Chapter 4
Motion in Two and Three Dimensions
In this chapter we will continue to study the motion of objects without the
restriction we put in Chapter 2 to move along a straight line. Instead we will
consider motion in a plane (two-dimensional motion) and motion in space
(three-dimensional motion).
The
following vectors will be defined for two- and three-dimensional motion:
Displacement
Average and instantaneous velocity
Average and instantaneous acceleration
We will consider in detail projectile motion and uniform circular motion as
examples of motion in two dimensions.
Finally, we will consider relative motion, i.e., the transformation of velocities
between two reference systems that move with respect to each other with
constant velocity.
(4-1)
Position Vector
The position vector r of a particle is defined as a vector whose tail is at
a reference point (usually the origin O) and its tip is at the particle at
point P.
Example: The position vector in the figure is
r  xˆi  yˆj  zkˆ


r  3iˆ  2jˆ  5kˆ m
P
(4-2)
Displacement Vector
For a particle that changes position vector from r1 to r2 we define the displacement
vector r as follows:
r  r2  r1.
The position vectors r1 and r2 are written in terms of components as
r  x ˆi  y ˆj  z kˆ
r  x ˆi  y ˆj  z kˆ
1
1
1
1
2
2
2
2
The displacement  r can then be written as
r   x2  x1  ˆi   y2  y1  ˆj   z2  z1  kˆ  xˆi  yˆj  zkˆ
x  x2  x1
y  y2  y1
z  z2  z1
t1
t2
(4-3)
Average and Instantaneous Velocity
Following the same approach as in Chapter 2 we define the average
velocity as
displacement
average velocity =
vavg
time interval
r xˆi  yˆj  zkˆ xˆi yˆj zkˆ





t
t
t
t
t
We define the instantaneous velocity (or
more simply the velocity) as the limit:
v  lim
t
t + Δt
r dr

t dt
t  0
(4-4)
If we allow the time interval t to shrink to zero, the following things happen:
1. Vector r2 moves toward vector r2 and r  0.
r
(and thus vavg ) approaches the direction
t
of the tangent to the path at position 1.
2. The direction of the ratio
3. vavg  v
v


d ˆ ˆ ˆ
dx
dy
dz
xi  yj  zk  ˆi  ˆj  kˆ  vx ˆi  v y ˆj  vz kˆ
dt
dt
dt
dt
The three velocity components are given by
the equations
vx 
t
t + Δt
dr
v
dt
(4-5)
dx
dt
dy
vy 
dt
vz 
dz
dt
Average and Instantaneous Acceleration
The average acceleration is defined as:
change in velocity
average acceleration =
time interval
aavg
v2  v1 v


t
t
We define the instantaneous acceleration as the limit:


dvx ˆ dv y ˆ dvz ˆ
v dv d
ˆ
ˆ
ˆ
a  lim


vx i  v y j  vz k 
i
j
k  ax ˆi  a y ˆj  az kˆ
t dt dt
dt
dt
dt
t  0
Note: Unlike velocity, the acceleration vector does not have any specific relationship
with the path.
The three acceleration components are given by
the equations
ax 
dvx
dt
ay 
dv y
dt
dvz
az 
dt
dv
a
dt
(4-6)
Projectile Motion
The motion of an object in a vertical plane under the influence of
gravitational force is known as “projectile motion.”
The projectile is launched with an initial velocity v0 .
The horizontal and vertical velocity components are:
v0 x  v0 cos 0
g
v0 y  v0 sin  0
Projectile motion will be analyzed in a
horizontal and a vertical motion along
the x- and y-axes, respectively. These
two motions are independent of each
other. Motion along the x-axis has
zero acceleration. Motion along the yaxis has uniform acceleration ay = -g.
(4-7)
Horizontal Motion: ax  0
vx  v0 cos  0
(eq. 1)
Vertical Motion:
v y  v0 sin  0  gt
ay   g
(eq. 3)
The velocity along the x-axis does not change:
x  xo   v0 cos 0  t
(eq. 2)
Along the y -axis the projectile is in free fall
gt 2
y  yo   v0 sin  0  t 
2
(eq. 4)
If we eliminate t between equations 3 and 4 we get v   v0 sin  0   2 g  y  yo  .
2
y
g
2
Here xo and yo are the coordinates
of the launching point. For many
problems the launching point is
taken at the origin. In this case
xo  0 and yo  0.
Note: In this analysis of projectile
motion we neglect the effects of
air resistance.
(4-8)
The equation of the path:
gt 2
x   v0 cos  0  t (eq. 2)
y   v0 sin  0  t 
2
If we eliminate t between equations 2 and 4 we get:
y   tan  0  x 
g
2  v0 cos  0 
2
(eq. 4)
x 2 . This equation describes the path of the motion.
The path equations has the form: y  ax  bx 2 . This is the equation of a parabola.
Note: The equation of the path seems too
complicated to be useful. Appearances can
deceive: Complicated as it is, this equation
can be used as a shortcut in many projectile
motion problems.
(4-9)
vx  v0 cos  0
(eq. 1)
x   v0 cos  0  t
(eq. 2)
sin
gt 2
O /2
v y  v0 sin  0  gt (eq. 3) y   v0 sin  0  t 
(eq. 4)
2
Horizontal Range: The distance OA is defined as the horizontal range R
At point A we have: y  0. From equation 4 we have:
3/2
gt 2
gt 

