Kinematics in Two-Dimensions1. Vectors

1.1. Vector Addition/Subtraction

1.2. Vector Multiplication:1.2. Vector Multiplication: × Cross Product• Dot Product

2. Motion in 2-Dimensions• Projectiles maximum height time range• Projectiles, maximum height, time, range

3. Relative velocity

22/01/2010 08:56FAP0015 PHYSICS I 1

Lesson OutcomesLesson OutcomesStudents should be able to:1 define the components of displacement velocity and1. define the components of displacement, velocity and

acceleration in both dimensions2 define projectile motion2. define projectile motion3. derive the projectile equations of motion4 appl the projectile eq ations of motion to determine4. apply the projectile equations of motion to determine

the maximum height and range and the total time of motionmotion

5. define the relative velocity

22/01/2010 08:56FAP0015 PHYSICS I 2

VectorsVectors

• A vector quantity is any quantity with magnitude and direction.

• Displacement, velocity, acceleration, d f l fmomentum and force are examples of vector

quantities.

22/01/2010 08:56FAP0015 PHYSICS I 3

Let’s say you have 2 vectors a & by y

bb

a

22/01/2010 08:56FAP0015 PHYSICS I 4

Adding Vectors

b Triangle Methoda b

a

a + b

a + bParallelogram Method

b

22/01/2010 08:56FAP0015 PHYSICS I 5

Subtracting Vectorsg

a - b = a + (-b)- b

b

( )

a b

22/01/2010 08:56FAP0015 PHYSICS I 6

Adding VectorsgMethod 3: Resolving the vectors

22

ai θ

2y

2x aaa +=

a1

θ

ay = a sin θx

y

aa1tanθ −=

θax = a cos θ

22/01/2010 08:56FAP0015 PHYSICS I 7

Adding VectorsMethod 3: Resolving the vectors

a a cos θ + b c = a + bax = a cos θa

ay = a sin θa

c = a + b cx = ax + bxcy = ay + by

22 ccc +=bx = b cos θb

y

yx

ctan

ccc

1−θ

+by = b sin θb

x

y

ctanc =θ

22/01/2010 08:56FAP0015 PHYSICS I 8

aay

ay by

c = a + b

cxbx

cyax

ax bxb

bby

22/01/2010 08:56FAP0015 PHYSICS I 9

Vector Multiplication

There are two type of vector multiplication:-

• dot multiplication scalar

and

× cross multiplication vector

22/01/2010 08:56FAP0014 PHYSICS I 10

aθ

a

b

Dot multiplication: a • b = ⎪a⎪⎪b⎪ cos θ

Scalar quantity

Dot multiplication: a • b = ⎪a⎪⎪b⎪ cos θ

22/01/2010 08:56FAP0014 PHYSICS I 11

Special cases: Case 1:θ = 0o

ab

a • b = a b cos 0 = a ba • b a b cos 0 a b

Case 2:θ = 90o a b

a • b = a b cos 90° = 0

22/01/2010 08:56FAP0015 PHYSICS I 12

aθ

a

b

Vector quantity

Cross multiplication: a × b = ab sin θ

q y

22/01/2010 08:56FAP0015 PHYSICS I 13

Special cases: Case 1:θ = 0o

ab

a × b = a b sin 0 = 0a × b a b sin 0 0

Case 2:θ = 90o a b

a × b = a b sin 90° = a b

22/01/2010 08:56FAP0015 PHYSICS I 14

y

ay

i • i = 1 i • j = 0j j 1 j k 0

xaj ax

y j • j = 1 j • k = 0k • k = 1 k • i = 0

xikaz i × j = k i × i = 0

z j × k = i j × j = 0k × i = j k × k = 0a = ax i + ay j + az kk × i j k × k 0

22/01/2010 08:56FAP0015 PHYSICS I 15

Displacement, velocity and acceleration

Displacement Δ 0rrr −=Δ

Average velocity g y

tr

ttrrv0

0av Δ

Δ=

−−

=0

Instantaneous velocitydΔdtdr

trv

0t=

ΔΔ

=→Δ

lim

22/01/2010 08:56FAP0015 PHYSICS I 16

Vector components of v are vx and vy

θθ id θθ sinvv,cosvv yx == and

Average acceleration

tv

ttvvaav Δ

Δ=

−−

=0

0

0

dvvΔInstantaneous acceleration

dtdv

tvlima

t=

ΔΔ

=→Δ 0

∗ The acceleration has a vector components ax and ay in x an y-directions.

