Chapter 3 Kinematics in Two or Three Dimensions; Vectors · 2017. 9. 26. · • Vectors and...

7/03/2012

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Lecture PowerPoints

Chapter 3

Physics for Scientists &Engineers, with Modern

Physics, 4th edition

Giancoli


Chapter 3

Kinematics in Two or ThreeDimensions; Vectors

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Opening Question

A box is dropped from a moving helicopter atpoint A as it flies in a horizontal direction. Whichpath in the drawing below best describes the pathof the box (ignore air resistance) as seen by aperson standing on the ground?


Units of Chapter 3

• Vectors and Scalars

• Addition of Vectors—Graphical Methods

• Subtraction of Vectors, and Multiplication of aVector by a Scalar

• Adding Vectors by Components

• Unit Vectors

• Vector Kinematics

• Projectile Motion

• Solving Problems Involving Projectile Motion

• Relative Velocity

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3-1 Vectors and Scalars

A vector has magnitude aswell as direction.

Some vector quantities:displacement, velocity, force,momentum

A scalar has only a magnitude.

Some scalar quantities: mass,time, temperature

The GREEN arrows representthe velocity vector at eachpoint.


Representing Vectors

Vectors can be represented in a variety of ways by

using Bold type and/or lines or arrows, etc. aboveor below the vector symbol. Eg.

value.averageormeanwith the

confusedbemaytheyasVandavoidbest toisIt

VV~

VV~

V

VVVVV

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3-2 Addition of Vectors—Graphical Methods

For vectors in onedimension, simpleaddition and subtractionare all that is needed.

You do need to be carefulabout the signs, as thefigure indicates.


3-2 Addition of Vectors—Graphical MethodsIf the motion is in two dimensions, the situation is somewhatmore complicated.

Here, the actual travel paths are at right angles to one another; wecan find the displacement by using the Pythagorean Theorem.

km11.2

510 22

2710

5tan

tan

1

1

21

D

D

EofN27atkm11.2 21R DDD

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Adding the vectors in the opposite order gives thesame result:

EofN27atkm11.212 DDDR

1221 VVVV



Even if the vectors are not at rightangles, they can be added graphically byusing the tail-to-tip method.

Note that (as in the previous example) the orderof adding the vectors is not important.

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Where there are two vectors the parallelogrammethod may also be used.


3-3 Subtraction of Vectors, andMultiplication of a Vector by a Scalar

In order to subtract vectors, wedefine the negative of a vector, whichhas the same magnitude but pointsin the opposite direction.

Then we add the negative vector.

1212 VVVV

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3-3 Subtraction of Vectors, andMultiplication of a Vector by a Scalar

A vector can be multiplied by a scalarc; the result is a vector c that has thesame direction but a magnitude cV. If c isnegative, the resultant vector points inthe opposite direction.

V

V


3-4 Adding Vectors by Components

Any vector can be expressed as the sumof two other vectors, which are called itscomponents. Usually the other vectors arechosen so that they are perpendicular toeach other.

x

y

yx

yx

V

Vtanθand

VVVV

:isvectorofThe

vectorofcomponentstheareand

1-

22V

|V|magnitude

VVV

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If the components areperpendicular, they can befound using trigonometricfunctions.



The components are effectively one-dimensional,so they can be added arithmetically.

yyyxxx 2121 VVVandVVV

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Adding vectors:

1. Draw a diagram; add the vectors graphically.

2. Choose x and y axes.

3. Resolve each vector into x and y components.

4. Calculate each component using sines and cosines.

5. Add the components in each direction.

6. To find the length and direction of the vector, use:

and .


3-4 Adding Vectors byComponents

Example 3-2: Mail carrier’sdisplacement.

A rural mail carrier leaves thepost office and drives 22.0 kmin a northerly direction. Shethen drives in a direction 60.0°south of east for 47.0 km. Whatis her displacement from thepost office?

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This image cannot currently be displayed.

E)51S(orS39Eatkm30ntDisplaceme

3923.5

18.7-

D

Dθ

km037.185.23DD

km.78140.722.0

km23.523.50

km40.760in7

km23.56047

km22

0

1

x

y1-

2222x

tantan

D

s

cos

y

2y1yy

2x1xx

2y

2x

1y

1x

DDD

DDD

4D

D

D

D


3-4 Adding Vectors byComponents

Example 3-3: Three short trips.

