Chapter 3: Motion in 2 or 3
Dimensions
Position & Velocity Vectors
Position Vector

To describe the motion of a particle in space, we first need
to describe the position of the particle.

Position vector of a particle is a vector that goes from the
origin of coordinate system to the point.

Position vector components are the Cartesian coordinates
of the particle.

ˆ
ˆ
ˆ
r  xi  yj  zk
Position & Velocity Vectors

As the particle moves through space, the path is a curve.

The change in position (the displacement) of a particle
during time interval t is:
2
1
2
1
2
1
2
1
  
r  r  r  ( x  x )iˆ  ( y  y ) ˆj  ( z  z )kˆ

Average velocity vector during this
time interval is the displacement
divided by the time interval:
 


r2  r1 r
vav 

t 2  t1 t
Position & Velocity Vectors

Instantaneous velocity vector is the limit of the average
velocity as the time interval approaches zero, and equals the
instantaneous rate of change of position with time:



r d r
v  lim

t  0  t
dt

Magnitude of the vector v at any
instant is the speed v of the particle at
that instant.

Direction of v at any instant is the
same as the direction in which particle
moves at that instant.
Position & Velocity Vectors

As t0, P1 and P2 move closer and in this limit vector r
becomes tangent to the curve.

Direction of r in the limit is the same as direction of
instantaneous velocity v.

At every point along the path, instantaneous velocity vector v
is the tangent to the path at that point.
dx
vx 
dt
dy
vy 
dt
dz
vz 
dt
Position & Velocity Vectors

Components of instantaneous velocity vector v :
dx
dy
dz
vx 
vy 
vz 
dt
dt
dt

 dr dx ˆ dy ˆ dz ˆ
v
 i
j k
dt dt
dt
dt

Magnitude of vector v by
Pythagorean theorem:

2
2
2
v  v  vx  v y  vz
Acceleration Vector
Acceleration Vector

Acceleration of a particle moving in space describes how the
velocity of particle changes.

Average acceleration is a vector change in velocity divided
by the time interval:
 


v2  v1 v
aav 

t 2  t1 t
Acceleration Vector

Instantaneous acceleration is the limit of the average
acceleration as the time interval approaches zero, and equals
the instantaneous rate of change of velocity with time:
dv x
ax 
dt
ay 
dv y
dt
dv z
az 
dt



v dv
a  lim

t 0 t
dt
Acceleration Vector

In terms of unit vectors:
d 2x
ax  2
dt
 dvx ˆ dv y ˆ dvz ˆ
a
i
j
k
dt
dt
dt
d2y
ay  2
dt
d 2z
az  2
dt
 d 2x ˆ d 2 y ˆ d 2z ˆ
a  2 i  2 j 2 k
dt
dt
dt
Acceleration Vector

Components of acceleration:

Acceleration vector can be resolved into a component parallel
to the path (and velocity), and a component perpendicular to
the path.
Acceleration Vector

Components of acceleration:

When acceleration vector is parallel to the path (and
velocity), the magnitude of v increases, but its direction
doesn’t change

When acceleration vector is perpendicular to the path
(and velocity), the direction of v changes, but magnitude
is constant
Acceleration Vector

Components of acceleration for a
particle moving along a curved path:
A. Constant speed
B. Increasing speed
C. Decreasing speed
Projectile Motion
Projectile Motion

A projectile is any object that is given an initial velocity and
then follows a path (trajectory) determined solely by gravity
and air resistance.

The motion of a projectile will take place in a plane (so, it is 2D motion).
For projectile motion we can analyze the x- and y-components
of the motion separately.




The horizontal motion (along the x-axis) will have zero
acceleration and thus have constant velocity.
The vertical motion (along the y-axis) will have constant
downward acceleration of magnitude g = 9.80 m/s2.
The initial velocity components, vox and voy, can be expressed
in terms of the magnitude vo and direction ao of the initial
velocity.
Projectile Motion
Projectile Motion

We analyze projectile motion as a
combination of horizontal motion with
constant velocity and vertical motion
with constant acceleration
ax  0  vx  v0 x , x  x0  v0 xt
a y   g  v y  v0 y  gt ,
y  y0  v0 y t  0.5 gt 2

Initial velocity is represented by its
magnitude and direction
v0 x  v0 cos a 0 , v0 y  v0 sin a 0
Projectile Motion

Trajectory of a body projected with initial velocity v0

h is maximum height of trajectory

R is horizontal range
x  (v0 cos a 0 )t
y  (v0 sin a 0 )t  0.5 gt 2
v x  v0 cos a 0
v y  v0 sin a 0  gt
Projectile Motion
Projectile Motion

