Chapter 3: Vectors Two-Dimensional Motion

Chapter 3: Vectors & Two-Dimensional Motion

Vectors and Their Properties All the physical quantities we will learn are classified either as a vector or scalar quantity. A scalar is specified by its magnitude, while a vector by its magnitude AND its direction.

Examples• Scalar : temperature, speed, mass, volume, length, etc.• Vector : displacement, velocity, acceleration, force, etc.

Vectors

Vectors and Their Properties

• Notation AAA

magnitude ;

• Equality of two vectors

BA

if their magnitudes and the directions are the same.

• Addition of vectors (geometrical method)

Commutative law of addition: ABBA

Vectors


• Negative of a vector A

A

is defined as the vector that gives zero when added to A

A

has the same magnitude but opposite direction of A

• Subtraction of vectors

)( BABA

Vectors


• Example 3.1: Taking a trip A car travels 20 km due northand then 35.0 km in a direction600 west of north.

What is the net effect of the car’strip?

39

km, 35.0km 20.0km 48

km 0.35km, 0.20

,

R

BA

RBA

Components of a vector

Components of a Vector

• A vector in two-dimension can be specified by a pair of coordinates

yxyx AAAAA

),(

xA

yA

x-component

y-component

x

y

x

y

yx

yx

AA

AA

AAA

AAAA

1

22

tan,tan

sin,cos

tan-1defined in (-900,900).Add 1800 when the vector is in 2nd or 3rd quadrant.

Components of a vector in different coordinates• A vector in two-dimension can be specified by a pair of coordinates

Bx

By

),(),( '' yxyx BBBBB

If a different coordinate system used, the components aredifferent to represent the samevector.

x

y

Vector addition by components:

),(

),(),(

yyxx

yxyx

BABA

BBAABA

But use the same coordinatesystem for both vectors


An example• Example 3.3: Take a hike

1st day : 25.0 km southeast2nd day : 40.0 km in a direction 60.0o north of east

A

B

Determine the components of the hiker’s displacements in the 1st and 2nd days.

km -17.7km)(0.707) 0.25(0.45sin

km 17.7km)(0.707) 0.25(0.45cos

AAAA

y

x

km 34.6km)(0.866) 0.40(0.60sinkm 20.0km)(0.500) 0.40(0.60cos

BBBB

y

x


1.24)/(tan km, 3.41

km 9.16km, 7.37122

xyyx

yyyxxx

RRRRR

BARBAR

Displacement, Velocity, & Acceleration in Two-Dimension

Displacement in 2D

• A position vector describes the position of an object at a time.

ii tr at timeector position v :

ff tr at timeector position v :

• An object’s displacement from ti to tf is defined by:

m :units SI if rrr


Velocity in 2D

• An object’s average velocity during a time interval t is:

m/s :units SI trvav

• An object’s instantaneous velocity is:

m/s :units SI lim 0 trv t


Acceleration in 2D

• An object’s average acceleration during a time interval t is:

2m/s :units SI tvaav

• An object’s instantaneous acceleration is:

20 m/s :units SI limtva t

Motion in Two-Dimension

Motion in 2D: horizontal and vertical direction

• In this chapter, we will learn movement of an object in both the x- and y-direction simultaneously under constant acceleration.

• An example: projectile motion under influence of gravity


Projectile motion under influence of gravity

• Let’s examine the motion of an object that is tossed into air but let’s neglect the effects of air resistance and the rotation of Earth.

• It was experimentally proven that the horizontal and vertical motions are completely independent of each other.

Motion in one direction has no effect on motion in the otherdirection.

So, in general, the equations of constant acceleration welearned in Lecture 2 follow separately for both the x-directionand the y-direction.


Projectile motion under influence of gravity

• Let’s assume that at t=0, the projectile leaves the origin with an initial velocity with an angle with the horizontal.0v

0

20

20

2

02

0

00

00000

221

,0;sin,cos

xxxx

xxx

xxxx

yxyx

vxavv

tvtatvx

vtavv

gaavvvv

x-direction:


Projectile motion under influence of gravity• Let’s assume that at t=0, the projectile leaves the origin with an initial velocity with an angle with the horizontal.0v

0

ygv

yavv

gttv

tatvy

gtvtavv

gaavvvv

y

yyy

y

yy

yyyy

yxyx

2

22121

,0;sin,cos

20

20

2

20

20

00

00000

y-direction:


Projectile motion under influence of gravity• Plug-in all the known quantities.

