1. Kinematics in two and three dimensions
Physics 221

2. Learning Goals for Chapter 3
Looking forward at …
how to use vectors to represent the position and velocity of a particle in two or three dimensions.
how to find the vector acceleration of a particle, and how to interpret the components of acceleration parallel to and perpendicular to a particle’s path.
how to solve problems that involve the curved path followed by a projectile.
how to analyze motion in a circular path, with either constant speed or varying speed.
how to relate the velocities of a moving body as seen from two different frames of reference.

3. Position vector
The position vector from the origin to point P has components x, y, and z.

4. Velocity
We define the average velocity as the displacement divided by the time interval:
Instantaneous velocity (a.k.a. “velocity”) is the instantaneous rate of change of position with time:

5. Average velocity
The average velocity between two points is the displacement divided by the time interval between the two points, and it has the same direction as the displacement.

6. Instantaneous velocity
The instantaneous velocity is the instantaneous rate of change of position vector with respect to time.
The components of the instantaneous velocity are vx = dx/dt, vy = dy/dt, and vz = dz/dt.
The instantaneous velocity of a particle is always tangent to its path.

7. Position and Velocity

8. Position-Time vs x-y Graphs
s-t Graph:
Slope = velocity
x-y Graph:
Slope = local trajectory

9. Acceleration
Acceleration describes how the velocity changes.

10. Acceleration
We define the average acceleration as the change in velocity divided by the time interval:
Instantaneous acceleration (a.k.a. “acceleration”) is the instantaneous rate of change of velocity with time:

11. Average acceleration
The change in velocity between two points is determined by vector subtraction.

12. Instantaneous acceleration
The velocity vector is always tangent to the particle’s path, but the instantaneous acceleration vector does not have to be tangent to the path.
If the path is curved, the acceleration points toward the concave side of the path.

13. Instantaneous acceleration

14. Acceleration in x-y Graphs

15. Parallel and perpendicular components of acceleration
Velocity and acceleration vectors for a particle moving through a point P on a curved path with constant speed
Velocity and acceleration vectors for a particle moving through a point P on a curved path with increasing speed

16. The ”dot” notation
Newton had a particular notation for time derivatives that is still used today. To indicate the time derivative of a quantity, he placed a “dot” over the symbol for that quantity
For instance, 𝑣= 𝑑𝑥 𝑑𝑡 = 𝑥 (which is read “x-dot”)
This works for vectors: 𝑣 = 𝑑 𝑠 𝑑𝑡 = 𝑠
And second derivatives: 𝑎 = 𝑑 2 𝑠 𝑑𝑡 2 = 𝑠 (read “x-double-dot”)

17. 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 by neglecting resistance and the curvature and rotation of the earth.

18. Projectile motion
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.

19. A simple case of projectile motion:
Here two heavy balls begin “free fall” at the same time.
The red one is dropped, so it moves straight downward.
The yellow ball is given some speed in the horizontal direction as it is released.
The horizontal lines show that they keep pace with each other in the vertical direction.
Why do they do that?
ANIMATED: Two mouse clicks. (The next two slides appear with a mouse click with almost the same figure but different text, making this one seem animated.)

20. They have the same downward acceleration, g, and they both started with zero speed in the downward direction.
The yellow ball’s horizontal speed is not affected by gravity, which acts only in the vertical direction
Second slide in the “animation”

21. The x- and y-motion are separable
The red ball is dropped at the same time that the yellow ball is fired horizontally.
The strobe marks equal time intervals.
We can analyze projectile motion as horizontal motion with constant velocity and vertical motion with constant acceleration:
Galileo came up with this idea first.

22. Cannonballs shot horizontally with different speeds from the ship travel different distances.
ANIMATED: One mouse click brings up the last sentence.
But each cannonball drops the same distance in the same amount of time, since the vertical acceleration is the same for each…that’s g !

23. Ignoring air resistance, this plane flies horizontally with constant speed and it drops a cannonball.
ANIMATED: one mouse click brings up the last sentence.
The cannonball initially has the horizontal speed of the plane. It keeps that horizontal speed as it falls, so it stays beneath the airplane.

24. Projectile motion – Initial velocity
The initial velocity components of a projectile (such as a kicked soccer ball) are related to the initial speed and initial angle.

25. The equations for projectile motion
If we set x0 = y0 = 0, the equations describing projectile motion are shown below:

26. Projectile Motion
The acceleration is always downward.
A projectile that is launched with an initial velocity (vix, viy) follows a parabolic trajectory.
Because there is no acceleration horizontally, the horizontal speed is always a constant. So,
Since there is a constant downward acceleration, the kinematic equations can be used to describe the vertical motion.

27. Reasoning about Projectile Motion
A hungry hunter wants to shoot down a coconut that is hanging from the branch of a palm tree. He aims the gun directly at the coconut, but, as luck would have it, the coconut falls from the branch at the exact instant that the hunter pulls the trigger.
Does the bullet hit the coconut? YES! Because the bullet and coconut both fall a distance ½gt2 in time t.

28. Projectile Motion PROBLEM-SOLVING STRATEGY Projectile motion problems
MODEL Make simplifying assumptions.
VISUALIZE Draw a picture. Establish a coordinate system with the x-axis horizontal and the y-axis vertical. Show important points in the motion on a sketch. Define symbols and identify what the problem is trying to find. Draw the horizontal and vertical initial velocity vectors.
SOLVE The acceleration is known: ax = 0 and ay = – g thus the problem becomes one of kinematics. The equations are:
Vertical Motion:
Horizontal Motion

29. Projectile Motion
Δt is the same for the horizontal and vertical components of the motion. Find Δt from one component, then use that value for the other component.
ASSESS Check that your result has the correct units, is reasonable, and answers the question.

