Motion Along Two or Three Dimensions

Review•Equations for Motion Along One

Dimension

dtdx

txv

txv

t

ave

0lim

dtdv

tva

tva

t

ave

0lim

Review•Motion Equations for Constant

Acceleration•1.

•2.

•3.

•4.

atvv 0

221

00 attvxx

20vvvave

xavv 220

2

Slow Down

Giancoli Problem 3-9•An airplane is traveling 735 km/hr in a

direction 41.5o west of north. How far North and how far West has the plane traveled after 3 hours?

Problem Solving Strategy•Define your origin•Define your axis•Write down the given (as well as what

you’re looking for)•Reduce the two dimensional problem into

two one dimensional problems. •Choose which of the four equations would

work best

Giancoli Problem 3-941.5

v

??

735

W

N

DD

kphv

Vector Addition

yx AAA cosAAx sinAAy

Giancoli Problem 3-941.5

vsinvvx

cosvvy yv

xv

)5.41sin(735xv)5.41cos(735yv

Giancoli Problem 3-941.5• In the Western direction

• In the Northern direction

v

yv

xv

kphvx 0.487)5.41sin(735

kphvy 5.550)5.41cos(735

Giancoli Problem 3-9

kmkmDkmD

vtDkphvkphv

W

N

W

N

146014611650

487550

41.

5D

ND

xD

X and Y components are independent• What happens along x

does not affect y• What happens along y

does not affect x

• We can break down 2 dimensional motion as if we’re dealing with two separate one dimensional motions,

Serway Problem 3-25• While exploring a cave, a

spelunker starts at the entrance and moves the following distances. She goes 75.0m N, 250m E, 125m at an angle 30.0 N of E, and 150m S. Find the resultant displacement from the cave entrance.

• NOT DRAWN TO SCALE

D

m0.75

m250 m125

m150

30

Serway Problem 3-25

• If you have the time and patience you can draw this system and solve the problem graphically.

Or

• Separate the vectors into their components.

• NOT DRAWN TO SCALE

D

m0.75

m250 m125

m150

30

4321 DDDDD

Serway Problem 3-25• NOT DRAWN TO SCALE

D

m0.75

m250 m125

m150

30

mDD

DDD

y

x

yx

0.750

1

1

111

mDmD

DDD

y

x

yx

0250

2

2

222

mDDmDD

DDD

y

x

yx

5.62)30sin()125()sin(25.108)30cos()125()cos(

33

33

333

mDmD

DDD

y

x

yx

1500

4

4

444

Serway Problem 3-25

• Where

• Substitute

• NOT DRAWN TO SCALE

D

m0.75

m250 m125

m150

30

4321 DDDDD

yx DDD

xxxxx DDDDD 4321

yyyyy DDDDD 4321

mDx 25.35825.108250 mDy 5.121505.6275

Serway Problem 3-25

• CAUTION

• NOT DRAWN TO SCALE

D

m0.75

m250 m125

m150

30yx DDD

mDx 25.35825.108250 mDy 5.121505.6275

yx DDD xD

yD

Serway Problem 3-25• NOT DRAWN TO SCALE

D

yx DDD

mDx 25.35825.108250 mDy 5.121505.6275

25.358xD

5.12yD

22yx DDD

mD

D360468.358

)5.12(25.358 22

2 degrees S of E

NOT DONE YET• NOT DRAWN TO SCALE

D

25.358xD

5.12yD

mDD

360468.358)5.12(25.358 22

0.2998.1

)(tan

tan

1

x

y

x

y

DDDD

AO

mD 360

Lets add another dimension• Serway 3-44• A radar station locates a

sinking ship at range 17.3 km bearing 136o clockwise from north. From the same station, a rescue plane is at horizontal range 19.6 km, 153o clockwise from north, with elevation 2.20 km. a) find position vector for the ship relative to the plane, letting i represent East, j represent north and k up. b) How far apart are the plane and the ship?

136

153 km3.17

km6.19

upkm2.2

Serway 3-44Vectors• S = Radar to ship• P=Radar to plane

• Vector of Plane to ship?

• Let D be plane to ship• Then

136

153 km3.17

km6.19

upkm2.2

P

S

PSD

SDP

Express vectors in terms of their components

kSjSiSS zyxˆˆˆ

kjSiSS ˆ0ˆ)(cosˆ)(sin

jiS ˆ)136cos(3.17ˆ)136sin(3.17

jiS ˆ44.12ˆ02.12

SimilarlykPjPiPP zyxˆˆˆ

kjPiPP ˆ2.2ˆ)(cosˆ)(sin

kjiP ˆ2.2ˆ)153cos(6.19ˆ)153sin(6.19

kjiP ˆ2.2ˆ46.17ˆ90.8

Vector Addition (Subtraction)

136

153 km3.17

km6.19P

S

PSD

D

jiS ˆ44.12ˆ02.12

kjiP ˆ2.2ˆ46.17ˆ90.8

kjiD ˆ2.2ˆ02.5ˆ12.3

Magnitude of Vector D

136

153 km3.17

km6.19P

S

D

kjiD ˆ2.2ˆ02.5ˆ12.3

222 )2.2()02.5()12.3( D

kmD 31.6

Car on a Curve

What is the Velocity?

