Kinematics in Two Dimensions

1
Kinematics in Two Dimensions

• Chapter 3

2
Expectations

• After Chapter 3, students will
• generalize the concepts of displacement,
  velocity, and acceleration from one to two
  dimensions
• use the projectile-motion principle to analyze
  two-dimensional constant-acceleration problems
• apply the kinematic equations to solve
  constant-acceleration problems in two dimensions
• add relative velocities in two dimensions

3
Two-Dimensional Displacement

• Sologdin the Turtle moves in the X-Y plane from
  the point (x0, y0) to the point (x1, y1).
• R1 and R2 are vectors from the origin to his
  starting and finishing point, respectively.

4
Two-Dimensional Displacement

• Sologdins displacement vector D is the vector
  that extends from the starting of his motion to
  its endpoint.
• In keeping with our Chapter 2 definition of
  displacement, we will define Sologdins
  displacement by

5
Two-Dimensional Displacement

• We can see the truth of this by solving for R1
• Our picture shows this addition being done,
  graphically.

6
Two-Dimensional Displacement

• Since both R0 and R1 start at the origin, it is
  very easy to write down the magnitudes of their X
  and Y components

7
Two-Dimensional Displacement

• Now we perform the subtraction by subtracting the
  component magnitudes
• The magnitude of D

8
Two-Dimensional Displacement

• The direction of D is more clearly seen if we
  translate D so that its starting point is at the
  origin. (Remember that vectors are not changed
  by translating them.)
• We can always write

9
Two-Dimensional Displacement

• We used the absolute value signs here because we
  want simply the lengths of the two legs of the
  right triangle. The sense of the angle is
  apparent from the drawing. We would describe the
  direction of D as 60 below the positive X
  axis, or 60 clockwise from X.

10
Two-Dimensional Velocity

• Lets return to the travels of Sologdin and
  this time, well use our stopwatch.
• Solgdins journey begins at time t0, and ends at
  a later time t1.

11
Two-Dimensional Velocity

• Sologdins average velocity is a vector. Its
  magnitude is the magnitude of his displacement,
  D, divided by the elapsed time
• Its direction is the same as the direction of the
  displacement vector.

12
Two-Dimensional Velocity

• Sologdins average speed is not merely the
  magnitude of his average velocity. Let S be the
  curvilinear length of his meandering path. Then
  his average speed is
• Note that his average speed exceeds his average
  velocity.

13
Two-Dimensional Velocity

• The critical difference between speed and
  velocity
• Consider the race car that wins the Indianapolis
  500.
• It completes 200 laps at 2.5 miles each in 2.75
  hours. Its average speed is 500 mi / 2.75 hr
  182 mi/hr.
• What is its average velocity?

14
Two-Dimensional Velocity

• What is its average velocity? Zero magnitude
  pick a direction.
• Its displacement is zero (the race both starts
  and ends at the start-finish line). And zero
  divided by any amount of time yields a magnitude
  of zero for the average velocity.

15
Two-Dimensional Acceleration

• A truck travels east at 12 m/s. Following a bend
  in the road, the truck turns so that it is headed
  north, still at 12 m/s. The truck spends 8.0 s
  negotiating the curve. What is its average
  acceleration?
• From the definition of acceleration
• Here, we have graphically added
• -v0 to v1.

16
Two-Dimensional Acceleration

• Having obtained v1 - v0, we need to multiply that
  vector by the scalar 1/t to obtain a

17
Two-Dimensional Acceleration

• The direction of a is the same as the direction
  of v1 - v0. Since the magnitudes of v1 and v0
  are equal, we can write
• So, the average acceleration is
• 2.1 m/s2, 45 west of north.
• Note that the speed did not change only the
  direction.

18
Projectile Motion

• A projectile is something that is launched or
  thrown.
• In the general case, it has a nonzero horizontal
  component of initial velocity.