30. Example: A Home Run
A baseball is hit so that it leaves the bat at 33.6 m/s making a 300 angle with the ground. It crosses a low fence at the boundary of the ballpark at the same height that it was struck. (Neglect air resistance.)
Find: the horizontal distance traveled to get to the fence, the maximum height it reached and the time the ball was in the park. i
Vi sin Θ = 16.8 m/s
Vi cos Θ = 29.1 m/s
Horizontal Motion
= 0 + (29.1 m/s)(Dt )
= (29.1 m/s)(3.43 s) = 99.8 m
Vertical Motion
Use
viy = 16.8 m/s
ay = -9,8 m/s2
vfytop = 0
Yi = 0
Yf = ? ttop = ?
Yf = 14.4 m = maximum
height
0 = (16.8 m/s)2 + 2 (-9.8 m/s2) yf
And use
ttop = s
So Dt = 3.43 s
0 = 16.8 m/s – 9.8 m/s2 (ttop)

31. Projectile Range
Horizontal Range = v02 sin (2Θ) / g. This formula only works when the projectile starts and stops at the same elevation. Note that sin(2Θ) = 1 at q = 450, so distance is maximum there.
Conclusion: in the absence of air resistance, a projectile launched at 450 will travel farthest before returning to the height of the launch.

32. Verifying the Parabolic Trajectory
Displacements
xf = vxi t = (vi cos q) t
yf = vyi t + ½ay t2 = (vi sin q)t - ½ gt2
Combining the equations gives:
This is in the form of y = ax – bx2 which is the standard form of a parabola

33. The effects of air resistance
Calculations become more complicated.
Acceleration is not constant.
Effects can be very large.
Maximum height and range decrease.
Trajectory is no longer a parabola.

34. Motion in a circle
Uniform circular motion is constant speed along a circular path.

35. Motion in a circle
Car speeding up along a circular path

36. Three cases for circular motion
For non-uniform circular motion use Pythagoreum Theorem to get the overall acceleration:

37. Acceleration in Uniform Circular Motion
A particle moves with constant speed in a circle. The length of the vector v is constant but its direction changes. Therefore, the acceleration vector a points inward toward the center of the circular path.
This is called centripetal or radial acceleration. Also we can define the speed of the object as
where T is the period—the time to go around the circle once.

38. Acceleration for uniform circular motion
For uniform circular motion, the instantaneous acceleration always points toward the center of the circle and is called the centripetal acceleration.
The magnitude of the acceleration is arad = v2/R.
The period T is the time for one revolution, and arad = 4π2R/T 2.

39. Uniform circular motion

40. Non-uniform circular motion
If the speed varies, the motion is nonuniform circular motion.
The radial acceleration component is still arad = v2/R, but there is also a tangential acceleration component atan that is parallel to the instantaneous velocity.

41. Example This is a tiny fraction of g. What would arad be at the poles?
What is the approximate radial acceleration of someone standing on the equator of the earth due to the earth’s spin?
We only have two equations to work with.
We want arad so we need T, v and r first.
In roughly 24 hours the earth spins once.
The earth’s radius at the equator is about 6.38 x 106 m. We can use these numbers to now find v and then arad :
This is a tiny fraction of g. What would arad be at the poles?

42. The effects of air resistance
Calculations become more complicated.
Acceleration is not constant.
Effects can be very large.
Maximum height and range decrease.
Trajectory is no longer a parabola.

43. Relative velocity
The velocity of a moving body seen by a particular observer is called the velocity relative to that observer, or simply the relative velocity.
A frame of reference is a coordinate system plus a time scale.
In many situations relative velocity is extremely important.

44. Relative velocity in one dimension
If point P is moving relative to reference frame A, we denote the velocity of P relative to frame A as vP/A.
If P is moving relative to frame B and frame B is moving relative to frame A, then the x-velocity of P relative to frame A is vP/A-x = vP/B-x + vB/A-x.

45. Relative velocity in two or three dimensions
We extend relative velocity to two or three dimensions by using vector addition to combine velocities.

46. Relative velocity in two or three dimensions

47. Relative Motion Example
A pilot wants to fly to an airport that is due north. Her plane has an airspeed of 80 m/s and there is a crosswind of 20 m/s that points due west. At what angle should she point her plane in order to end up at the airport to the north? What is her speed as measured from the ground? q
Va/g
Vp/a
Vp/g N W E S
Vp/g = Vp/a + Va/g
Vp/g = ( )1/2 m/s = 77.5 m/s
q = tan-1 (20/77.5) = 14.5o E of N

48. Example: Ball Toss
Mike throws a ball upward at a 63° angle with a speed of 22 m/s. Nancy rides past Mike on her bicycle at 10 m/s at the instant he releases the ball.
What trajectory does Mike see?
What trajectory does Nancy see?

49. Object and Frame Velocities Add Vectorally

50. Galilean Relativity
Now consider how the acceleration transforms from S to S’, which is moving with constant velocity V relative to S. Velocities add, so
So both the observer at rest in frame S and the observer moving along with frame S’ see the same acceleration for an object. This is only true if both observers aren’t accelerating (we say they are in inertial reference frames).

51. Galileo vs. Einstein
The laser beam moves along the x axis away from Tom at the speed of light, vx= 3 x 108 m/s.
Sue flies by in her space ship, moving along the x axis at Vx= 2 x 108 m/s. From her point of view, how fast is the laser beam moving?
Galileo: vx’ = vx – Vx = 1 x 108 m/s
Einstein: vx’ = vx = 3 x 108 m/s
Wrong! Galileo’s method only works when speeds are much less than the speed of light.
The speed of light is the same in all inertial reference frames.