22/01/2010 08:56FAP0015 PHYSICS I 17

ExampleA spacecraft has initial velocity component ofv0x= +22m/s and acceleration component ax = +24m/s2. Inv0x 22m/s and acceleration component ax 24m/s . Inthe y-direction it has v0y = +14m/s and ay = +12m/s2. Thedi ti t th i ht d d h b hdirection to the right and upward have been chosen aspositive directions.Find (a) x and vx, (b) y and vy and (c) the final velocity(magnitude and direction) of the spacecraft at(magnitude and direction) of the spacecraft attime, t = 7.0 s.

22/01/2010 08:56FAP0015 PHYSICS I 18

The spacecraft motion is two–dimensional motion

X-part of the motion is independent of the y-part. Similarlythe y-part is independent of x-part of the motion.

22/01/2010 08:56FAP0015 PHYSICS I 19

y p p p

Solution v0x = +22m/s ax = +24m/s2

v0 =+14m/s a = +12m/s2

( )// 7427241722 2 +××+×

a) x and vx

v0y +14m/s ay +12m/s

1 2+ tatvx ( ) mss/mss/m 7427242

722 +=××+×=

m/ss.s/ms/mtavv xxx 1900724220 +=××+=+=20 =+= tatvx xx

1b) y and vy

121 2

0 =+= tatvy yy

( ) m/sss/ms/mtavv 9871214 +++

( ) mss/mss/m 39271221714 2 +=××+×=

( ) m/sss/ms/mtavv yyy 98712140 +=+=+=

22/01/2010 08:56FAP0015 PHYSICS I 20

The magnitude and the direction of spacecraft areThe magnitude and the direction of spacecraft are

s/m)s/m()s/m(vvv yx 21098190 2222 =+=+=

°=⎟⎠⎞

⎜⎝⎛== 2719098tanor 1-θθ

x

y

vv

tan⎠⎝x

+x+x

22/01/2010 08:56FAP0015 PHYSICS I 21

Projectile Motion• Projectile motion is a motion of object in the 2-D plane

under the influence of gravity, as shown in Fig. 2.

• To analyze a projectile motion we need to consider thecomponents of the motion in the x- and y- directionsp yseparately.

• We note that the x-component of the acceleration is zerop(ax= 0), and the y-component is constant and equal to – g org, (ay= g), depending on whether we take upwards to be the+ di ti di ti ti l+ve y-direction or –ve y-direction, respectively.

22/01/2010 08:56FAP0015 PHYSICS I 22

Fig. 2: The trajectory of a body projected with an initial velocity vo at an angle θ above thehorizontal. The distance R is the horizontal range, and h is the maximum height to whichthe particle risesthe particle rises.

vx= vox

v = 0y

vy 0

vy

hvx

vovy= voy

Oθ

vx= vox

vx xR

O x ox

vy= - voyθ=θ= sin and cos 0y00x0 vvvv

22/01/2010 08:56FAP0015 PHYSICS I 23

Projectile MotionIn projectile motion we can express all the vector

relationships in terms of separate equations for the

j

relationships in terms of separate equations for thehorizontal and vertical components. The components ofthe acceleration are:

ax = 0 and ay = - g

Therefore we can still directly use the previousequations of motion with constant acceleration.

22/01/2010 08:56FAP0015 PHYSICS I 24

The components of initial velocity, v0 are

θandθ 0000 sinvv cosvv yx ==

x-direction y-direction

A l ti 0Acceleration

V l it gtvv

ay = −gax = 0

Velocity xx vv 0= gtvv yy −= 0

tΔ 21ΔDisplacement tvx x0=Δ 2

0 2gttvy y −=Δ

22/01/2010 08:56FAP0015 PHYSICS I 25

P ojectile MotionProjectile Motion

• U i th i f ti i i th t bl b d th• Using the information given in the table above and theequations of motion, you can solve any problem dealingwith motion in a plane provided you make an assumptionwith motion in a plane, provided you make an assumptionthat there is no air resistance.

