3. Motion in Two
and Three Dimensions
2
0
( )
v t v
a
t
Recap:
Constant
Acceleration
0
0
( )
t
x x v t dt
Area under
the function
v
(
t
).
3
Recap:
Constant
Acceleration
a
d
t
v
d
0
v
a
v t
2
1
0 0
2
x x v t
a
t
2 2
0 0
2 ( )
v x
a
v
x
4
Recap: Acceleration due to
Gravity (Free Fall)
In the absence of air resistance
all
objects
fall with the
same
constant
acceleration
of about
g = 9.8 m/s
2
near
the Earth’s surface.
5
Recap: Example
A ball is thrown upwards at
5 m/s,
relative to the ground
,
from a height of 2 m.
We need to choose a
coordinate system
.
2 m
5 m/s
6
Recap: Example
Let’s measure
time from when
the ball is launched.
This defines
t
= 0.
Let’s choose
y
= 0 to be ground level
and
up
to be the
positive
y
direction.
y
0
= 2 m
v
0
= 5 m/s
y
7
Recap: Example
1
. How high above
the ground will the
ball reach?
with
a
=
–
g
and
v
= 0.
2 2
0 0
2 ( )
v y
a
v
y
use
y
0
= 2 m
v
0
= 5 m/s
y
8
Recap: Example
2
1
0 0
2
y v t
a
y
t
Use
2.
How long does it
take the ball to
reach the ground?
y
0
= 2 m
v
0
= 5 m/s
y
with
a
=
–
g
and
y
= 0.
9
Recap: Example
2 2
0 0
2 ( )
y
a
v v y
Use
3.
At what speed
does the ball hit
the ground?
y
0
= 2 m
v
0
= 5 m/s
y
with
a
=
–
g
and
y
= 0.
Vectors
11
A
vector
is a mathematical quantity
that has two properties:
direction
and
magnitude
.
Vectors
One way to
represent
a vector
is as an
arrow
: the arrow gives
the direction and its length the
magnitude.
12
Position
A position
p
is a vector: its
direction
is from
o
to
p
and its
length
is the distance from
o
to
p
.
A vector is usually
represented by a
symbol like .
p
r
13
Displacement
A
displacement
is another example of
a vector.
r
14
Vector Addition
The order in which the
vectors are added does
not matter, that is, vector
addition is
commutative
.
15
Vector Scalar Multiplication
A
a
A
a
and
–
q
are
scalars
(numbers).
A
q
16
( 1)
C A B
Vector Subtraction
If we multiply a vector by
–
1
we reverse its
direction, but keep its magnitude the same.
Vector subtraction
is really vector
addition with
one vector reversed.
17
Vector Components
A
cos
q
is the
component
, or
the
projection
, of
the vector
A
along
the vector
B
.
q
A
B
cos
A
q
18
Vector Components
19
Vector Addition using
Components
x x x
y y y
C A B
C A B
20
Unit Vectors
If the vector
A
is multiplied
by the scalar
1/A
we get a new
vector of
unit length
in the
same direction
as vector
A
;
that is, we get a
unit vector
.
From the components,
A
x
,
A
y
, and
A
z
, of a vector,
we can compute its length,
A
, using
2 2 2
x y z
A A A A
A
ˆ
1
A
A
A
21
Unit Vectors
It is convenient to define
unit vectors
parallel to the
x
,
y
and
z
axes, respectively.
ˆ
ˆ ˆ
x y z
A Ai A j Ak
ˆ
ˆ ˆ
,,
i j k
Then, we can write a
vector
A
as follows:
Velocity and Acceleration
Vectors
23
Velocity
0
lim
t
dr r
v
dt t
v
2 1
r r r
24
Acceleration
0
lim
t
dv v
a
dt t
2 1
v v v
Relative Motion
26
Relative Motion
Velocity of plane
relative to air
Velocity of air
relative to ground
Velocity of plane
relative to ground
v
V
V
v
v
N
S
W
E
v
V
v
q
27
Example
–
Relative Motion
A pilot wants to fly plane due north
Airspeed:
200 km/h
Windspeed
:
90 km/h
direction:
W to E
1. Flight heading?
2. Groundspeed?
N
S
W
E
v
V
v
q
Coordinate system
:
î
points from
west to east and
ĵ
points from south to north.
28
ˆ ˆ
90 0 km/h
ˆ
cos(/2 )
ˆ
sin(
20
/2 ) k/
0
2
h
00
m
V
i
i
j
v
j
q
q
ˆ ˆ
0 km/h
v i
V
v j
v
Example
–
Relative Motion
N
S
W
E
v
V
v
q
29
200
ˆ
90 km/h
ˆ ˆ
( sin cos ) km/h
V
i
v
i
j
q q
ˆ ˆ
( sin 90) cos
km
20
h
0
/
200
v i j
q q
Example
–
Relative Motion
N
S
W
E
v
V
v
q
30
Equate
x
components
0 =
–
200 sin
q
90
q
= sin

