Two-Dimensional Motion
and Vectors
3.1 Introduction to Vectors
Scalars & Vectors
•
•
•
•
Scalar quantities have magnitude only
Speed, volume, # students
Vectors have magnitude & direction
Velocity, force, weight, displacement
Representing Vectors
• Boldface type: v is a vector; v is not
• Symbol with arrow:
• Arrow drawn to scale:
•
----->
5m/s, 0° (east)
•
---------->
10 m/s, 0° (east)
•
<---------------15 m/s, 180° (west)

v
Resultant Vectors
• Are the sum of two or
more vectors
• Are “net” vectors
• Can be determined by
various methods
– Graphical addition
– Mathematically
• Pythagorean theorem
• Trigonometry
Graphical Addition of Vectors
• Parallelogram method
• Head-to-tail method
• Draw vectors to scale, with direction, in
head-to-tail fashion
• Resultant drawn from tail of first vector to
head of last vector
• Measure length and angle to determine
magnitude and direction
Head-to-Tail Addition of Vectors
Head-to-Tail Addition of Vectors
Other Properties of Vectors
• Vectors may be moved parallel to
themselves when constructing vector
diagrams
• Vectors may be added in any order
• Vectors are subtracted by adding its
opposite
• Vectors multiplied by scalars are still
vectors
Vectors may be added in any order
Direction of Vectors
• Degrees are measured counterclockwise
from x-axis
Direction of Vectors
• Direction may be
described with
reference to N, S, E, or
West axis
• 30° West of North
• 60° North of West
• Or 120°
3.2 Vector Operations
• Adding parallel
vectors
• Simple arithmetic
Perpendicular Vectors & the
Pythagorean Theorem
• If two vectors are
perpendicular, then you
can use the Pythagorean
theorem to determine
magnitude only
• Example: if
Δx = 40 km/h East
Δy = 100 km/h North
Then
d = (402 + 1002)½
d ≈ 108 km/h
But what is the direction?
Basic Trig Functions
hyperphysics.phy-astr.gsu.edu
Tangent Function
When two vectors are
perpendicular:
Use Pythagorean theorem to
find the magnitude of the
resultant vector
Use the tangent function to find
the direction of the resultant
vector
Sample Problem
• Indiana Jones climbs a
square pyramid that is
136 m tall. The base is
2.30 x 102 m wide. What
was the displacement of
the archeologist?
• What is the angle of the
pyramid
d
136m2  115m2
d  178m
 136 
tan   

 115 
1  136 
  tan 

 115 
  49.8
Resolving Vectors
• Vectors can be “resolved” into x- and y- components
• Resolve = Decompose = Break down
• Trig functions are used to resolve vectors
Resolving Vectors into X & Y
Components
• For the vector A
• Horizontal component =
Ax = A·cos θ
• Vertical component =
Ay = A·sin θ
hyperphysics.phy-astr.gsu.edu
Resolving a Vector
• A helicopter travels
95 km/h @ 35º above
horizontal. Find the
x- and y- components
of its velocity.
adj vx
cos  