• And always remember that there is no acceleration in thex-direction.

22/01/2010 08:56FAP0015 PHYSICS I 26

Time taken for a projectile to reach the p jmaximum height.

• You can see that the object in the projectile changed itsdirection from going upwards to coming downwards.

• That clearly indicates that it reached a point whereby thevertical velocity was zero.

• The maximum height occurs when vy is equals to zero.

• But the horizontal velocity remained constantBut the horizontal velocity remained constant.

22/01/2010 08:56FAP0015 PHYSICS I 27

Projectile Motionj• Time taken to reach maximum height, when vy=0,

h i b d i• The equation to be used is

gtvv −= and v =00 gtvv yy = and vy 0

0 gtv =

vv y θsin00

0 gtv y

gv

gt y θsin00 ==

22/01/2010 08:56FAP0015 PHYSICS I 28

Projectile Motionj• Maximum height when vy=0

Th i b d i• The equation to be used is

( )ygvv Δ= 222 and v =0( )ygvv yy Δ−= 20and vy 0

Thusvv

hy y

2sin

2

220

20 θ

===ΔThus,

gg 22

22/01/2010 08:56FAP0015 PHYSICS I 29

Projectile Total Flight timeWhen the projectile reach the ground, the y-component of the displacement is zerothe displacement is zero.

01 20 =−=Δ gttvy y 0

20 gttvy y

010 =⎟

⎠⎞

⎜⎝⎛ − gtvt y

021or ,0 0 =−= gtvt y

20 ⎟⎠

⎜⎝

gy 2y

vvt y θsin22 00

ggt y 0==

22/01/2010 08:56FAP0015 PHYSICS I 30

Projectile Total Flight time

• The total flight time will be 2t.

gv22T 0 θ

==sintg

• The time taken to reach the maximum height isthe same as the time taken to reach the groundafter achieving the maximum height.

22/01/2010 08:56FAP0015 PHYSICS I 31

NOTE

• You are advised not to memorize the generalequations above Instead you must trainequations above. Instead you must trainyourself to derive the equations to be usedwhenever it is necessarywhenever it is necessary.

22/01/2010 08:56FAP0015 PHYSICS I 32

Example 1A handball is thrown with an initial vertical velocity componentof 18.0 m/s and a horizontal velocity component of 25.0 m/s.(a) How much time is required for the handball to reach thehighest point of the trajectory?(b) How high is this point?(c) How much time (after being thrown) is required for theh db ll t t t it i i l l l? H d thihandball to return to its original level? How does this comparewith the time calculated in part (a)?(d) How far has it traveled horizontally during this time?(d) How far has it traveled horizontally during this time?

22/01/2010 08:56FAP0015 PHYSICS I 33

Solution to Example 1a) The time required to reach the maximum height is obtained from

s 81m/s819m/s 1800

Thus, 20 .

gv

t y =−−

=−

= 00 ,gtvv yy =−=m/s819.g

m516m/s) (18-00yThus

20v y ==

−=

b) At max height, vy = 0, 0220

2 ,gyvv y =−=

m 516m/s 8192-2

y Thus 2 ..g

=×

=−

=

181 22

c) The time taken to reach its original level y=0,

s635

18Thus518210 22

0 .tttgttvy y ==−=−==

(which is twice with that calculated in (a).

m. 90s 63m/s 250 =×==Δ= .tvxR x

d) It traveled horizontally during this time a distance of

22/01/2010 08:56FAP0015 PHYSICS I 34

Projectile Range• Range, R, is actually x-displacement just as the maximum

height is the vertical displacement.h i d f l l i h i• The equation used for calculating the range is

tvRxxx x00 ==−=Δ

• Range R, here represents the maximum horizontaldisplacement, the time will be the total flight time 2t.

( )g

vg

vvR θθθ 2sinsin2cos200

0 =⎟⎟⎠

⎞⎜⎜⎝

⎛=

θθθ cossin22sin =• Using the trigonometric identity

22/01/2010 08:56FAP0015 PHYSICS I 35

Which launch angle, 30, 45 and 60° gives t t ?greatest range?