Velocity on a Curve

0t

t

dtDd

tDv

tDv

t

ave

0lim

D

avev

Lets make Δt smaller

0t

1t

dtDd

tDv

tDv

t

ave

0lim

1D

avev1

2t

2D

avev2

Lets make Δt smaller and smaller

0t

1t

dtDd

tDv

tDv

t

ave

0lim 1D

avev1

2t

2D

avev2

3t

3D4t

4D

avev3

avev4

Velocity on a Curve

• Velocity is tangent to the path

0t

dtDd

tDv

t

0lim

v

Velocity on a Curve

• We can find direction of

• velocity at any point in time

• • Velocity is changing

0t

dtDd

tDv

t

0lim

0v

1t

1v

Acceleration on a Curve

•

0t

0v

1t

1v

dtvd

tva

tva

t

ave

0lim

0v

1v

v

avea

Acceleration on a Curve

•

0t

0v

2t

2v

dtvd

tva

tva

t

ave

0lim

0v

2v

v2avea

Acceleration on a Curve

• Average Acceleration is changing

• Acceleration is not constant

0t

0v

2t

2v

dtvd

tva

tva

t

ave

0lim

1t

1v

Special Cases•We’re not yet equipped to deal with non-

constant acceleration.•So lets first examine some situations

where acceleration is constant.

Projectile Motion•A projectile is any body that is given an

initial velocity and then follows a path determined entirely by gravity and air resistance.

•For simplicity lets ignore air resistance first.

•The trajectory is the path a projectile takes.

•We don’t care about how the projectile was launched or how it lands. We only care about the motion when it’s in free fall.

Projectile Motion - Trajectory• Follows Parabolic path

(proof algebra) • Velocity is always tangent

to the path• Since acceleration is

purely downwards, motion is constrained to two dimension.

Projectile Motion - Trajectory

Projectile Motion - Trajectory

Projectile Motion - Components•Reduce the velocity vector to its

components.•These components are orthogonal to each

other so they have no effect on each other.

•Motion along each axis is independent.•We can then use the equations of motion

in one direction.

• x-axis

• 1.

• 2.

• 3.

• 4.

Equations for Motion with constant Acceleration

tavv xxx 0

221

00 tatvxx xx

20xx

avexvvv

xavv xxx 220

2

• y-axis

• 1.

• 2.

• 3.

• 4.

tavv yyy 0

221

00 tatvyy yy

20 yy

avey

vvv

yavv yyy 220

2

But Wait•In projectile motion, only gravity is acting

on the object•a=-g=-9.80m/s2•What are the components of this

acceleration

But Wait•In projectile motion, only gravity is acting

on the object•a=-g=-9.80 m/s2

•What are the components of this acceleration

•ay=-9.80 m/s2

•ax= 0 there is NO x-component

• x-axis (ax=0)

• 1.

• 2.

• 3.

• 4.

Equations of Motion for Projectile Motion

tavv xxx 0

221

00 tatvxx xx

20xx

avexvvv

xavv xxx 220

2

• y-axis (ay=-g)

• 1.

• 2.

• 3.

• 4.

tavv yyy 0

221

00 tatvyy yy

20 yy

avey

vvv

yavv yyy 220

2

• x-axis (ax=0)

• 1.

• 2.

Equations of Motion for Projectile Motion

xx vv 0

tvxx x00

• y-axis (ay=-g)

• 1.

• 2.

• 3.

• 4.

gtvv yy 0

221

00 gttvyy y

20 yy

avey

vvv

ygvv yy 220

2

Example•A motorcycle stuntman rides over a cliff.

Just at the cliff edge his velocity is completely horizontal with magnitude 9.0 m/s. Find the motorcycles position, distance from the cliff edge, and velocity after 0.50s.

