Motion Along Two or Three
Dimensions
Review
• Equations for Motion Along One Dimension
v ave 
x
v  lim
a ave 
t
t  0
x
t

dx
dt
v
a  lim
t
t  0
v
t

dv
dt
Review
• Motion Equations for Constant Acceleration
• 1.
v  v 0  at
• 2. x  x 0  v 0 t 
1
• 3. v 2  v 02  2 a  x
• 4. v ave 
v  v0
2
2
at
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 3-9
v  735 kph
DN  ?
DW  ?

v
41.5
Vector Addition
A  Ax  A y
A x  A cos 
A y  A sin 
Giancoli Problem 3-9
v x  v sin 

v
v y  v cos 

vx
v x  735 sin( 41 . 5 )
v y  735 cos( 41 . 5 )
41.5

vy
Giancoli Problem 3-9
v x  735 sin( 41 . 5 )  487 . 0 kph

v
• In the Western direction
v y  735 cos( 41 . 5 )  550 . 5 kph
• In the Northern direction

vx
41.5

vy
Giancoli Problem 3-9
v N  550 kph
vW  487 kph
D  vt
D N  1650 km
D W  1461 km  1460 km

D

Dx

41.5 D N
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
125 m
250 m
30 
75 . 0 m

D
150 m
Serway Problem 3-25





D  D1  D 2  D 3  D 4
• 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
125 m
250 m
30 
75 . 0 m

D
150 m
Serway Problem 3-25



D1  D1 x  D1 y
D1 x  0
D 1 y  75 . 0 m



D2  D2x  D2 y
D 2 x  250 m
D2 y  0m



D3  D3 x  D3 y
D 3 x  D 3 cos(  )  (125 ) cos( 30 )  108 . 25 m
D 3 y  D 3 sin(  )  (125 ) sin( 30 )  62 . 5 m



D4  D4x  D4 y
D4x  0m
D 4 y   150 m
• NOT DRAWN TO SCALE
125 m
250 m
30 
75 . 0 m

D
150 m
Serway Problem 3-25





D  D1  D 2  D 3  D 4



D  Dx  Dy
• Where
D x  D1 x  D 2 x  D 3 x  D 4 x
D y  D1 y  D 2 y  D 3 y  D 4 y
• Substitute
D x  250  108 . 25  358 . 25 m
D y  75  62 . 5  150   12 . 5 m
• NOT DRAWN TO SCALE
125 m
250 m
30 
75 . 0 m

D
150 m
Serway Problem 3-25
D x  250  108 . 25  358 . 25 m
D y  75  62 . 5  150   12 . 5 m



D  Dx  Dy
• CAUTION
D  Dx  Dy
• NOT DRAWN TO SCALE
125 m
250 m
75 . 0 m

Dx

D
30 

Dy
150 m
Serway Problem 3-25
D x  250  108 . 25  358 . 25 m
• NOT DRAWN TO SCALE
D y  75  62 . 5  150   12 . 5 m



D  Dx  Dy
D 
D 
D D
2
x

D x  358 . 25
2
y
358 . 25  (  12 . 5 )
2
D  358 . 468  360 m
2


D

D y   12 . 5
NOT DONE YET
D 
358 . 25  (  12 . 5 )
2
2
• NOT DRAWN TO SCALE
D  358 . 468  360 m
tan  
O

Dy
A
  tan
1
(
Dx
Dy
)
Dx
   1 . 998   2 . 0

D  360 m
2 degrees S of E

D x  358 . 25


D

D y   12 . 5
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 
17 . 3 km
19 . 6 km
2 . 2 km
up
Serway 3-44
Vectors
• S = Radar to ship
• P=Radar to plane
136 
• Vector of Plane to ship?
153 
• Let D be plane to ship
• Then

P



PD S
 

D S P

S
17 . 3 km
19 . 6 km
2 . 2 km
up
Express vectors in terms of their
components

S  S x iˆ  S y ˆj  S z kˆ
S  S (sin  ) iˆ  S (cos  ) ˆj  0 kˆ

S  17 . 3 sin( 136 ) iˆ  17 . 3 cos( 136 ) ˆj

S  12 . 02 iˆ  12 . 44 ˆj
Similarly

P  Px iˆ  Py ˆj  Pz kˆ
P  P (sin  ) iˆ  P (cos  ) ˆj  2 . 2 kˆ

P  19 . 6 sin( 153 ) iˆ  19 . 6 cos( 153 ) ˆj  2 . 2 kˆ

P  8 . 90 iˆ  17 . 46 ˆj  2 . 2 kˆ
Vector Addition (Subtraction)
 

