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

2. Goals for Chapter 3
To study position, velocity, and acceleration vectors
To apply position, velocity, and acceleration insights to projectile motion
To extend our linear investigations to uniform and non-uniform circular motion
To investigate relative velocity

3. Position relative to the origin—Figure 3.1
An overall position relative to the origin can have components in x, y, and z dimensions.
The path for a particle is generally a curve.

4. Average velocity—Figure 3.2
The average velocity between two points will have the same direction as the displacement.

5. Average velocity vector
During a time interval t the particle moves from P1 , where its position vector is r1, to P2, where its position vector is r2.
vav =
(x2 – x1)i (y2 – y1)j (z2 – z1)k ∆t +
One dimension

6. Instantaneous velocity
(2D)

7. v = dr dt

8. Example 3.1: calculating average and instantaneous velocity

10. Test your understanding
In which of these situations would the average velocity vector vav over an interval be equal to the instantaneous velocity v at the end of interval:
A body moving along a curved path at constant speed
A body moving along a curved path and speeding up
A body moving along a straight line at constant speed
A body moving along a straight line and speeding up.

11. example v = dr/dt = 2bt i + 3ct2 j If r = bt2i + ct3j t =
Where b and c are positive constants, when does the velocity vector make an angle of 45.0o with the x- and y-axes?
v = dr/dt = 2bt i + 3ct2 j
= 1 2b 3c
t =

12. The acceleration vector—Figure 3.6
The acceleration vector can result in a change in either the magnitude OR the direction of the velocity.
Consider the race car in Figure 3.6.

13. aav =
(v2x – v1x)i (v2y – y1y)j (v2z – v1z)k ∆t +
One dimension

14. Instantaneous Acceleration
a = i j k
dvx dt
dvy
dvz
a = i j k
d2x dt
d2y
d2z
tanθ = ay ax
|a| = √ax2 + ay2 + az2

15. All about acceleration
Equal to time rate of change of velocity
≠ 0 if velocity changes in magnitude or direction.
It does not necessarily have same direction as velocity vector
Acceleration vector lies on concave side of curved path.

16. Example 3.2
Given:
Find the components of the average acceleration in the interval from t = 0.0 s to t = 2.0 s.
Find the instantaneous acceleration at t = 2.0 s, its magnitude and its direction.

17. Parallel and perpendicular components of acceleration
The acceleration vector a for a particle can describe changes in the particle’s speed, its direction of motion, or both.
The component of acceleration parallel to a particle’s path (parallel to the velocity) tells us about changes in the particle's speed.
The acceleration component perpendicular to the path (perpendicular to the velocity) tells us about changes in the particle’s direction of motion.

19. Consider the different vectors—Figure 3.12
Notice the acceleration vector change as velocity decreases, remains the same, or increases.

20. Parallel and perpendicular components of acceleration
Refer to the worked Example 3.3.
Figure 3.13 illustrates the calculation in the example.

21. Example 3.3
Given:
Find the parallel and perpendicular components, relative to velocity, of the instantaneous acceleration at t = 2.0 s

22. A skier moves along a ski-jump ramp as shown in the figure
A skier moves along a ski-jump ramp as shown in the figure. The ramp is straight from point A to point C onward. The skier picks up speed as she moves downhill from point A to point E. Draw the direction of the acceleration vector at points B, D, E and F.
Example 3.4

23. Test your understanding 3.2
A sled travels over the crest of a snow-covered hill. The sled slows down as it climbs up one side of the hill and gains speed as it descends on the other side. Which of the vectors (1 though 9) in the figure correctly shows the direction of the sled’s acceleration at the crest? Choice 9 is that the acceleration is zero.)

24. 3.3 Projectile motion
A projectile is any body given an initial velocity that then follows a path determined by the effects of gravity and air resistance.
Begin neglecting resistance.

25. X and Y motion are separable—Figure 3.16
The red ball is dropped, and the yellow ball is fired horizontally as it is dropped.
The strobe marks equal time intervals.