An airplane trip involves threelegs, with two stopovers. Thefirst leg is due east for 620 km;the second leg is southeast(45°) for 440 km; and the thirdleg is at 53° south of west, for550 km, as shown. What is theplane’s total displacement?

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S50.4Eatkm980ntDisplacemeTotal

50.4620

750

D

Dθ

km980750620DDD

km-750439-311-0

km620311-311620

km39435s055

km31135055

km31145s440

km31145440

0

km620

1

x

y1-

222y

2x

3

3

tantan

in

cos

in

cos

3y2y1yy

3x2x1xx

y

x

2y

2x

1y

1x

DDDD

DDDD

D

D

D

D

D

D


3-5 Unit Vectors

Unit vectors have magnitude 1.

Using unit vectors, any vectorcan be written in terms of itscomponents:

V

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Eg. Using Unit Vectors to Solve the Mail CourierProblem – (Example 3-2)

kmˆ18.7ˆ23.5

kmˆ40.7-22.0ˆ23.50

kmˆ40.7-ˆ23.5

kmˆ22.0ˆ0

km40.7km,23.5

andkm22.00,

ji

jiD

jiD

jiD

DD

DD

2

1

2y2x

1y1x

.anglesareiseAnticlockw

axis x ve thealong 0θ definitionby that Note

axis.xthebelow39atkm30ntDisplaceme:Ans

3923.5

18.7-θ

km3018.723.5D

1-

22

positive

D

tan


3-6 Vector Kinematics

In two or threedimensions, thedisplacement is avector:

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As Δt and Δr becomesmaller and smaller, theaverage velocityapproaches theinstantaneous velocity.



v

v

The instantaneousacceleration is in thedirection of Δ = 2 – 1,and is given by:

v

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Using unit vectors,



Generalizing the one-dimensional equationsfor constant acceleration:

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3-7 Projectile Motion

A projectile is anobject moving in twodimensions under theinfluence of Earth'sgravity; its path is aparabola.


It can be understoodby analyzing thehorizontal and verticalmotions separately.


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The speed in the x-directionis constant; in the y-direction the object moveswith constant acceleration g.

This photograph shows two ballsthat start to fall at the same time.The one on the right has an initialspeed in the x-direction. It can beseen that vertical positions of thetwo balls are identical at identicaltimes, while the horizontalposition of the yellow ballincreases linearly.



If an object is launched at an initial angle of θ0

with the horizontal, the analysis is similar exceptthat the initial velocity has a vertical component.

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3-8 Solving Problems InvolvingProjectile Motion

Projectile motion is motion with constantacceleration in two dimensions, where theacceleration is g and is down.



1. Read the problem carefully, and choose theobject(s) you are going to analyze.

2. Draw a diagram.

3. Choose an origin and a coordinate system.

4. Decide on the time interval; this is the same inboth directions, and includes only the time theobject is moving with constant acceleration g.

5. Examine the x and y motions separately.

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6. List known and unknown quantities.Remember that vx never changes, and thatvy = 0 at the highest point.

7. Plan how you will proceed. Use theappropriate equations; you may have tocombine some of them.



Example 3-6: Driving off acliff.

A movie stunt driver on amotorcycle speedshorizontally off a 50.0 m highcliff. How fast must themotorcycle leave the cliff topto land on level ground below,90.0 m from the base of thecliff where the cameras are?Ignore air resistance.

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x direction:)1(

x

xv

xt

t

xv

?constant

m90

xv

x

y direction:)2(

2

1 200 attvyy y

m/s9.8

0

0

m50

0y

0

ga

v

y

y

22

22

2

2

2

m/s28.2100

908.9

908.9100

2

908.90050

(2)into(1)Substitute

x

x

x

x

2

x

0y

0

v

v

v

2v

ax

v

xvyy



Example 3-7: A kicked football.

A football is kicked at an angle θ0 = 37.0° with a velocityof 20.0 m/s, as shown. Calculate (a) the maximumheight, (b) the time of travel before the football hits theground, (c) how far away it hits the ground, (d) thevelocity vector at the maximum height, and (e) theacceleration vector at maximum height. Assume theball leaves the foot at ground level, and ignore airresistance and rotation of the ball.