• This equation shows R varies with angle as sin2θ. q g

θ2sin20

gvR =g

• Thus R is largest when sin 2θ is largest, that is when sin 2θ=1.

v2

• Since sin 90 = 1, its follows that 2θ = 90 °, thus θ =45°gives the maximum range.

gvR 0

max =

22/01/2010 08:56FAP0015 PHYSICS I 36

Projectile MotionjThe distance r of the projectile from the origin at any time (themagnitude of the position vector r) is given by:g p ) g y

22 yxr +=22The projectiles speed at any time is

Th di i f h l i i i b

2y

2x vvv +=

yvθThe direction of the velocity is given by

The velocity vector v is tangent to the trajectory at each pointx

y

v=θtan

The velocity vector v is tangent to the trajectory at each point.

22/01/2010 08:56FAP0015 PHYSICS I 37

Conceptual QuestionConceptual Question

• A wrench is accidentally dropped from the top of• A wrench is accidentally dropped from the top ofthe mast on a sailboat. Will the wrench hit at thesame place on the deck whether the sailboat is atsame place on the deck whether the sailboat is atthe rest or moving with a constant velocity? Justifyyour answeryour answer.

22/01/2010 08:56FAP0015 PHYSICS I 38

REASONING AND SOLUTION• The wrench will hit at the same place on the deck of the ship

regardless of whether the sailboat is at rest or moving with ag gconstant velocity.

• If the sailboat is at rest, the wrench will fall straight downhitting the deck at some point Phitting the deck at some point P.

• If the sailboat is moving with a constant velocity, the motionof the wrench will be two dimensions. However, thehorizontal component of the velocity of the wrench will be thehorizontal component of the velocity of the wrench will be thesame as the velocity of the sailboat.

• Therefore, the wrench will always remain above the samepoint P as it is fallingpoint P as it is falling.

22/01/2010 08:56FAP0015 PHYSICS I 39

DiscussionDiscussion • Suppose you are driving in convertible with the top• Suppose you are driving in convertible with the top

down. The car is moving to the right at a constantvelocity. You point a gun straight upward and fire it. Iny p g g pthe absence of air resistance, where would the bulletland- behind you, ahead of you, or in the barrel of thegun?

22/01/2010 08:56FAP0015 PHYSICS I 40

Conceptual Questionp• A stone is thrown horizontally from the top of a cliff

d ll hi h d b l A dand eventually hits the ground below. A second stoneis dropped from rest from the same cliff, fallsthrough the same height and also hits the groundthrough the same height, and also hits the groundbelow. Ignore air resistance. Discuss whether each ofthe following quantities is different or the same inthe following quantities is different or the same inthe two cases; if there is difference, describe thedifference: (a) displacement, (b) speed just before( ) p , ( ) p jimpact with the ground and (c ) time of flight.

22/01/2010 08:56FAP0015 PHYSICS I 41

REASONING AND SOLUTION

a) The displacement is greater for the stone that is thrownhorizontally, because it has the same vertical component as thedropped stone and in addition has a horizontal componentdropped stone and, in addition, has a horizontal component.

b) The impact speed is greater for the stone that is thrownhorizontally. The reason is that it has the same verticalo o ta y. e easo s t at t as t e sa e ve t cavelocity component as the dropped stone but, in addition, alsohas a horizontal component that equals the throwing velocity.

c) The time of flight is the same in each case, because the verticalpart of the motion for each stone is the same. That is, eachstone has an initial vertical velocity component of zero andstone has an initial vertical velocity component of zero andfalls through the same height.

22/01/2010 08:56FAP0015 PHYSICS I 42

Conceptual Example

22/01/2010 08:56FAP0015 PHYSICS I 43

22/01/2010 08:56FAP0015 PHYSICS I 44

The package falling from the plane is a projectile motion

22/01/2010 08:56FAP0015 PHYSICS I 45

Relative Velocity y• Relative velocity is the velocity of an object relative to the

b h i ki th tobserver who is making the measurement.

