# Kinematics in Two Dimensions

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```					Kinematics in Two
Dimensions

Chapter 3
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
Two-Dimensional Displacement

Sologdin the Turtle moves in     y
the X-Y plane from the
point (x0, y0) to the point         (x0, y0)
(x1, y1).
R0
R1 and R2 are vectors from the
origin to his starting and
finishing point,
respectively.                           R1
(x1, y1)
x
Two-Dimensional Displacement

Sologdin’s displacement          y
vector D is the vector that
extends from the starting of        (x0, y0)
his motion to its endpoint.
R0
In keeping with our Chapter 2                  D
definition of displacement,
we will define Sologdin’s
displacement by                        R1
(x1, y1)
  
D  R1  R0                                            x
Two-Dimensional Displacement

We can see the truth of this by
solving for R1:                  y
                   
D  R1  R0  R1  R0  D            (x0, y0)

Our picture shows this
graphically.                                  D

R1
(x1, y1)
x
Two-Dimensional Displacement

Since both R0 and R1 start at
y
the origin, it is very easy to
write down the magnitudes
of their X and Y                      (x0, y0)
components:
R0
R0 X  x0       R0Y  y0                       D

R1 X  x1      R1Y  y1
R1
(x1, y1)
x
Two-Dimensional Displacement

Now we perform the                 y
subtraction by subtracting
the component magnitudes:             (x0, y0)
  
D  R1  R0
R0
DX  R1 X  R0 X  x1  x0
D
DY  R1Y  R0Y  y1  y0
The magnitude of D:
R1
(x1, y1)
D  DX  DY
2         2

x
D    x1  x0    y1  y0 
2             2
Two-Dimensional Displacement
x1 – x0
The direction of D is more clearly
seen if we translate D so that its                              +X
starting point is at the origin.

(Remember that vectors are not
changed by translating them.)                         y1 – y0
D

We can always write
DY   y1  y0
tan      
DX   x1  x0
-Y
 y1  y0   
  arctan
 x x


 1 0       
Two-Dimensional Displacement
 y1  y0   
  arctan
 x x


x1 – x0
 1 0                                    +X
We used the absolute value signs           
here because we want simply the
lengths of the two legs of the                  y1 – y0
D
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°        -Y
clockwise from +X.”
Two-Dimensional Velocity
y
t = t0
Sologdin … and this time,           (x0, y0)
we’ll use our stopwatch.

R0
Solgdin’s journey begins at
D
time t0, and ends at a later
t = t1
time t1.
R1        (x1, y1)
x
Two-Dimensional Velocity
y
Sologdin’s average velocity is               t = t0
a vector. Its magnitude is
the magnitude of his
displacement, D, divided by                    V
the elapsed time:                    R0
D
D
v                                           t = t1
t1  t0
R1
Its direction is the same as the
x
direction of the
displacement vector.
Two-Dimensional Velocity
y
Sologdin’s average speed is             t = t0
not merely the magnitude
of his average velocity. Let
V
S be the curvilinear length
R0
of his meandering path.
Then his average speed is                 D
t = t1
S
ave. speed 
t1  t0              R1
Note that his average speed
x
exceeds his average
velocity.
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?
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.
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?                        y
(north)

From the definition of acceleration:                    -V0
 
 v1  v0
a                                  V1 – V0
V1

t
V0
-v0 to v1.                                                           x
(east)
Two-Dimensional Acceleration

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

a
v1  v0
2    2


12 m/s2  12 m/s2        y
t               8.0 s            (north)

a  2.1 m/s2                                                  -V0

a        V1
V1 – V0         

V0

x
(east)
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
y
v0                 (north)
tan    1
v1                                     -V0

  arctan1  45
a        V1
V1 – V0         
So, the average acceleration is
2.1 m/s2, 45° west of north.                           V0

x
(east)
Note that the speed did not change – only the direction.
Projectile Motion

A projectile is something
that is launched or
thrown.
v0y
v0
In the general case, it has a
nonzero horizontal
component of initial                    v0x
velocity.
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.
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?
Projectile Motion: Example
y
First step: resolve the
ball’s initial velocity
into horizontal and
vertical components.

v0 x  v0 cos                    v0

v0 y  v0 sin                         
v0y

30

v0x
x
Projectile Motion: Example
y
Keep firmly in mind
that v0y is not going
v0 x  v0 cos
to remain constant …             v0 y  v0 sin 
but that v0x will.

v0

v0y

30

v0x
x
Projectile Motion: Example
During the time the ball      y
takes to rise vertically,            v0 x  v0 cos
stop vertically, and fall
vertically back to the               v0 y  v0 sin 
ground, the ball moves
horizontally with
constant velocity.                     v0

v0y

We must determine just            30

how long that time of                        v0x
x
flight is.
Projectile Motion: Example
Vertically, the ball obeys the first kinematic equation:
v  v0  at
In our case, the initial vertical velocity is:   v0 y  v0 sin

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:

vy  v0 y  gt  v0 sin   gt
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:
1                  1 
v y  v0 y  g  t   v0 sin   g  t   0
2                  2 
Projectile Motion: Example
Solving for t, the total time of flight:

1               g
v0 sin   g  t   0          t  v0 sin 
2               2
2v0 sin  245 m/ssin 31
t                          2
 4.730 s
g             9.8 m/s
Projectile Motion: Example
The ball’s horizontal motion is governed by another
kinematic equation:
1 2
x  v0t  at
2

In the horizontal, however, the acceleration is zero, and the
initial velocity is v  v cos
0x    0

so we can rewrite our equation:
x  v0 cos  t  45 m/s cos 31 4.730 s   182 m
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?
Projectile Motion: Example
We can calculate this maximum height directly from the
third kinematic equation:
1 2
y  v0 y t  gt
2
Notice that the initial y velocity (for the falling part of the
trajectory) is zero, reducing the equation to:
2
1 2    1
y   gt   9.8 m/s
2
 4.730 
      s   27 m
2      2                   2      

The time we used was half the time-of-flight, t.
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
projectile’s trajectory.

Suppose our object is launched from the origin: x = y = 0.
Then:       x  v cos  t
0                           (1)

y  v0 sin  t  gt
1 2
(2)
2
Parametric Representation of a Trajectory

Solve eq’n (1) for t:
x  v0 cos  t
x
       t
v0 cos
Substitute this result for t in eq’n. 2:

v0 sin  x  1 g 
2
x 
y                             
v0 cos       2  v0 cos 
         
gx 2
y  x tan  
2v0 cos2 
2
Parametric Representation of a Trajectory
trajectory

30

25

20
y, m

15

10

5

0
0   20   40   60   80         100    120   140   160   180   200
x, m
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.
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 airplane’s velocity over the
ground … magnitude and direction?)
Relative Motion
Principle of relative
                                 
velocity:     vplane - ground    vplane - air  vair - ground
y (north)

va-g

Suppose the airplane’s
speed through the air is                        vp-a          vp-g
55 m/s, and the air moves
over the ground with a
wind speed of 8.9 m/s.
x (east)
Relative Motion
Velocity of plane over ground                     y (north)
Magnitude (speed):
va-g
vp  g  vp  a  va  g
2           2

vp  g    55 m/s 2  8.9 m/s 2    56 m/s

vp-g

Direction:                                           vp-a
 va  g 

  arctan          arctan 8.9 m/s   9.2
         
          55 m/s 
 vp  a 
(9.2° east of north)                                                  x (east)

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