hyp v
vx  v cos 
 95 cos35 km/h
 78 km/h
opp v y
sin  

hyp v
v y  v sin 
v y  95 sin 35 km/h
v y  55 km/h
Adding Non-perpendicular Vectors
A+B=R
Resolving Vectors into x and y
Components
hyperphysics.phy-astr.gsu.edu
Adding vectors that are not
perpendicular
• Two or more vectors can
be added by
decomposing each vector
• Add all x components to
determine Rx
• Add all y components to
determine Ry
• Determine magnitude of
the resultant R using
Pythagorean theorem
• Determine direction angle
θ of resultant using tan-1
x  R
x
y  R
y
R  Rx  R y
2
 Ry 
  tan  
 Rx 
1
2
Vector Analysis
Vector Analysis Worksheet
Magnitude
Direction
(units)
(deg)
Decomposition of Vectors
Vector A
6
25
Ax =
5.44
Ay =
2.54
Vector B
3
60
Bx =
1.50
By =
2.60
Vector C
Cx =
0.00
Cy =
0.00
Vector D
Dx =
0.00
Dy =
0.00
Rx =
6.94
Ry =
5.13
Resultant
8.6
36.5
3.3 Projectile Motion
phet.colorado.edu
• Objectives
• Recognize examples of projectile motion
• Describe the path of a projectile as a
parabola
• Resolve vectors into components and
apply kinematic equations to solve
projectile motion problems
Projectile Motion
• Motion of objects moving in two dimensions
under the influence of gravity
• Baseball, arrow, rocket, jumping frog, etc.
• Projectile trajectory is a parabola
Projectile motion
• Assumptions of our
problems
• Horizontal velocity is
constant, i.e.
• Air resistance is
ignored
• Projectile motion is
free fall with a
horizontal velocity
Motion of a projectile
Equations relating to vertical motion:
∆y = vyi(∆t) + ½ag(∆t)2
vyf = vyi + ag∆t
vyf2 = vyi2 + 2ag∆y
Equations relating to horizontal motion:
∆x = vx∆t
vx = vxi = constant
(an assumption relating to Newton’s 1st law of
motion)
Horizontal Projectile
vx = 5.0 m/s
t (s)
vy
vx
Δy
Δx
vy =
agt
0
vx =
1
Δy =
1/2agt2
2
Δx =
vxΔt
3
4
5
Horizontal Launch
Comparison of vx & vy vectors
Driving off a Cliff
• A stunt driver on a motorcycle speeds
horizontally of a 50.0m high cliff. How fast
must the motorcycle leave the cliff in order
to land on level ground below, 90.0m from
the base of the cliff? Ignore air resistance.
• Sketch the problem
• List knowns & unknowns
• Apply relevant equations
Driving off a Cliff
• Known: ∆x = 90.0m; ∆y =
-50.0m; ax = 0; ay = -g = 9.81 m/s2; vyi = 0
• Unknown: vx; ∆t
• Strategy: vx = ∆x/∆t
• Since ∆tx must = ∆ty
determine ∆t from the
Now solve for vx using ∆t
vertical drop
1 2
1 2
vx
y  v yi t  a y t y  a y t
2
2
2y
2(50.0m)
t

 3.19s
2
ay
 9.81m / s
x 90.0m


 28.2m / s
t 3.19s
Effect of Gravity on Ballistic Launch
[physicsclassroom.com]
Projectile Motion:
Horizontal vs Angled
Horizontal Launch
Ballistic Launch
www.ngsir.netfirms.com/englishhtm/ThrowABall.htm
Sample Projectile MotionProblem
• A ball is thrown with
an initial velocity of
50.0 m/s at an angle
of 60º.
• How long will it be in
the air?
• How high will it go?
• How far will it go?
Sample Problem
• A ball is thrown with an
initial velocity of 50.0 m/s
at an angle of 60º.
• How long will it be in the
air?
• Known: vi = 50.0 m/s, θ =
60º, a = -g = -9.81 m/s2;
• Find: Δt, total time in the
air
v f  vi  at
since v f  -vi
 vi  vi  at
 2vi  at
 2vi
 t
a
t 
 250 sin( 60) 
 9.81
t  8.83s
m
s2
m
s
Sample Problem
• A ball is thrown with
an initial velocity of 50
m/s at an angle of 60.
• How high will it go?
v fy  viy  2ay
2
2
v fy  viy
2
y 
y 
2
2a
2
v fy  (vi sin  ) 2
2a
0  (50 sin 60) 2
y 
2(9.81)
y  95.7 m
Sample Problem
• A ball is thrown with an initial velocity of 50 m/s
at an angle of 60º.
• How far will it go?
x  v x  t
x  vi cos60 t
x  50 cos60 m/s  8.83 s
x  25.0 m/s  8.83 s
x  221 m
3.4 Relative Motion
Objectives
• Describe motion in terms of frames of
reference
• Solve problems involving relative velocity
Frames of Reference
Motion is relative to frame of reference
To an observer in the
plane, the ball drops
straight down (vx = 0)
To an observer on
the ground, the ball
follows a parabolic
projectile path
(vx ≠ 0)
Frame of reference: a coordinate (defined by the observer) system for
specifying the precise location of objects in space
A frame of reference is a “point of view” from which motion is
described
Relative Velocity
• Relative velocity of one object to another
is determined from the velocities of each
object relative to another frame of
reference
Example
• See problem 1, page 109
• Use subscripts to indicate relative
velocities
• vbe = vbt + vte
• vbe = -15 m/s + 15 m/s
• vbe = 0 m/s
Example
• Car A travels 40 mi/h north; Car B travels
30 mi/h south. What is the velocity of Car
A relative to Car B?
• vae = 40; vbe = -30; veb = +30
• Find vab
• vab = vae + veb
• vab = 40 + 30
• vab = 70 mi/h North
Relative Velocity
• One car travels 90 km/h north, another
travels 80 km/h north. What is the speed
of the fast car relative to the slow car?
• vfe = 90 km/h north; vse = 80 km/h north
• vfs = vfe + ves
• vfs = 90 + (-)80
• vfs = 10 km/h north