Two-Dimensional Motion and Vectors

44
Two-Dimensional Motion and Vectors

description

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 - PowerPoint PPT Presentation

Transcript of Two-Dimensional Motion and Vectors

Page 1: Two-Dimensional Motion  and Vectors

Two-Dimensional Motion and Vectors

Page 2: Two-Dimensional Motion  and Vectors

3.1 Introduction to Vectors

Page 3: Two-Dimensional Motion  and Vectors

Scalars & Vectors

• Scalar quantities have magnitude only

• Speed, volume, # students

• Vectors have magnitude & direction

• Velocity, force, weight, displacement

Page 4: Two-Dimensional Motion  and Vectors

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

Page 5: Two-Dimensional Motion  and Vectors

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

Page 6: Two-Dimensional Motion  and Vectors

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

Page 7: Two-Dimensional Motion  and Vectors

Head-to-Tail Addition of Vectors

Page 8: Two-Dimensional Motion  and Vectors

Head-to-Tail Addition of Vectors

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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

Page 10: Two-Dimensional Motion  and Vectors

Vectors may be added in any order

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Direction of Vectors

• Degrees are measured counterclockwise from x-axis

Page 12: Two-Dimensional Motion  and Vectors

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°

Page 13: Two-Dimensional Motion  and Vectors

3.2 Vector Operations

• Adding parallel vectors

• Simple arithmetic

Page 14: Two-Dimensional Motion  and Vectors

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?

Page 15: Two-Dimensional Motion  and Vectors

Basic Trig Functions

hyperphysics.phy-astr.gsu.edu

Page 16: Two-Dimensional Motion  and Vectors

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

Page 17: Two-Dimensional Motion  and Vectors

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

Page 18: Two-Dimensional Motion  and Vectors

Resolving Vectors• Vectors can be “resolved” into x- and y- components• Resolve = Decompose = Break down• Trig functions are used to resolve vectors

Page 19: Two-Dimensional Motion  and 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

Page 20: Two-Dimensional Motion  and Vectors

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

Page 21: Two-Dimensional Motion  and Vectors

Adding Non-perpendicular VectorsA + B = R

Page 22: Two-Dimensional Motion  and Vectors

Resolving Vectors into x and y Components

hyperphysics.phy-astr.gsu.edu

Page 23: Two-Dimensional Motion  and Vectors

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

Page 24: Two-Dimensional Motion  and Vectors

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

Page 25: Two-Dimensional Motion  and Vectors

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

Page 26: Two-Dimensional Motion  and Vectors

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

Page 27: Two-Dimensional Motion  and Vectors

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

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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)

Page 29: Two-Dimensional Motion  and Vectors

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        

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Horizontal LaunchComparison of vx & vy vectors

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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

Page 32: Two-Dimensional Motion  and Vectors

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

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Effect of Gravity on Ballistic Launch

[physicsclassroom.com]

Page 34: Two-Dimensional Motion  and Vectors

Projectile Motion: Horizontal vs Angled

Horizontal Launch Ballistic Launch

www.ngsir.netfirms.com/englishhtm/ThrowABall.htm

Page 35: Two-Dimensional Motion  and Vectors

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?

Page 36: Two-Dimensional Motion  and Vectors

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

Page 37: Two-Dimensional Motion  and Vectors

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

Page 38: Two-Dimensional Motion  and Vectors

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

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3.4 Relative MotionObjectives

• Describe motion in terms of frames of reference

• Solve problems involving relative velocity

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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

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Relative Velocity

• Relative velocity of one object to another is determined from the velocities of each object relative to another frame of reference

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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

Page 43: Two-Dimensional Motion  and Vectors

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

Page 44: Two-Dimensional Motion  and Vectors

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