t' ) in S', where
That's how positions transform - these are known as the Galilean transformations.

64. What about velocities ? The velocity in S' in the x' direction
This is just the addition of velocities formula

65. How does acceleration transform?

66. Since v is constant we have
the acceleration is the same in both frames. This again is obvious - the acceleration is the rate of change of velocity, and the velocities of the same particle measured in the two frames differ by a constant factor - the relative velocity of the two frames.

67. Relative motion Particle’s velocity depends on reference frame 1D
Frame moves at constant velocity
Observers on different frames of reference measure the same acceleration
for a moving particle if their relative velocity is constant.

68. 75. A sled moves in the negative x direction at speed vs while a ball of ice is shot from the sled with a velocity v0= v0xi+ v0yj relative to the sled. When the ball lands, its horizontal displacement Δxbg relative to the ground (from its launch position to its landing position) is measured. The figure gives Δxbg as a function of vs. Assume it lands at approximately its launch height. What are the values of (a) v0x and (b) v0y? The ball’s displacement Δxbs relative to the sled can also be measured. Assume that the sled’s velocity is not changed when the ball is shot. What is Δxbs when vs is (c) 5m/s and (d) 15m/s?

69. 75. A sled moves in the negative x direction at speed vs while a ball of ice is shot from the sled with a velocity v0= v0xi+ v0yj relative to the sled. When the ball lands, its horizontal displacement Δxbg relative to the ground (from its launch position to its landing position) is measured. The figure gives Δxbg as a function of vs. Assume it lands at approximately its launch height. What are the values of (a) v0x and (b) v0y? The ball’s displacement Δxbs relative to the sled can also be measured. Assume that the sled’s velocity is not changed when the ball is shot. What is Δxbs when vs is (c) 5m/s and (d) 15m/s?
Δxbg= a+ b vs

70. Displacements relative to ground
75. A sled moves in the negative x direction at speed vs while a ball of ice is shot from the sled with a velocity v0= v0xi+ v0yj relative to the sled. When the ball lands, its horizontal displacement Δxbg relative to the ground (from its launch position to its landing position) is measured. The figure gives Δxbg as a function of vs. Assume it lands at approximately its launch height. What are the values of (a) v0x and (b) v0y? The ball’s displacement Δxbs relative to the sled can also be measured. Assume that the sled’s velocity is not changed when the ball is shot. What is Δxbs when vs is (c) 5m/s and (d) 15m/s?
Δxbg= a+ b vs
Displacements relative to ground a b

71. 75. A sled moves in the negative x direction at speed vs while a ball of ice is shot from the sled with a velocity v0= v0xi+ v0yj relative to the sled. When the ball lands, its horizontal displacement Δxbg relative to the ground (from its launch position to its landing position) is measured. The figure gives Δxbg as a function of vs. Assume it lands at approximately its launch height. What are the values of (a) v0x and (b) v0y?
The ball’s displacement Δxbs relative to the sled can also be measured. Assume that the sled’s velocity is not changed when the ball is shot. What is Δxbs when vs is (c) 5m/s and (d) 15m/s?
Δxbg= a+ b vs a b

72. Displacements relative to the sled
75. A sled moves in the negative x direction at speed vs while a ball of ice is shot from the sled with a velocity v0= v0xi+ v0yj relative to the sled. When the ball lands, its horizontal displacement Δxbg relative to the ground (from its launch position to its landing position) is measured. The figure gives Δxbg as a function of vs. Assume it lands at approximately its launch height. What are the values of (a) v0x and (b) v0y?
The ball’s displacement Δxbs relative to the sled can also be measured. Assume that the sled’s velocity is not changed when the ball is shot. What is Δxbs when vs is (c) 5m/s and (d) 15m/s?
Δxbg= a+ b vs
Displacements relative to the sled
Relative to the sled, the displacement does not depend on the sled’s speed Answer (c)= Answer (d)

73. 120. A hang glider is 7.5 m above ground level with a velocity of 8m/s at an angle of 30º below the horizontal and a constant acceleration of 1m/s2, up. (a) Assume t=0 at the instant just described and write an equation for the elevation y of the hang glider as a function of t, with y=0 at ground level. (b) Use the equation to
determine the value of t when y=0.
(c) Explain why there are two
solutions to part (b). Which one
represents the time it takes the
hang glider to reach ground level?
(d) how far does the hang glider
travel horizontally during the interval between t=0 and the time it reaches the ground? For the same initial position and velocity, what constant acceleration will cause the hang glider to reach ground level with zero velocity? Express your answer in terms of unit vectors. y
30º
v0= 8m/s
h=7.5m x

74. 120. A hang glider is 7.5 m above ground level with a velocity of 8m/s at an angle of 30º below the horizontal and a constant acceleration of 1m/s2, up. (a) Assume t=0 at the instant just described and write an equation for the elevation y of the hang glider as a function of t, with y=0 at ground level. (b) Use the equation to determine the value of t when y=0. (c) Explain why there are two solutions to part B. Which one represents the time it takes the hang glider to reach ground level? (d) how far does the hang glider travel horizontally during the interval between t=0 and the time it reaches the ground? For the same initial position and velocity, what constant acceleration will cause the hang glider to reach ground level with zero velocity? Express your answer in terms of unit vectors. y
30º
v0= 8m/s
h=7.5m x
y(m)
If the ground were not solid, the glider would swoop down, passing through the surface, then back up again, with the two times of passing being t=3s, t=5s.
7.5
t(s) 1 2 3 4 5

75. 120. A hang glider is 7.5 m above ground level with a velocity of 8m/s at an angle of 30º below the horizontal and a constant acceleration of 1m/s2, up. (a) Assume t=0 at the instant just described and write an equation for the elevation y of the hang glider as a function of t, with y=0 at ground level. (b) Use the equation to determine the value of t when y=0. (c) Explain why there are two solutions to part B. Which one represents the time it takes the hang glider to reach ground level? (d) how far does the hang glider travel horizontally during the interval between t=0 and the time it reaches the ground? For the same initial position and velocity, what constant acceleration will cause the hang glider to reach ground level with zero velocity? Express your answer in terms of unit vectors. y
30º
v0= 8m/s
h=7.5m x

76. 40. A ball is to be shot from level ground with certain speed
40. A ball is to be shot from level ground with certain speed. The figure below shows the range R it will have versus the launch angle θ0 at which it can be launched. The choice of θ0 determines the flight time; let tmax represent the maximum flight time. What is the least speed the ball will have during its flight if θ0 is chosen such as that the flight time is 0.5tmax?
R(m)
240
200
100 θ0

77. 40. A ball is to be shot from level ground with certain speed
40. A ball is to be shot from level ground with certain speed. The figure below shows the range R it will have versus the launch angle θ0 at which it can be launched. The choice of θ0 determines the flight time; let tmax represent the maximum flight time. What is the least speed the ball will have during its flight if θ0 is chosen such as that the flight time is 0.5tmax?
R(m)
200
100 θ0

78. 40. A ball is to be shot from level ground with certain speed
40. A ball is to be shot from level ground with certain speed. The figure below shows the range R it will have versus the launch angle θ0 at which it can be launched. The choice of θ0 determines the flight time; let tmax represent the maximum flight time. What is the least speed the ball will have during its flight if θ0 is chosen such as that the flight time is 0.5tmax?
R(m)
200
100 θ0
The lowest speed occurs at the top of the trajectory (half of total time of flight), when the velocity has simply an x-component.