26. Projectiles move in TWO dimensions
Horizontal and Vertical
The path of a projectile is called a trajectory

27. Horizontal Component Velocity is constant: vx0 = vi0cosθ
Acceleration: ax = 0
Displacement = vx0∙t
In other words, the horizontal velocity is CONSTANT. BUT WHY?
Gravity DOES NOT work horizontally to increase or decrease the velocity. θ

28. Vertical Component Acceleration: ay = -g
Velocity: vy = vy0 – gt = vi0sinθ - gt
Displacement: y = y0 + vy0∙t - ½ gt2 θ

29. If air resistance is negligible, the trajectory of a projectile is a combination of horizontal motion with constant velocity and vertical motion with constant acceleration. The path of the projectile is parabolic.

30. The effects of wind resistance—Figure 3.20
Cumulative effects can be large.
Peak heights and distance fall.
Trajectories cease to be parabolic.

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 = -10 m/s/s
t = 10 seconds
x = 1000 m

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

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

34. 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. You will still use kinematic equations, 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 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

38. A special case…
What if the projectile was launched from the ground at an angle and did not land at the same level height from where it started? In other words, what if you have a situation where the “y-displacement” DOES NOT equal zero?
Assuming it is shot from the ground. We see we have one squared term variable, one regular term variable, and a constant number with no variable. What is this?
A QUADRATIC EQUATION!

39. A special case example
An object is thrown from the top of a 100m high cliff at an initial speed of 20 m/s. How long does it take to reach the ground?
Well, here is what you have to realize. As it goes up its speed decreases, reaches zero, the increasing back to the ground. When the ball reaches the cliff face it is now traveling at 20 m/s again , but in the opposite direction. DOWNWARD.
Thus the speed is considered
to be -20 m/s.

40. Example What I know What I want to know voy=-20.0 m/s t = ? y= 0 m
g = m/s/s
GOTO F2 ( which is the Algebra screen) CHOOSE SOLVE.
Type in the equation.
In this example our equation and set the it equal to zero. Then add a comma after the equation and tell it that you want it to solve for, in this case "t".

41. Example
It will respond by showing you exactly what you typed and the TWO ROOTS!
Since we are solving for time we choose the positive root as
TIME cannot be negative. Thus it took 2.92 seconds for the ball to hit the ground.

42. If at t = 0, x0 = y0 = 0 then we can find the x, y coordinates and the x, y velocity at time t:
The magnitude of the position:
The magnitude of the velocity:
The direction of the velocity:

43. Example 3.5
Let’s consider again the skier in example 3.4. What is her acceleration at points G, H, and I after she flies off the ramp? Neglect air resistance.
The acceleration at points G, H, I are the same:
ax = 0; ay = -9.8 m/s2

44. Example 3.6:a body projected horizontally
A motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal, with magnitude 9.0 m/s. find the motorcycle’s position, distance and velocity from the edge of the cliff, and velocity at 0.50 s.

45. Example: 3.7 – height and range of a projectile I – a batted baseball

46. Example 3.8 – height and range of a projectile II: maximum height, maximum range
For a projectile launched with speed v0 at initial angle α0 (between 0o and 90o), derive general expressions for the maximum height h and horizontal range R. For a given v0, what value of α0 gives maximum height? What value gives maximum horizontal range?

47. Maximum height: vy = 0 0 = v0sinα0 - gt t = v0sinα0 / g
h = (v0sinα0)(v0sinα0 /g) – ½ g(v0sinα0 /g)2
h = (v0sinα0)2 / 2g
When α0 is 90o, sin α0 is 1, h has maximum value.

48. Maximum range: t = 2v0sinα0 / g R = (v0cosα0)(2v0sinα0/g)
The time for the ball to return to the ground is twice the time for the ball to reach the maximum height.
t = 2v0sinα0 / g
R = (v0cosα0)(2v0sinα0/g)
R = v02sin2α0/g
When 2α0 is 90o, or when α0 is 45o, R has maximum value.

49. You toss a ball from your window 8. 0 m above the ground
You toss a ball from your window 8.0 m above the ground. When the ball leaves your hand, it is moving at 10.0 m/s at an angle of 20o below the horizontal. How far horizontally from your window will the ball hit the ground? Ignore air resistance.