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m7.359.802

12.0y

0y9.80212.00

0y2avv

m/s 12.037.020.0θ

2

2

20y

2y

0

sinsin00y vv

(a) Maximum Height

Only need to consider vertical motion andcalculate time to maximum height.


(b) Time for ball to hit ground

Consider vertical motion and calculate time tomaximum height.

s2.452tgroundhittotimeTotal

s1.229.80

12.0t

9.80t12.00

t

avv 0yy

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(c) How far away it hits the ground

Horizontal motion at constant velocity and time offlight = 2.45 s.

m39.22.4516.0t

m/s 16.037.020.0θ0

x

0xx

vx

vv

coscos


(d) The velocity vector at the maximum height

m/sˆ16.0horizontalisMotion ivv 0x

(e) The acceleration vector at the maximum height

2m/sˆ-9.80

pointsallat

ja

ga

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A child sits upright in a wagonwhich is moving to the right atconstant speed as shown. Thechild extends her hand and throwsan apple straight upward (from herown point of view), while thewagon continues to travel forwardat constant speed. If air resistanceis neglected, will the apple land (a)behind the wagon, (b) in thewagon, or (c) in front of thewagon?

Conceptual Example 3-8: Where does theapple land?



Conceptual Example 3-9: The wrong strategy.

A boy on a small hill aims his water-balloon slingshothorizontally, straight at a second boy hanging from atree branch a distance d away. At the instant the waterballoon is released, the second boy lets go and fallsfrom the tree, hoping to avoid being hit. Show that hemade the wrong move. (He hadn’t studied physicsyet.) Ignore air resistance.

Both water-balloonand second boy havethe same vertical(vertical) acceleration

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Example 3-10: Level horizontal range.

(a) Derive a formula for thehorizontal range R of a projectile interms of its initial speed v0 and angleθ0. The horizontal range is defined asthe horizontal distance the projectiletravels before returning to itsoriginal height (which is typically theground); that is, y(final) = y0. (b)Suppose one of Napoleon’s cannonshad a muzzle speed, v0, of 60.0 m/s.At what angle should it have beenaimed (ignore air resistance) tostrike a target 320 m away?


022

0

2

0

0

022

0

2

00

00

2

21

002

21

0y0

000y

00

00x

000xx

θcos2v

gx

θ

θx

θ2v

gx

θv

xθsinvy


(2) gttθsinv0gttvyy

θvv

(1)θv

xttθvtvx

θvvv

cos

sin

coscos

sin

coscos

cos

:MotionVertical(b)

:MotionHorizontal(a)

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θθ22θθ2

R

θθ2R

θ

θ

θ2v

gR0R

θ2v

gR

θ

θR

θ2v

gR

θ

θR0

R)for x(SolveRor x0x0,yWhen

020

0020

0

0

022

0

022

00

0

022

0

2

0

0

cossinsing

sinv

g

cossinv

cos

sin

cosor

coscos

sin

coscos

sin


rangesamethegivewillanglesBoth

solutionstwoaretheresoquadraticaisRforequationthethatNote

away m 320 is (R)target and m/s 60.0v if θ Find(b) 00

.79530.390θ30.3θ

30.3θ60.62θ

871.00.60

80.9320

v

Rgθ2

g

θ2vR

00

00

220

00

20

or

sinsin

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Example 3-11: A punt.

Suppose the football in Example 3–7 waspunted and left the punter’s foot at a height of1.00 m above the ground. How far did thefootball travel before hitting the ground? Setx0 = 0, y0 = 0.


NB: Cannot use horizontal range

v0 = 20.0 m/s0 = 37.0y0 = 0y = -1 m

v0x = 20.0 cos 37.0 = 16.0 m/sv0y = 20.0 sin 37.0 = 12.0 m/sa = -g = -9.8 m/s2

m1.00ywhenSolve

0.0191x0.750x16.0

4.90x

16.0

12.0xy


(2)4.9t12.0t0y

attvyy

(1)16.0

x

v

xttvx

2

2

2

2

2

21

0y0

0x

0x

:Vertically

:lyHorizontal

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ve)bemust(xm1.28orm40.5x

0382.0

799.0750.0

0191.02

00.10191.04750.0750.0x

01.000.750x0.0191x

2

2


3-8 Solving Problems Involving Projectile Motion

Example 3-12: Rescue helicopter drops supplies.A rescue helicopter wants to drop a package of supplies to isolatedmountain climbers on a rocky ridge 200 m below. If the helicopter istraveling horizontally with a speed of 70 m/s (250 km/h), (a) how far inadvance of the recipients (horizontal distance) must the package bedropped? (b) Suppose, instead, that the helicopter releases the packagea horizontal distance of 400 m in advance of the mountain climbers.What vertical velocity should the package be given (up or down) so thatit arrives precisely at the climbers’ position? (c) With what speed doesthe package land in the latter case?

