Motion in two and three
dimensions
• We now apply the concepts on motion and
vectors which were previously reviewed and
introduced.
• As me move our discuss into two and three
dimensions we start by introducing the
concept of the position vector.
• As me move our discuss into two and three
dimensions we start by introducing the
concept of the position vector.
• The position vector gives the coordinate (the
position) of an object relative to the origin.

r  xiˆ  yˆj  zkˆ
y-axis
(Position vector)
z-axis
y
x
z
x-axis
Example
y-axis
• For example,
4
z-axis
2
5

r  5iˆ  4 ˆj  2kˆ
x-axis
y-axis
• The
displacement
in
unit
vector
notation
is,
  
r  r2  r1  x2  x1 iˆ   y2  y1  ˆj  z2  z1 kˆ

r2
z-axis

r

r1
x-axis
y-axis
• The
displacement
in
unit
vector
notation
is,
  
r  r2  r1  x2  x1 iˆ   y2  y1  ˆj  z2  z1 kˆ

r2

r

r1
x-axis
z-axis
• Alternatively,

r  xiˆ  yˆj  zkˆ
• Example: Determine the displacement of a car

which is initially at r1   3.0miˆ  2.0m ˆj  5.0mkˆ and

later at r2  9.0miˆ  2.0m ˆj  8.0mkˆ .
• Example: Determine the displacement of a car

which is initially at r1   3.0miˆ  2.0m ˆj  5.0mkˆ and

later at r2  9.0miˆ  2.0m ˆj  8.0mkˆ .

• Sol: r  9  (3)iˆ  2  2 ˆj  8  5kˆ
 12miˆ  0 ˆj  3mkˆ
Average and instantaneous velocity
• Average velocity in unit vector notation,

x  y ˆ z ˆ
r


i
j
k
vavg 
t
t
t
t
• Instantaneous velocity,

 dr dx  dy ˆ dz ˆ
v
 i
j k
dt
dt
dt
dt
• Recall: the direction of the velocity is at a tangent to the
position of the particle.
• Example: The coordinates of a spy satellite
orbiting the Earth are given by the position
function (x & y in km):
x  0.3t 2  7t  12
y  0.2t 2  10t  20
• Find the velocity and acceleration at time t  15 s
in unit vector notation.
x  0.3t 2  7t  12
• Sol:
v
dx  dy ˆ
i
j
dt
dt
vx  0.6t  7
vy  0.4t  10
y  0.2t 2  10t  20
x  0.3t 2  7t  12
y  0.2t 2  10t  20
• Sol:
v
dx  dy ˆ
i
j
dt
dt
vx  0.6t  7
vy  0.4t  10
vx 15  0.615  7  2ms1
vy 15  0.415 10  16ms1

 


Thus, v  2ms1 i  16ms1 ˆj
vx  0.6t  7
vy  0.4t  10
• Continuing,
dvx  dvy ˆ
a
i
j
dt
dt
ax  0.6
a y  0.4

2
a  0.6ms
 


i  0.4ms2 ˆj
• In addition the velocity and acceleration in
magnitude angle format.
v  v x2  v y2  22  162  260ms1
v
and   tan1  y
v
 x
 83

16
  tan1  
 2

Similarly,
a  ax2  a y2  0.62  0.42  0.72ms2
a
and   tan1  y
a
 x
 34

0.4 
  tan1 

 0.6 

Projectile Motion
A special case of motion in two
dimensions
• A projectile is an object where the only force
acting on it is gravity (which acts downward).
• A projectile is an object where the only force
acting on it is gravity (which acts downward).
• Consider a particle which is release with an
initial velocity v0.
vy
vy
vx
v
v
vx
vx
vx
vy
v
• The horizontal and vertical motion is
independent.
• The horizontal and vertical motion is
independent.
• What this means is that we can convert the
two dimensional projectile problem into two
one dimensional problems.
• ie: One 1D equation for the x-direction and
another for the y-direction. These equations
can be solved separately!
• For our analysis we assume that there is no air
resistance.
• Consider the vertical motion:
2
1
• Recall: the equation of motion, y  y0  v0t  2 at
• Through the motion,a   g
g
2
1
 y  y0  v0 yt  2 gt
2
1
y