D  S P

136 
S  12 . 02 iˆ  12 . 44 ˆj


P  8 . 90 iˆ  17 . 46 ˆj  2 . 2 kˆ
S 17 . 3 km

153

D  3 . 12 iˆ  5 . 02 ˆj  2 . 2 kˆ
19 . 6 km


P
D
Magnitude of Vector D

D  3 . 12 iˆ  5 . 02 ˆj  2 . 2 kˆ
D 
136 
( 3 . 12 )  ( 5 . 02 )  (  2 . 2 )
2
D  6 . 31 km
2
2
153 
19 . 6 km

P

S 17 . 3 km

D
Car on a Curve
What is the Velocity?
Velocity
on a Curve


D
v ave 
t

v  lim
t  0

D
t

t

dD

D
dt

v ave
t0
Lets make Δt smaller

D
v ave 
t

v  lim
t  0

D
t


dD

D2

v 2 ave
dt

v1 ave
t0
t2

D1
t1
Lets make Δt smaller and smaller

D
v ave 
t

v  lim
t  0

D
t


dD
dt

v 4 ave
t0

D4
t4

D3
t3

D2

v 3 ave v
2 ave 
v1 ave
t2

D1
t1
Velocity on a Curve

v  lim
t  0

D
t


dD
dt

v
•
Velocity is tangent to the path
t0
Velocity on a Curve

v  lim
t  0
•
•
•
•

D
t


dD
dt

v1
t1
We can find direction of
velocity at any point in time

v0
Velocity is changing
t0
Acceleration
on a Curve



v
a ave 
t

a  lim
t  0
•
v1

v
t


dv
t1

a ave
dt

v0

v0

v

v1
t0
Acceleration
on a Curve


v
a ave 
t

a  lim
t  0
•

v
t


v2

dv
t2
dt

v0

v0
t0

v

a ave 2

v2
Acceleration
on a Curve



v
a ave 
t

a  lim
t  0

v
t

v2
v1


dv
t1
t2
dt

v0
•
•
Average Acceleration is changing
Acceleration is not constant
t0
Special Cases
• We’re not yet equipped to deal with nonconstant 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.
Equations for Motion with constant
Acceleration
• x-axis
• y-axis
• 1.
v x  v0 x  a x t
• 2.
x  x0  v0 xt 
• 3.
v x  v0 x  2 a x  x
v x  v0 x
v avex 
2
• 4.
2
2
1
2 a xt
2
• 1.
v y  v0 y  a yt
• 2.
y  y0  v0 yt 
• 3.
v v
• 4.
2
y
2
0y
v avey 
1
2 a yt
 2a yy
v y  v0 y
2
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
Equations of Motion for Projectile
Motion
• x-axis (ax=0)
• y-axis (ay=-g)
• 1.
v x  v0 x  a x t
• 2.
x  x0  v0 xt 
• 3.
v x  v0 x  2 a x  x
v x  v0 x
v avex 
2
• 4.
2
2
1
2 a xt
2
• 1.
v y  v0 y  a yt
• 2.
y  y0  v0 yt 
• 3.
v v
• 4.
2
y
2
0y
v avey 
1
2 a yt
 2a yy
v y  v0 y
2
2
Equations of Motion for Projectile
Motion
• x-axis (ax=0)
• 1.
• 2.
v x  v0 x
x  x0  v0 xt
• y-axis (ay=-g)
• 1.
v y  v 0 y  gt
• 2.
y  y0  v0 yt 
• 3.
v v
• 4.
2
y
v avey 
2
0y
1
 2 gy
v y  v0 y
2
2
gt
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 
x0  0
y0  0

v 0  9 .0 m s
• At time t=0.50s
d ?

v ?