1
(90/200)
=
26.7
o
west of north.
0
( sin 90) cos
200
km
ˆ
200
ˆ
ˆ
ˆ
/h
v
j
v V
j
v
i
i
q q
Example
–
Relative Motion
N
S
W
E
v
V
v
q
31
Example
–
Relative Motion
Equate
y
components
v
= 200 cos
q
=
179 km/h
N
S
W
E
v
V
v
q
200
0
( sin 90
ˆ
) cos
ˆ
ˆ
ˆ
200
v
v i
i
V v
j
j
q q
Projectile Motion
33
0
r
r
Projectile Motion under
Constant
Acceleration
Coordinate system
:
î
points to the right,
ĵ
points upwards
ˆ
j
ˆ
i
34
0
r
)
ˆ
0 (
ˆ
i
j
a g
2
1
0
2
v t at
r
2
2
0
1
0
0
v
r r
v v
t
t
t
a
a
R
= Range
Impact point
Projectile Motion under
Constant
Acceleration
35
Projectile Motion under
Constant
Acceleration
0 0
0 0
cos
sin
x
y
v v
v v
q
q
Strategy: split motion into
x
and
y
components.
2
1
0 0
2
0 0
y
x
y y v t gt
x x v t
R
= Range
R
=
x

x
0
h
=
y

y
0
36
Projectile Motion under
Constant
Acceleration
2
0
2 2 0
y
gt v t h
2
0 0
2
y y
v v gh
t
g
Find time of flight by solving
y
equation:
0
x
R v t
And find
range from:
37
Projectile Motion under
Constant
Acceleration
Special case:
y
=
y
0
, i.e.,
h
= 0
0 0
2
0
2
sin 2
x y
v v
R
g
v
g
q
R
y
0
y
(
t
)
Uniform Circular Motion
39
Uniform Circular Motion
ˆ
c
)
ˆ
(
s
os
in
i
r r
j
q
q
r
= Radius
r
ˆ
i
ˆ
j
(,)
x y
y
x
O
40
Uniform Circular Motion
ˆ
co
(
ˆ
sin
)
s
dr
v
dt
d
r
d
j
i
t
q
q
Velocity
41
Uniform Circular Motion
ˆ
cos
ˆ
cos
ˆ
sin
ˆ
sin
ˆ
sin
ˆ
c
( )
o
( )
(
s
)
d
v r
dt
i
i
i
j
d
d
j
j
d
r
d
r
t
d
dt
q
q
q
q
q
q
q
q
q
d
q
/
dt
is called the
angular
velocity
42
Uniform Circular Motion
2
ˆ
co
( )
( )
s
ˆ
sin
ˆ
co
s
s
ˆ
in
dv
a
dt
d
r
dt
d
dt
i
i
j
j
d
dt
r
q
q
q
q
q
q
Acceleration
For uniform motion
d
q
/
dt
is
constant
43
Uniform Circular Motion
2
2
(
ˆ
cos
ˆ
sin
)
ˆ
a r
r
d
d
j
r
d
t
i
t
d
q
q
q
q
Acceleration
is towards
center
Centripetal
Acceleration
44
Uniform Circular Motion
2
v
a
r
Magnitudes
of velocity
and centripetal
acceleration are
related
as follows
45
Uniform Circular Motion
2
r
v
T
Magnitude
of velocity
and period
T
related as
follows
r
46
Summary
In general, acceleration changes both
the magnitude and direction of the
velocity.
Projectile motion results from the
acceleration due to gravity.
In uniform circular motion, the
acceleration is
centripetal
and has
constant magnitude
v
2
/
r
.
47
How to Shoot a Monkey
x
= 50 m
h
= 10 m
H
= 12 m
Compute
minimum
initial
velocity
H
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