• The velocity of object A relative to object B is written d l i f bj l i i ivAB, and velocity of object B relative to C is written as

vBC. The velocity of A relative C is (note the ordering of subscript)subscript)

BCABAC VVV +=

• While the velocity of object A relative to object B is vAB, the velocity of B relative A is vBA=- vAB

22/01/2010 08:56FAP0015 PHYSICS I 46

TGPTPG VVV += Vector sum of the two- velocity-vectors.TGPTPG VVV + y

VPT= velocity of the Passenger relative to the Train.V = velocity of the Train relative to the GroundVTG= velocity of the Train relative to the Ground.VPG= velocity of the Passenger relative to the Ground.

22/01/2010 08:56FAP0015 PHYSICS I 47

Relative VelocityTGPTPG VVV +=

• VPT= velocity of the Passenger relative to the Train.

• V = velocity of the Train relative to the Ground• VTG= velocity of the Train relative to the Ground.

• VPG= velocity of the Passenger relative to the Ground.

• Each velocity symbol contains two-letter subscript, the1st for the moving body, the 2nd indicates the objectiverelative to it the velocity is measured.

22/01/2010 08:56FAP0015 PHYSICS I 48

Example 1• On a pleasure cruise a boat is traveling to the water at

a speed of 5 0 m/s due south Relative to the boat aa speed of 5.0 m/s due south. Relative to the boat, apassenger walks toward the back of the boat at a speedof 1.5 m/s. (a) What is the magnitude and direction of( ) gthe passenger’s velocity relative to the water? (b) Howlong does it take for the passenger to walk a distance of27 m on the boat? (c) How long does it take for thepassenger to cover a distance of 27 m on the water?

22/01/2010 08:56FAP0015 PHYSICS I 49

REASONING• The time it takes for the passenger to walk the distance on the

boat is the distance divided by the passenger’s speed vboat is the distance divided by the passenger s speed vPBrelative to the boat.

• The time it takes for the passenger to cover the distance on theThe time it takes for the passenger to cover the distance on thewater is the distance divided by the passenger’s speed vPWrelative to the water.

• The passenger’s velocity relative to the boat is given.However, we need to determine the passenger’s velocity

l ti t th trelative to the water.

22/01/2010 08:56FAP0015 PHYSICS I 50

ANSWERS

+= vvv BWPBPW

a) passenger’s velocity relative to the water, vpw

b) The time it takes for the passenger to walk a distance of 27 m

south m/s, 3.5south 5.0m/s,north m/s, 51 =+= .

b) The time it takes for the passenger to walk a distance of 27 m on the boat is

s 18m/s5.1m 27m 27

===PBv

tPB

27 27

c) The time it takes for the passenger to cover a distance of 27 m on the water is

P W

27 m 27 m 7.7 s3 .5 m /s

tv

= = =o e wa e s

22/01/2010 08:56FAP0015 PHYSICS I 51

Example 2p

• The engine of a boat drives it across a river that isg1800 m wide. The velocity vBW of the boat relative tothe water is 4.0 m/s, directed perpendicular to thecurrent, as shown in the Fig. The velocity vws of thewater relative to the shore is 2.0 m/s. a) What is thevelocity vBS of the boat relative to the shore? b) Howlong does it take for the boat to cross the river?

22/01/2010 08:56FAP0015 PHYSICS I 52

WSBwBS vvv +=

m/s 0.4sin == BWBS vv θ

m/s 0.2cos == WSBS vv θ04

°=== 632tan Thus 0204 1-θθ ,

.

.tan

( ) ( ) m/s 5.4m/s 0.2m/s 0.4 2222 =+=+= WSBwBS vvva)

s 450m/s 0.4

m 1800sinv

Width time)BS

===θ

b

22/01/2010 08:56FAP0015 PHYSICS I 53

SummaryThe components of initial velocity, v0 are

θsinv vθ cosvv yx 0000 and ==

X-Direction Y-DirectionX-Direction Y-Direction

AccelerationAcceleration ay = -gax = 0

VelocityVelocity xx vv 0= gtvv yy −= 0

1DisplacementDisplacement tvx x0=Δ 2

0 21 gttvy y −=Δ

22/01/2010 08:56FAP0015 PHYSICS I 54