v0

v
Split into components

v 0 x  9 .0 m s

v0 y  0



D  Dx  Dy
Dx  x
Dy  y
 

v  vx  vy

v0

v
Calculate components independently
x  x0  v0 xt

v0
x  v 0 x t  ( 9 . 0 )( 0 . 5 )

D
x  4 .5 m
y  y0  v0 yt 
y 
1
2
gt
2
1

y   1 . 225 m
2
1
gt
2
2 ( 9 . 8 )( 0 . 5 )
2

v
Calculate distance
d 
x  y
d 
( 4 . 5 )  (  1 . 225 )
2

v0
2
2
d  4 . 66  4 . 7 m
2

D

v
Calculate components independently
v x  v 0 x  9 .0 m s
v y  v 0 y  gt

v0

vx
v y   gt  (  9 . 8 )( 0 . 5 )
v y   4 .9 m s

vy

v
Calculate velocity
v x  9 .0 m s
v y   4 .9 m s
v
v v
v
(9 .0 )  (  4 .9 )
2
x
v  10 . 25

v0

vx
2
y
2
m
s

vy
2
 1 . 0 x10
m
s

v
Don’t forget direction
v x  9 .0 m s
v y   4 .9 m s

v0
v  1 . 0 x10 m s
vy
 4 .9
tan  

 0 . 544
vx
9
   28 . 56    29 
v  1 . 0 x10
m
s
• 29o below the horizontal

vx

vy


v
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

x0  0
y0  0

v 0  11 . 0 m s   20 
• Find R (horizontal range)
• Find h (maximum height)
Horizontal Range

v 0  11 . 0 m s   20 


v 0 x  v 0 cos   (11 ) cos( 20 )

v 0 x  10 . 34 m s
• No horizontal acceleration
x  x0  v0 xt
x  v0 xt
R  v ox t f
• Where tf is time of flight
How do we find time of flight


v 0 y  v 0 sin   (11 ) sin( 20 )

v 0 y  3 . 762 m s
a y   9 . 80
m
s
2
• y-axis (ay=-g)
• 1.
v y  v 0 y  gt
• 2.
y  y0  v0 yt 
• 3.
v v
• 4.
2
y
v avey 
2
0y
1
 2 gy
v y  v0 y
2
2
gt
2
Many solutions, this is just one
y  y0  v0 yt 
y  v0 yt 
( 3 . 762 ) t 
1
1
2
gt
1
2
2
gt
2
0
0
2 (9 .8 )t
2
t [( 3 . 762 )  ( 4 . 9 ) t ]  0
t0
( 3 . 762 )  ( 4 . 9 ) t  0
t  0 . 7678 s
Range
• t=0, 0.7678
• t=0 is when the runner begins
his flight
• tf=0.7678s
R  v ox t f
R  (10 . 34 )( 0 . 7678 )
R  7 . 94 m
Max Height?
• At peak,

vy  0

v 0 y  3 . 762
a y   9 . 80
m
s
m
s
2
Max Height?
• At peak,

vy  0

v 0 y  3 . 762
a y   9 . 80
m
s
m
s
2
v y  v0 y  2 g y
2
2
0  ( 3 . 762 )  2 ( 9 . 8 )  y
2
 y  0 . 722 m
h  0 . 722 m
General Equations about h and R
• Time to reach max height
v y  v 0 y  gt
• At peak vy=0
• Time to reach max range
y  y0  v0 yt 
0  v0 yt 
0  v 0 y  gt
gt  v 0 y
gt  v 0 sin 
t
v 0 sin 
g
1
2
1
gt
2
gt
2
2
• t= 0 is a trivial solution
0  v0 y 
1
2
1
2
gt
gt  v 0 y  v 0 sin 
t
2 v 0 sin 
g
General Equations about h and R
• Max height
y  v0 yt 
1
2
gt
2
0
 v 0 sin  
 
y  v 0 sin  
g


h
( v 0 sin  )
2g
2
1
2
 v 0 sin  

g 
g


2
General Equations about h and R
• Range
R  v ox t f
R  v 0 cos 
R 
2 v 0 sin 
v 2 cos  sin 
2
0
g
R 
g
v sin 2
2
0
g
CAUTION
• Only valid when Δy=0
h
( v 0 sin  )
2g
R 
v sin 2
2
0
g
2
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?