v
sin

t


0

t
v
sin


0
0
0
 0
  0. This equation has two solutions:
2
2

Solution 1. t  0. This solution corresponds to point O and is of no interest.
gt
 0. This solution corresponds to point A.
2
2v0 sin  0
From solution 2 we get t 
. If we substitute t in eq. 2 we get
g
2v02
v02
R
sin  0 cos  0  sin 2 0 .
g
g
t
A
O
R has its maximum value when  0  45 :
Solution 2. v0 sin  0 
R
2sin A cos A  sin 2 A
Rmax
v02

g
(4-10)

tA
g
Maximum Height H
v02 sin 2 0
H
2g
H
The y -component of the projectile velocity is v y  v0 sin  0  gt.
At point A: v y  0
v0 sin  0
 v0 sin  0  gt  t 
g
H  y (t )   v0 sin  0  t 
v02 sin 2  0
H
2g
2
v sin  0 g  v0 sin  0 
gt
  v0 sin  0  0
 
 
2
g
2
g

2
(4-11)
Maximum Height H (encore)
tA
g
H
v02 sin 2 0
H
2g
We can calculate the maximum height using the third equation of kinematics
2
for motion along the y -axis: v y 2  v yo
 2a  y  yo  .
In our problem: y0  0, y  H , v yo  v0 sin  0 , v y  0 , and a   g 
v02 sin 2  0
v  2 gH  H 

.
2g
2g
2
yo
2
v yo
(4-12)
Uniform Circular Motion:
A particle is in uniform circular motion if it moves on a circular path of
radius r with constant speed v. Even though the speed is constant, the
velocity is not. The reason is that the direction of the velocity vector
changes from point to point along the path. The fact that the velocity
changes means that the acceleration is not zero. The acceleration in uniform
circular motion has the following characteristics:
1. Its vector points toward the center C of the circular path, thus the name
“centripetal.”
v2
2. Its magnitude a is given by the equation a  .
r
Q
r
C r
P
The time T it takes to complete a full revolution is
known as the “period.” It is given by the equation
r
R
2 r
T
.
v
(4-13)
yP
xP
sin  
cos 
r
r
Here xP and yP are the coordinates of the rotating particle.
v  vx ˆi  v y ˆj   v sin   ˆi   v cos   ˆj
dv  v dyP  ˆ  v dxP  ˆ
 y P  ˆ  xP  ˆ
v   v  i   v  j. Acceleration a 
= 
i 
 j.
r   r 
dt  r dt   r dt 

dyP
dxP
We note that
 v y  v cos  and
 vx  v sin  .
dt
dt
 v2
 ˆ  v2
ˆ
a    cos   i    sin   j
 r
  r

tan  
vx  v sin 
v y  v cos 
ay
ax

  v 2 / r  sin 
  v / r  cos 
2
2
v
a  ax2  a y2 
r
 cos    sin  
2
2
v2

r
 tan       a points toward C.
P
C
C
A
 cos     sin  
2
(4-14)
2
1
Relative Motion in One Dimension:
The velocity of a particle P determined by two different observers A and B varies from
observer to observer. Below we derive what is known as the “transformation
equation” of velocities. This equation gives us the exact relationship between the
velocities each observer perceives. Here we assume that observer B moves with a
known constant velocity vBA with respect to observer A. Observers A and B
determine the coordinates of particle P to be xPA and xPB , respectively.
xPA  xPB  xBA . Here xBA is the coordinate of B with respect to A.
d
d
d
We take derivatives of the above equation:
x

x

 PA 
 PB 
 xBA  
dt
dt
dt
vPA  vPB  vBA
If we take derivatives of the last equation and take
into account that
dvBA
0
dt
aPA  aPB
Note: Even though observers A and B
measure different velocities for P,
they measure the same acceleration.
(4-15)
Relative Motion in Two Dimensions:
Here we assume that observer B moves with a known constant velocity vBA with
respect to observer A in the xy-plane.
Observers A and B determine the position vector of particle P to be
rPA and rPB , respectively.
rPA  rPB  rBA . We take the time derivative of both sides of the equation
d
d
d
rPA  rPB  rBA  vPA  vPB  vBA
vPA  vPB  vBA
dt
dt
dt
If we take the time derivative of both sides of the last equation we have:
dvBA
d
d
d
vPA  vPB  vBA . If we take into account that
 0  aPA  aPB .
dt
dt
dt
dt
Note: As in the one-dimensional
case, even though observers A and B
measure different velocities for P,
(4-16)
they measure the same acceleration.