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: ifΔx = 40 km/h EastΔy = 100 km/h North
Thend = (402 + 1002)½
d ≈ 108 km/hBut 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
m178
m115m136 22
d
d
8.49
115
136tan
115
136tan
1
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.
km/h 55
km/h 35sin95
sin
sin
y
y
y
y
v
v
vv
v
v
hyp
opp
km/h 78
km/h 35cos95
cos
cos
vv
v
v
hyp
adj
x
x
Adding Non-perpendicular VectorsA + 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
22yx RRR
xRx
yRy
x
y
R
R1tan
Vector Analysis
Vector Analysis Worksheet
Magnitude (units)
Direction (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
Resultant 8.6 36.5 Rx = 6.94 Ry = 5.13
3.3 Projectile Motionphet.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 = vyi
2 + 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 LaunchComparison 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 vertical drop
ssm
m
a
yt
taytatvy
y
yyyi
19.3/81.9
)0.50(22
2
1
2
1
2
22
Now solve for vx using ∆t
sms
m
t
xvx /2.28
19.3
0.90
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
ta
v
tav
tavv
-vv
tavv
i
i
ii
if
if
2
2
since
stsm
sm
t
83.8
81.9
)60sin(502
2
Sample Problem
• A ball is thrown with an initial velocity of 50 m/s at an angle of 60.
• How high will it go?
my
y
a
vvy
a
vvy
yavv
ify
iyfy
iyfy
7.95
)81.9(2
)60sin50(0
2
)sin(
2
2
2
22
22
22
Sample Problem
• A ball is thrown with an initial velocity of 50 m/s at an angle of 60º.
• How far will it go?
m 221
s 83.8m/s 0.25
s 83.8m/s 60cos50
60cos
x
x
x
tvx
tvx
i
x
3.4 Relative MotionObjectives
• Describe motion in terms of frames of reference
• Solve problems involving relative velocity
Frames of ReferenceMotion 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
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