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Example 3-12: Rescue helicopter drops supplies.

Try this one yourself

Answer: (a) 447 m(b) 7 m/s downwards(c) 94 m/s

Worked solution in text book.



Projectile motion is parabolic:

Taking the equations for x and y as a functionof time, and combining them to eliminate t, wefind y as a function of x:

This is the equation for a parabola.

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Examples of projectilemotion. Notice theeffects of air resistance.


3-9 Relative Velocity

We have already considered relative speed inone dimension; it is similar in two dimensionsexcept that we must add and subtract velocitiesas vectors.

Each velocity is labeled first with the object,and second with the reference frame in whichit has this velocity.

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Here, vWS is the velocityof the water in the shoreframe, vBS is the velocityof the boat in the shoreframe, and vBW is thevelocity of the boat in thewater frame.


The relationship betweenthe three velocities is:



Example 3-14: Heading upstream.

A boat’s speed in stillwater is vBW = 1.85 m/s. Ifthe boat is to traveldirectly across a riverwhose current has speedvWS = 1.20 m/s, at whatupstream angle must theboat head?

upstream40.41.85

1.20θ 11

sinsin

BW

WS

v

v

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Example 3-15: Heading across the river.

The same boat (vBW = 1.85 m/s) nowheads directly across the riverwhose current is still vWS = 1.20 m/s.

(a) What is the velocity (vBS)(magnitude and direction) of theboat relative to the shore?

(b) If the river is 110 m wide, howlong will it take to cross and howfar downstream will the boat bethen?


vBW = 1.85 m/s, vWS =1.20 m/s.

(a) What is the velocity (vBS)(magnitude and direction) ofthe boat relative to the shore?

downstream33.0atm/s2.21velocity v

33.01.85

1.20

v

vθ

m/s2.211.201.85vvv

BS

1

BW

WS1

222WS

2BWBS

tantan

WSBWBS vvv

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(b) If the river is 110 m wide, howlong will it take to cross and howfar downstream will the boat bethen?

m71.459.51.20tvd

s59.5inat vtravelleddistancedownstreamDistance

s59.5v

Dt

t

Dv:rivercrosstoTime

m/s 1.8533.02.21θv

Dofdirectioninvelocitysboat'ofComponent

riverofwidthm110D

WS

WS

BW

BW

BS

coscos



Example 3-16: Car velocities at 90°.

Two automobiles approach a street corner atright angles to each other with the same speedof 40.0 km/h (= 11.1 m/s), as shown. What is therelative velocity of one car with respect to theother? That is, determine the velocity of car 1 asseen by car 2.

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The diagram at left shows the situation from a referenceframe fixed to the Earth.

We need to view from a reference frame in which car 2 is atrest which is shown in the diagram at right.

Car 2 sees both the Earth and Car 1 moving towards it.


2Cartowards45atkm/h56.64040v

anglesrightmovingaretheyand

km/h40iscarsbothofspeedtheSince

is2Carbyseenas1CarofvelocityThe

or

km/h)40(speed2CartorelativeEarthofvelocity

km/h)40(speedEarthtorelative2Carofvelocity

2212

2E1EE21E12

2EE2E22E

E2

2E

vvvvv

vvvv

v

v

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Summary of Chapter 3

• A quantity with magnitude and direction is avector.

• A quantity with magnitude but no direction isa scalar.

• Vector addition can be done either graphicallyor by using components.

• The sum is called the resultant vector.

• Projectile motion is the motion of an objectnear the Earth’s surface under the influence ofgravity.

Chapter 3 Kinematics in Two or Three Dimensions; Vectors · 2017. 9. 26. · • Vectors and...

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