List the given• Origin is cliff edge• a=-g=-9.80m/s2

• At time t=0s

• At time t=0.50s

0v

v00 x 00 y

?d?v

smv 0.90

Split into components

0v

v

yDxD

DDD

y

x

yx

yx vvv

sm

xv 0.90

00 yv

Calculate components independently

0v

vmxtvx

tvxx

x

x

5.4)5.0)(0.9(0

00

mygty

gttvyy y

225.1)5.0)(8.9( 2

212

21

221

00

D

Calculate distance

0v

vmdd

yxd

7.466.4)225.1()5.4( 22

22

D

Calculate components independently

0v

vxv

yv

sm

xx vv 0.90

sm

y

y

yy

v

gtv

gtvv

9.4

)5.0)(8.9(0

Calculate velocity

0v

vxv

yv

sm

xv 0.9s

myv 9.4

sm

sm

yx

xvv

vvv

100.125.10)9.4()0.9( 22

22

Don’t forget direction

• 29o below the horizontal

0v

v

xv

yv

sm

xv 0.9s

myv 9.4

smxv 100.1

544.099.4tan

x

y

vv

2956.28

smxv 100.1

Another Example• A long jumper leaves the

ground at an angle 20o above the horizontal and at a speed of 11.0 m/s. a) How far does he jump in the horizontal direction? (assume his motion is equivalent to a particle) b) What is the maximum height reached?

Origin• Origin is at point jumper

leaves the ground• At t=0

• Find R (horizontal range)• Find h (maximum height)

00 x 00 y

smv 0.110

20

Horizontal Range

• No horizontal acceleration

• Where tf is time of flight

smv 0.110

20

sm

x

x

vvv34.10

)20cos()11(cos

0

00

fox

x

x

tvRtvx

tvxx

0

00

How do we find time of flight

280.9

762.3

)20sin()11(sin

0

00

sm

y

sm

y

y

a

v

vv

• y-axis (ay=-g)

• 1.

• 2.

• 3.

• 4.

gtvv yy 0

221

00 gttvyy y

20 yy

avey

vvv

ygvv yy 220

2

Many solutions, this is just one2

21

00 gttvyy y 02

21

0 gttvy y

0])9.4()762.3[(0)8.9()762.3( 2

21

tttt

stt

t

7678.00)9.4()762.3(

0

Range• t=0, 0.7678• t=0 is when the runner

begins his flight• tf=0.7678s

mRR

tvR fox

94.7)7678.0)(34.10(

Max Height?• At peak,

280.9

762.3

0

0

sm

y

sm

y

y

a

v

v

Max Height?• At peak,

280.9

762.3

0

0

sm

y

sm

y

y

a

v

v

mhmy

y

ygvv yy

722.0722.0

)8.9(2)762.3(0

22

20

2

General Equations about h and R• Time to reach max height

• At peak vy=0

• Time to reach max range

• t= 0 is a trivial solution

gtvv yy 0

gvt

vgt

vgt

gtv

y

y

sinsin

0

0

0

0

0

221

00 gttvyy y 2

21

00 gttv y

gvt

vvgt

gtv

y

y

sin2

sin

0

0

0021

21

0

General Equations about h and R• Max height

0221

0 gttvy y

20

2

0210

0

2)sin(

sinsinsin

gvh

gvg

gvvy

General Equations about h and R• Range

foxtvR

gvR

gvR

gvvR

2sin

sincos2

sin2cos

20

20

00

CAUTION• Only valid when Δy=0

20

2)sin(

gvh

gvR 2sin20

Young & Freedman Problem 3.10•A military helicopter is flying horizontally at a

speed 60.0 m/s. and accidentally drops a bomb (not armed) at an elevation of 300m. Ignoring air resistance, find a) How much time is required for the bomb to reach the earth? b)How far does it travel horizontally while falling? c) Find horizontal and vertical components just before hitting the earth. d) if velocity of the helicopter remains constant, where is the helicopter when the bomb hits the ground.

Giancoli 3-20•Romeo is chucking pebbles at Juliet’s

window, and wants to hit the window with only a horizontal component of velocity. He is standing at the edge of a rose garden 4.5 m below her window and 5.0m from the base of the wall. How fast are the pebble when they hit the window?

Young & Freedman 3.24• Firemen are shooting a stream of water at a burning

building using high pressure hose that shoots water at a speed at 25.0 m/s. Once it leaves the hose the water travels in projectile motion. Firemen adjust the angle of elevation α of the hose such that it takes 3.00 seconds to reach a building 45 m away. Ignore air resistance and assume hose is at ground level. (a) find angle of elevation α. (b) Find speed and acceleration of water at its highest point. (c) How high above the ground does the water hit the building and how fast is it moving just before it hits the building?

Giancoli 3-35• A rescue plane wants to drop supplies to isolated

mountain climbers on a rocky ridge 235m below. If the plane is travelling horizontally with a speed of 250 km/h (69.4 m/s), (a) how far in advanced (horizontal range) must the goods be dropped? (b) Suppose, instead the plane releases the supplies a horizontal distance of 425m in advance of the mountain climbers. What vertical velocity should supplies be given so that they arrive precisely at the climbers position? (c) With what speed do the supplies land in the latter case?