1. Motion in Two Dimensions
Chapter 3
Motion in Two Dimensions

3. 3.1 Position and Displacement
The position of an object is described by its position vector,
The displacement of the object is defined as the change in its position
Fig 3.1

4. 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,
Fig 3.2

5. Average Velocity, cont
The average velocity between points is independent of the path taken
This is because it is dependent on the displacement, which is also independent of the path
If a particle starts its motion at some point and returns to this point via any path, its average velocity is zero for this trip since its displacement is zero

6. Instantaneous Velocity
The instantaneous velocity is the limit of the average velocity as ∆t approaches zero

7. Instantaneous Velocity, cont
The direction of the instantaneous velocity vector at any point in a particle’s path is along a line tangent to the path at that point and in the direction of motion
The magnitude of the instantaneous velocity vector is the speed

8. 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.

9. Average Acceleration, cont
As a particle moves, can be found in different ways
The average acceleration is a vector quantity directed along
Fig 3.3

10. Instantaneous Acceleration
The instantaneous acceleration is the limit of the average acceleration as approaches zero

11. 3.2 Producing An Acceleration
Various changes in a particle’s motion may produce an acceleration
The magnitude of the velocity
The direction of the velocity
Even if the magnitude remains constant
Both change simultaneously

12. 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

13. Kinematic Equations, 2 Position vector Velocity
Since acceleration is constant, we can also find an expression for the velocity as a function of time:

14. Kinematic Equations, 3
The velocity vector can be represented by its components
is generally not along the direction of either or
Fig 3.4(a)

15. Kinematic Equations, 4
The position vector can also be expressed as a function of time:
This indicates that the position vector is the sum of three other vectors:
The initial position vector
The displacement resulting from

16. Kinematic Equations, 5 is generally not in the same direction as or as
and are generally not in the same direction
Fig 3.4(b)

17. 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

18. Active Figure
3.4
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20. Solution
Conceptualize Establish the mental representation and thinking about what the particle is doing.
Categorize Consider that because the acceleration is only in the x direction, the moving particle can be modeled as one under constant acceleration in the x direction and one under constant velocity in the y direction.

21. Analyze Identify vxi = 20 m/s and ax = 4. 0 m/s2
Analyze Identify vxi = 20 m/s and ax = 4.0 m/s2. The equations of kinematics for the x direction,

22. Solution At t = 5.0 s, the velocity expression from part A gives

23. 拋 射 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. 假 設 Assumptions of Projectile Motion
The free-fall acceleration 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
This is the simplification model that will be used throughout this chapter

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 and vyi = vi sin

26. Projectile Motion – Velocity Equations
The velocity components for the projectile at any time t are:
vxf = vxi = vi cosi = constant
vyf = vyi – gt = vi sini – gt

28. Projectile Motion – Position
Displacements
Combining the equations gives:
This is in the form of y = ax – bx2 which is the standard form of a parabola

29. Analyzing Projectile Motion
Consider the motion as the superposition of the motions in the x- and y-directions
The x-direction has constant velocity
ax = 0
The y-direction is free fall
ay = -g
The actual position at any time is given by:

30. Projectile Motion Vectors
Fig 3.6

31. Projectile Motion Diagram
Fig 3.5

32. Active Figure
3.5
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33. Projectile Motion – Implications
The y-component of the velocity is zero at the maximum height of the trajectory
The accleration stays the same throughout the trajectory

34. Horizontal Range and Maximum Height of a Projectile
When analyzing projectile motion, two characteristics are of special interest
The range, R, is the horizontal distance of the projectile
The maximum height the projectile reaches is h
Fig 3.7

35. Height of a Projectile, equation
The maximum height of the projectile can be found in terms of the initial velocity vector:

36. Range of a Projectile, equation
The range of a projectile can be expressed in terms of the initial velocity vector:
This is valid only for symmetric trajectory

37. More About the Range of a Projectile

38. Range of a Projectile, final
The maximum range occurs at qi = 45o
Complementary angles will produce the same range
The maximum height will be different for the two angles
The times of the flight will be different for the two angles

39. Active Figure
3.8
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40. Projectile Motion – Problem Solving Hints
Conceptualize
Establish the mental representation of the projectile moving along its trajectory
Categorize
Confirm that the problem involves the particle in free fall (in the y-direction) and air resistance can be neglected
Select a coordinate system

41. Problem Solving Hints, cont.
Analyze
Resolve the initial velocity into x and y components
Remember to treat the horizontal motion independently from the vertical motion
Treat the horizontal motion using constant velocity techniques
Analyze the vertical motion using constant acceleration techniques
Remember that both directions share the same time
Finalize
Check your results

42. Non-Symmetric Projectile Motion
Follow the general rules for projectile motion
Break the y-direction into parts
up and down or
symmetrical back to initial height and then the rest of the height
May be non-symmetric in other ways