Trajectory of a cow

A cow is launched from the top of a hill with an initial velocity
vector that makes an angle of 45 degrees with the horizontal.
The projectile lands at a point that is 10 m vertically below the
launch point and 300 m horizontally away from the launch point.
A.
B.
Determine the time the cow was in the air.
Determine the initial speed of the cow.
Projectile Motion

Initial speed of the cow
g
2
y  (tan a 0 ) x  2
x
2v0 cos 2 a 0
Projectile Motion

Flight time of the cow
x  (v0 cos a 0 )t
y  (v0 sin a 0 )t  0.5gt 2
Motion in a Circle
Motion in a Circle

When a particle moves along a curve, direction of its
velocity vector changes.

Particle must have component of acceleration  to the
curved path even if the speed is constant.

Motion in a circle is a special case of motion along a
curved path.

Uniform circular motion - when a particle moves with
constant speed

Non-uniform circular motion - if the speed of a particle
varies.
Uniform Circular Motion

No component of acceleration parallel (tangent) to the
path. Otherwise, speed would change.

Non-zero component of acceleration is perpendicular to
the path.
Uniform Circular Motion

A particle that is undergoing motion in such a manner that its direction
is changing is experiencing a radial acceleration that has magnitude
equal to the square of its velocity divided by the instantaneous radius
of curvature of its motion. The direction of this radial, or centripetal,
acceleration is toward the center of circular path of particle's motion.

v
s

v1
R
v1 s v1
s
 lim
t 0 R t
R t 0 t
a  lim
arad
v2

R
 v1
v  s
R
aav 

v
t

v1 s
R t
Uniform Circular Motion

In uniform circular motion, the
magnitude a of instantaneous
acceleration is equal to the
square of the speed v divided by
the radius R of the circle.

Its direction is  to v and inward
along the radius.

Centripetal  “seeking the
center” (Greek)
Uniform Circular Motion


Period of the motion T  the time of one revolution, or one
complete trip around the circle.
2R
v
In time T, particle travels the distance 2R
of the circle, so its speed can be expressed as
arad
v2

R
arad
4R
 2
T
T
Motion in a Circle: Example
Centripetal acceleration on a curved road

A car has a “lateral acceleration” of 0.87g, which is
(0.87)/(9.8m/s2)=8.5m/s2. This represents the maximum centripetal
acceleration that can be attained without skidding out of the circular
path. If the car is traveling at a constant speed 40m/s (~89mi/h, or
144km/h), what is the max radius of curve it can negotiate?
IDENTIFY and SET UP

Car travels along a curve, speed is constant  apply equation of
circular motion to find the target variable R.
EXECUTE

We know arad and v, so
we can find R:
arad
v2

R
(40 ms ) 2
v2
R

 190m
m
arad
8.5 s 2
Non-Uniform Circular Motion

An object that is undergoing non-uniform circular motion, or motion
where the magnitude and the direction of the velocity is changing, will
experience an acceleration that can be described by two components:

A radial or centripetal acceleration equal to
the square of speed divided by radius of
curvature of motion directed toward the
center of curvature of the motion, and
Tangential component of acceleration that is
equal to the rate of change of the particle's
speed and is directed either parallel (in the
case of speeding up) or anti-parallel (in the
case of slowing down) to the particle's
velocity.

arad
v2

R
atan 

dv
dt
Relative Velocity
Relative Velocity

The velocity seen by particular observer is called relative
to that observer, or relative velocity.
Relative Velocity in 1-D

Woman walks with a velocity of 1.0m/s along the aisle of a train that is
moving with a velocity of 3.0m/s. What is the woman’s velocity?

For passenger sitting in a train: 1.0m/s

For bicyclist standing: 1.0m/s + 3.0m/s = 4.0m/s

Frame of reference is a
coordinate system +
time scale
 


r2  r1 r
vav 

t 2  t1 t
Relative Velocity in 1-D

Cyclist: frame of reference A

Moving train: frame of reference B

In 1-D motion, position of P relative to frame of
reference A is given by distance XP/A

Position of P relative to frame of reference B is given
by distance XP/B

Distance from origin A to
origin B is given by XB/A
xP / A  xP / B  xB / A
Relative Velocity in 1-D

Velocity VP/A of P relative to frame A is the derivative of XP/A with respect
to time
dxP / A dxP / B dxB / A


dt
dt
dt
vP / B  1.0 ms vB / A  3.0 ms
vP / A  1.0 ms  3.0 ms  4.0 ms
vP / A  vP / B  vB / A