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?