43. A stone is thrown from the top of a building at an angle of 30
A stone is thrown from the top of a building at an angle of 30.0° to the horizontal and with an initial speed of 20.0 m/s, as in Figure 3.9.
Fig 3.9

44. To find t, we use the vertical motion, in which we model the stone as a particle under constant acceleration. We use Equation 3.13 with yf = 45.0 m and vyi = 10.0 m/s (we have chosen the top of the building as the origin):
t = 4.22 s

45. Solution The y component of the velocity just before the stone strikes the ground can be obtained using Equation 3.11, with t = 4.22 s:

46. An Alaskan rescue plane drops a package of emergency rations to a stranded party of explorers, as shown in the pictorial representation in Figure If the plane is traveling horizontally at 40.0 m/s at a height of 100 m above the ground, where does the package strike the ground relative to the point at which it is released?
Fig 3.10

47. Solution Ignore air resistance Model this problem as a particle in two-dimensional free-fall,
a particle under constant velocity in the x direction and
a particle under constant acceleration in the y direction
Define the initial position xi = 0 right under the plane at the instant the package is released.

48. If we know t, the time at which the package strikes the ground, we can determine xf. To find t, we turn to the equations for the vertical motion of the package.
At the instant the package hits the ground, its coordinate is 100 m.
The initial component of velocity vyi of the package is zero.
From Equation 3.13,
xf = (40.0 m/s)(4.52 s) = 181 m

49. An athlete throws a javelin a distance of 80
An athlete throws a javelin a distance of 80.0 m at the Olympics held at the equator, where g = 9.78 m/s2. Four years later the Olympics are held at the North Pole, where g = 9.83 m/s2. Assuming that the thrower provides the javelin with exactly the same initial velocity as she did at the equator, how far does the javelin travel at the North Pole?

50. = 79.6 m

51. 3.4 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

52. Fig 3.11

53. 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
Centripetal means center-seeking

54. Changing Velocity in Uniform Circular Motion
The change in the velocity vector is due to the change in direction
The vector diagram shows
Fig 3.11

56. Centripetal Acceleration, cont
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

57. Period
The period, T, is the time interval 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

58. What is the centripetal acceleration of the Earth as it moves in its orbit around the Sun?
Solution We model the Earth as a particle and approximate the Earth’s orbit as circular Although we don’t know the orbital speed of the Earth, with the help of Equation 3.18 we can recast Equation 3.17 in terms of the period of the Earth’s orbit, which we know is one year:

60. 3.5 Tangential Acceleration
The magnitude of the velocity could also be changing
As well as the direction
In this case, there would be a tangential acceleration

61. Total Acceleration
Fig 3.12
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

62. Total Acceleration, equations
The tangential acceleration:
The radial acceleration:
The total acceleration:
Magnitude
The direction of the acceleration is the same or opposite that of the velocity

63. Active Figure
3.12
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64. 3.6 Relative Velocity
Two observers moving relative to each other generally do not agree on the outcome of an experiment
For example, the observer on the side of the road observes a different speed for the red car than does the observer in the blue car
Fig 3.13

65. Relative Velocity, generalized
Reference frame S is stationary
Reference frame S’ is moving
Define time t = 0 as that time when the origins coincide
Fig 3.14

66. Relative Velocity, equations
The positions as seen from the two reference frames are related through the velocity
The derivative of the position equation will give the velocity equation
This can also be expressed in terms of the observer O’

67. A boat heading due north crosses a wide river with a speed of 10
A boat heading due north crosses a wide river with a speed of 10.0 km/h relative to the water. The river has a current such that the water moves with uniform speed of 5.00 km/h due east relative to the ground.

71. 3.7 Acceleration of Autos
The lateral acceleration 橫向加速度is the maximum possible centripetal acceleration the car can exhibit without rolling over in a turn
The lateral acceleration depends on the height of the center of mass of the vehicle and the side-to-side distance between the wheels

72. Beamon於1968在Mexico創下8. 90 m的跳遠紀錄，由照片得知，他起跳時重心高度約為1. 0 m，在最高點處重心高度為1
Beamon於1968在Mexico創下8.90 m的跳遠紀錄，由照片得知，他起跳時重心高度約為1.0 m，在最高點處重心高度為1.90 m，落地時重心高度0.15 m。由這些數據，求(a)他在空中停留時間。(b)起跳時速度之水平分量及垂直分量。(c)起跳角度。

73. An enemy ship is on the east side of a mountain island, as shown in Figure P3.61. The enemy ship has maneuvered to within 2500 m of the 1800-m-high mountain peak and can shoot projectiles with an initial speed of 250 m/s. If the western shoreline is horizontally 300 m from the peak, what are the distances from the western shore at which a ship can be safe from the bombardment of the enemy ship?

74. Problem 1, 3, 10, 13, 15, 19, 20, 27, 31, 36, 38, 40, 41, 48, 51, 55, 60