19
Projectile Motion

• In most cases, the projectile experiences a
  negative vertical acceleration (g, due to
  gravity).
• In most cases (assuming we can ignore resistance
  due to the air), the projectile experiences zero
  horizontal acceleration. Its horizontal velocity
  is constant.
• The general approach to all problems of this
  kind apply the kinematic equations separately to
  the vertical and horizontal motion.

20
Projectile Motion Example

• Illustrative example the golfer in the picture
  has launched his ball with an initial velocity of
  45 m/s, 31 above the horizontal. We will
  analyze the flight of the ball, in order to be
  able to say
• How far will the ball carry over level ground?
• How high above the ground will the ball reach at
  its highest point?
• What is the total time the ball will be in the
  air?

21
Projectile Motion Example

• First step resolve the balls initial velocity
  into horizontal and vertical components.

22
Projectile Motion Example

• Keep firmly in mind that v0y is not going to
  remain constant but that v0x will.

23
Projectile Motion Example

• During the time the ball takes to rise
  vertically, stop vertically, and fall vertically
  back to the ground, the ball moves horizontally
  with constant velocity.
• We must determine just how long that time of
  flight is.

24
Projectile Motion Example

• Vertically, the ball obeys the first kinematic
  equation
• In our case, the initial vertical velocity is
• The acceleration is downward g, the acceleration
  due to gravity. If we define our coordinate
  system with Y pointing up, we rewrite the first
  kinematic equation

25
Projectile Motion Example

• In terms of its vertical motion, the ball stops
  momentarily at the top of its trajectory. We
  also showed, in chapter 2, that its vertical
  travel is symmetric it rises for half the time
  of flight, and falls for half the time. So, if t
  is the full time of flight, we can rewrite our
  kinematic equation for the upward travel by
  setting the final (y) velocity to zero

26
Projectile Motion Example

• Solving for t, the total time of flight

27
Projectile Motion Example

• The balls horizontal motion is governed by
  another kinematic equation
• In the horizontal, however, the acceleration is
  zero, and the initial velocity is
• so we can rewrite our equation

28
Projectile Motion Example

• How high does the ball go at the top of its
  trajectory?
• Recall that we already calculated the total time
  of flight
• t 4.730 s. Recall also that the ball spent
  half that time
• falling (with zero initial vertical velocity)
  from its highest
• point. How far did it fall?

29
Projectile Motion Example

• We can calculate this maximum height directly
  from the third kinematic equation
• Notice that the initial y velocity (for the
  falling part of the trajectory) is zero, reducing
  the equation to
• The time we used was half the time-of-flight, t.

30
Parametric Representation of a Trajectory

• So far, we have written both x and y as
  independent functions of time, t. We can combine
  these descriptions to express y as a function of
  x the equation of the projectiles trajectory.
• Suppose our object is launched from the origin x
  y 0.
• Then

31
Parametric Representation of a Trajectory

• Solve eqn (1) for t
• Substitute this result for t in eqn. 2

32
Parametric Representation of a Trajectory
33
Projectile Motion Things to Remember

• Basic principle the vertical motion and the
  horizontal motion are treated separately.
• In the vertical motion, there is acceleration
  (usually, g).
• In the horizontal motion, acceleration is zero
  and velocity is constant.
• Resolve the initial velocity into horizontal and
  vertical components.
• Apply the kinematic equations to each motion,
  separately.
• If g appears in an equation with x, slap your
  own hand.

34
Relative Motion

• Consider an airplane flying through air, when the
  air is in motion relative to the ground (wind).
• The wind is blowing eastward. The pilot points
  his airplane straight north. In what direction
  does the airplane move over the ground? How fast
  does it go?
• (In other words, what is the airplanes velocity
  over the ground magnitude and direction?)

35
Relative Motion

• Principle of relative velocity
• Suppose the airplanes speed through the air is
  55 m/s, and the air moves over the ground with a
  wind speed of 8.9 m/s.

36
Relative Motion

• Velocity of plane over ground
• Magnitude (speed)
• Direction
• (9.2 east of north)