Physics 204A Class Notes
4-1
Section 4 - Motion in More Than One Dimension
Outline
1. The Definitions of Position, Displacement, Velocity and Acceleration
2. The Special Case of Projectile Motion
3. The Special Case of Uniform Circular Motion
What do objects do?
They move in more than one dimension.
In this section we will describe motion in
two dimensions by extending the same ideas of position, displacement, velocity and acceleration.
The
“good news” is we won’t change their definitions at all.
The “bad news” is that we’ll let them become
vectors.
So we will trade off having to define more concepts for increasingly complex math.
This is the
first example of the way that physics tries to minimize the number of physical concepts and, at the same
time, explain the widest possible variety of phenomena.
1. The Definitions of Position, Displacement, Velocity and Acceleration
The definitions of position, displacement, velocity and acceleration all stay the same, they just
become vectors.
Position: The location of an object with respect to a coordinate system.
Mathematically, position is now a vector
r
r
=
x
ˆ
i
+
y
ˆ
j
.
Displacement: A change in position.
Now displacement is a vector as well,
!
r
r
"
r
r
f
#
r
r
i
=
(x
f
ˆ
i
+
y
f
ˆ
j )
#
i
ˆ
i
+
y
i
ˆ
j )
=
(x
f
#
x
i
)
ˆ
i
+
(y
f
#
y
i
)
ˆ
j
!
r
r
=
!
x
ˆ
i
+
!
y
ˆ
j
Average Velocity: The average rate of displacement.
Mathematically,
r
v
!
"
r
r
"
t
.
Using the displacement,
r
v
=
!
x
ˆ
i
+
!
y
ˆ
j
!
t
=
!
x
!
t
ˆ
i
+
!
y
!
t
ˆ
j
=
v
x
ˆ
i
+
v
y
ˆ
j
.
The definition of speed is now a bit more complex because the velocity is a vector.
Speed: The magnitude of the velocity vector.
Mathematically,
v
!
r
v
=
x
2
+
v
y
2
y
x
r
r
y
x
∆
y
∆
x
y
f
r
r
r
∆
r
x

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4-2
Example 4.1: A hiker walks from a position of 3.00km due north of a ranger station to 1.00km
east of the ranger station in 0.500h.
Using an origin at the ranger station, find (a)the initial
position, (b)the final position, (c)the displacement, (d)the average velocity and (e)the average
speed of the hiker.
Given: initial position 3.00km due north, final position 1.00km east,
and
∆
t = 0.500h
Find:
r
r
i
,
r
r
f
,
"
r
r
,
r
v
, and v.
(a)Using the coordinates shown at the right,
r
r
i
=
3.00
ˆ
j
and
(b)
r
r
f
=
1.00
ˆ
i
.
(c)Using the definition of displacement,
!
r
r
"
r
r
f
#
r
r
i
=
1.00
ˆ
i
#
(3.00
ˆ
j )
$
"
r
r
=
1.00
ˆ
i
#
3.00
ˆ
j
.
(d)Using the definition of average velocity,
r
v
!
"
r
r
"
t
=
1.00
ˆ
i
#
3.00
ˆ
j
0.500
$
r
v
=
2.00
ˆ
i
!
6.00
ˆ
j
in km/h.
(e)Using the definition of speed,
v
!
r
v
=
x
2
+
v
y
2
=
2
+
6
2
"
v
=
6.32km/h
.
Average Acceleration: The rate of change of velocity.
Mathematically,
r
a
!
"
r
v
"
t
.
Using the velocity,
r
a
=
r
v
f
!
r
v
i
"
t
=
(v
xf
ˆ
i
+
v
yf
ˆ
j )
!
(v
xi
ˆ
i
+
v
yi
ˆ
j )
"
t
=
(v
xf
!
v
xi
)
ˆ
i
+
yf
!
v
yi
)
ˆ
j
"
t
=
"
v
x
"
t
ˆ
i
+
"
v
y
"
t
ˆ
j
=
a
x
ˆ
i
+
a
y
ˆ
j
.
The extension from two dimensions to three dimensions you could probably guess.
The definitions stay
the same, but the vectors now have a z-component.

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