Equations that describe the motion in the y-direction:

Equations that describe the motion in the x-direction:

tvtvxvvv

x

xx

)cos(cosntant cos

000

000

ygvv

gttvy

gtvv

y

y

2)sin(21)sin(

sin

200

2

200

00

Velocity :

x

yyx v

vvvv 122 tan,


Projectile motion under influence of gravity• Trajectories as a function of the projection angle

Note : Given the displacement in x there are two corresponding projection angles.

o

o

o

o

o


Examples• Problem 3.5: Stranded explorers(a) What is the range of the package?

range of package

200 2

1 gttvyyy y

m 100)m/s 90.4( 22 ty

s 52.4t

tvxxx x00

m 181s) m/s)(4.52 0.40( x

s 52.4 m/s, 0.40m, 00.0 00 tvx x


Examples• Problem 3.5: Stranded explorers(b) What are the velocity components of the package at impact?

range of package

m/s 40.0m/s)cos0 0.40(cos0

vvx

m/s 3.44s) m/s)(4.52 80.9(00.0

sin0

gtvvy

x component:

y component:


Examples• Problem 3.6: The long jump(a) How long does it take for the jumper to reach the max. height?

s 384.0m/s 80.9

)20.0m/s)(sin 0.11(

sin

0sin

2

o

00max

max00

gvt

gtvvy

y component:

=20.0ov 0

= 11.0 m/s

at max.height vy =0


Examples• Problem 3.6: The long jump(b) Find the maximum height he reaches.

m 722.0s) 384.0)(m/s 80.9)(2/1(

s) 384.0)(0.20sinm/s)( 0.11()()2/1()sin(

22

2maxmax00max

tgtvy

y component:

=20.0ov 0

= 11.0 m/s

From part (a) s 384.0max t

m 7.94s) 768.0)(0.20cosm/s)( 0.11(

)cos(

s 0.768s) 384.0(22

00

max

tvx

tt


Examples• Problem 3.6: The long jump(c) Find the horizontal distance he jumps.

Displacement in x:

=20.0ov 0

= 11.0 m/s

s 384.0max t

m/s 386s) 3.14)(m/s 0.20(

m/s 1000.12

2

0

tavv xxx


Examples• Problem 3.8: The rocket

Eq.3.14c:

m/s 1040.1

m) 1000.1)(m/s 08.9(20

2

2

322

20

2

y

y

yy

v

v

ygvv

(b) Find the rocket’s velocity in x direction. Eq.3.12a:

s 3.14)m/s 80.9(0m/s 1040.1 22

0

tt

tavv yyy

Eq.3.11a:

(a) Find the rocket’s velocity in y direction.


Examples• Problem 3.8: The rocket(c) Find the magnitude and direction of the rocket’s velocity.

9.19m/s 386

m/s 1040.1tantan

m/s 411m/s) 386(m/s) 1040.1(

211-

222

22

x

y

yx

vv

vvv

Relative Velocity

What is relative velocity?

• Relative velocity relates velocities measured by two different observers, one moving with respect to the other.

•The measured velocity of an object depends on the velocity of the observer with respect to the object.

• Measurements of velocity depend on the reference frame of the observer where the reference frame is a just coordinate system used to measure physical quantities such as velocity, acceleration etc. Most of time, we will use a stationary frame of reference, relative to earth, but occasionally we will use a moving frame of reference.

Relative Velocity

What is relative velocity? (cont’d)• More elaborate definition: - Let’s define E as a stationary observer with respect to Earth - A and B as two moving cars

BCar in observer an by measured asA Car ofposition theEby measured as BCar ofposition theEby measured asA Car ofposition the

AB

BE

AE

rrr

)()()()()()(

222

111

trtrtrtrtrtr

BEAEAB

BEAEAB

BEAEAB

ABABAB

rrrtrtrr

etc. )()( 12

Relative Velocity

What is relative velocity? (cont’d)

BCar in observer anby measured asA Car of velocity theEby measured as BCar of velocity theEby measured asA Car of velocity the

AB

BE

AE

vvv

tr

tr

tr BEAEAB

BEAEAB vvv

12 ttt BEAEAB rrr

relative to; with respect to

tr

tr

tr BE

t

AE

t

AB

t

000limlimlim

Examples• Example 3.10: Crossing a river

Relative Velocity

determined be toEarth, to relativeboat theofvelocity :

km/h 10.0 water, the torelativeboat theofvelocity :

km/h 5.00 Earth, torelative flowriver theof velocity :

BE

BR

RE

v

v

v

REBRBE vvv

Examples

Relative Velocity

),( REyBRyRExBRxREBRBE vvvvvvv

km/h) km/h,0 00.5(),(

)km/h 0.10,km/h 0(),(

REyRExRE

BRyBRxBR

vvv

vvv

km/h) 10.0 km/h, 00.5()km/h 00.00.10,km/h 00.500.0(

BEv

6.26tan

km/h 2.11

1

22

BEy

BEx

BEyBExBE

vv

vvv

• Example 3.10: Crossing a river (cont’d)

Chapter 3: Vectors Two-Dimensional Motion

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Transcript of Chapter 3: Vectors Two-Dimensional Motion