50. A monkey escapes from the zoo and climbs a tree
A monkey escapes from the zoo and climbs a tree. After failing to entice the monkey down, the zookeeper fires a tranquilizer dart directly at the monkey. The clever monkey lets go at the same instant the dart leaves the gun barrel, intending to land on the ground and escape. Show that the dart always hits the monkey, regardless of the dart’s muzzle velocity (provided that it gets to the monkey before he hits the ground).

51. Fire at the monkey with fast speed

52. Fire at the monkey with slow speed

53. tanα0 = h / d

54. If dart hits the monkey, the position of the dart and the position of the monkey must be the same at time t.
For the dart:
For the monkey:
xdart = d = (v0cosα0)t
ydart = (v0sinα0)t – ½ gt2
xmonkey = d
ymonkey = h – ½ gt2
For the dart:
xdart = d = (v0cosα0)t t = d / (v0cosα0)
ydart = (v0sinα0)(d /(v0cosα0)) – ½ gt2
ydart = (dtanα0) – ½ gt2
Since tanα0 = h / d
ydart = h – ½ gt2 = ymonkey

55. Check your understanding 3.3
In example 3.10, suppose the tranquilizer dart has a relatively low muzzle velocity so that the dart reaches a maximum height at a point before striking the monkey. When the dart is at point P, will the monkey be
At point A (higher than P)
At point B (at the same height as P)
At point C (lower than P)?

56. 3.4 motion in a circle
When a particle moves along a curved path, the direction of its velocity changes. This means that the particle must have a component of acceleration perpendicular to the path, even if its speed is constant.

57. Finding motion information—Figure 3.28
Velocity change, average acceleration, and instantaneous acceleration may be found.

58. Uniform circular motion
When a particle moves in a circle with constant speed, the motion is called uniform circular motion.
Example:
a car rounding a curve with constant radium at constant speed
A satellite moving in a circular orbit,
An ice skater skating in a circle with constant speed

59. There is no component of acceleration parallel (tangent) to the path; the acceleration vector is perpendicular (normal) to the path and hence directed inward toward the center of the circular path.

60. The subscript “rad” is a reminder that the direction of the instantaneous acceleration at each point is always long a radius of the circle, toward its center.

61. Uniform circular motion vs. projectile motion
Same: the magnitude of acceleration is the same at all times.
Difference: the direction of acceleration in uniform circular motion changes continuously so that it always points toward the center of the circle. However, the direction of the acceleration in projectile always points down;

62. Example 3.11

63. Example 3.12

64. Non uniform circular motion
If the speed of the particle varies while it is moving in a circular path, the motion is non uniform circular motion.
The acceleration in a non uniform circular motion has two components: a (tangent) and a (radial).

65. a: Radial component:
arad is always perpendicular to the instantaneous velocity and directed toward the center of the circle. But since v is changing, arad is not constant. arad is greatest at the point in the circle where the speed is greatest.

66. a: Tangent component
The component of acceleration that is parallel to the instantaneous velocity is the atan because it is tangent to the circle.
atan is equal to the rate of change of speed.
In uniform circular motion, there is no change in speed, atan = 0

67. caution The two quantities are not the same.
The first, equal to the tangential acceleration, is the
rate of change of speed; it is zero whenever a particle moves with constant speed, even when its direction of motion changes.
The second, is the magnitude of the vector
acceleration; it is zero only when the particle’s acceleration vector is zero – motion in a straight line with constant speed.
In uniform circular motion
In non uniform circular motion

68. Test your understanding 3.4
Suppose that the particle experiences 4 times the acceleration at the bottom of the loop as it does at the top of loop. Compared to its speed at the top the loop, is its speed the bottom of the loop:
√2 times as great
2 times as great
2√2 times as great
4 times as great
16 times as great.

69. 3.5 relative velocity
The velocity seen by a particular observer is called the velocity relative to that observe, or simply relative velocity. What is the planes’ speed?
Relative to each other, the planes are almost at rest
Relative to the observers on the ground, the planes are flying at a great speeds.

70. Relative velocity in one dimension

71. Relative velocity in two or three dimensions
The acceleration of a (plane to earth) is identical to a (plane to air) because the v (air to earth) is assumed to be constant.

73. R: 150 km/h
150 km/h