y

v
sin

t

gt
• Alt:
0
0
0
2
• Similarly, v y2  v0 sin 0 2  2 g  y  y0 
+ve
direction
• Consider the horizontal motion:
• Again the general equation, x  x0  v0t  12 at2
• However the acceleration in the x-dir is zero
and vx is constant.
+ve
2
direction
 x  x0  v0 xt  12 0t
g
• Thus,  x  x0  v0 xt
 x  x0  v0 cos0t
Other equations
• Equation of the path (trajectory):
gx2
y  t an0 x 
2
2v0 cos0 
The trajectory gives the path of the projectile. For a given horizontal
distance we can calculate its height.
• Horizontal Range:
v02
R
sin 2 0
g
The range gives the maximum horizontal distance of the projectile for a given
initial velocity and launch angle.
Example
• A missile launcher on a cliff fires a missile at a
velocity of 25m/s at an angle of 53° above the
horizontal.
v
53 
25m
Example
• A missile launcher on a cliff fires a missile at a
velocity of 25m/s at an angle of 53° above the
horizontal.
(a) The initial position of the object along
the x-axis is x0  0 .
v
53 
+a
25m
g
Example
• A missile launcher on a cliff fires a missile at a
velocity of 25m/s at an angle of 53° above the
horizontal.
(a) The initial position of the object along
the x-axis is x0  0 .
v
53 
+a
(b) The initial position of the object along
the y-axis is y0  0 .
25m
g
Example
• A missile launcher on a cliff fires a missile at a
velocity of 25m/s at an angle of 53° above the
horizontal.
(a) The initial position of the object along
the x-axis is x0  0 .
v
53 
+a
(b) The initial position of the object along
the y-axis is y0  0 .
(c) Find the initial velocity along the xaxis and y-axis.
25m
g
Example
• A missile launcher on a cliff fires a missile at a
velocity of 25m/s at an angle of 53° above the
horizontal.
(a) The initial position of the object along
the x-axis is x0  0 .
v
53 
+a
(b) The initial position of the object along
the y-axis is y0  0 .
(c) Find the initial velocity along the xaxis and y-axis.
Recall: v0 x  v0 cos and v0 y  v0 sin 
v0 x  25cos53  15ms1
v0 y  25sin 53  20ms1
25m
g
Example
• A missile launcher on a cliff fires a missile at a
velocity of 25m/s at an angle of 53° above the
horizontal.
(d) Find the time at which it hits the
ground.
v
53 
+a
25m
g
Example
• A missile launcher on a cliff fires a missile at a
velocity of 25m/s at an angle of 53° above the
horizontal.
(d) Find the time at which it hits the
ground.
v
53 
+a
When the object hits the ground y=-25m.
Using,  y  y0   v0 yt  12 gt2
Then,  25  20t  12 9.81t 2
Solving gives us, t  5s and t  1s
The former is the solution as time can’t
be negative.
25m
g
Circular motion
• When an object moves in a circle or circular
arc at a constant speed we describe this
motion as uniform circular motion.

v

a

v

v

a

a
(a) The direction of the
velocity changes but not
its magnitude.
(b) The direction of the
acceleration is always
towards the centre.
• The acceleration is called the centripetal
acceleration.

v

a

v

v

a

a
(a) The direction of the
velocity changes but not
its magnitude.
(b) The direction of the
acceleration is always
towards the centre.
• For uniform motion:
 v2
ˆ  v2
ˆ
v2

which in unit vector notion is: a   


 j
a
cos

i


sin




r
 r

 r

and
T 
2r
(The period)
v

v

a

v

v

a

a
Example
• A small object of mass m rotates anticlockwise
on a horizontal frictionless plane at the end of
a string of length r = 0.4 m with constant
speed v = 2.0 m/s. Draw the magnitude and
direction of (a) the velocity and (b) the
centripetal acceleration at points A, B, and C.
(c) Are the velocity and the acceleration
constant?
A
B
r
C
A
v
a
B
v
a
r
a
C
v
• The velocity is not constant.
• The magnitude of the acceleration is:
v2
a
r
2

2.0
a 
0.4
 10m s2
• The velocity is not constant.
• The magnitude of the acceleration is:
v2
a
r
2

2.0
a 
0.4
 10m s2
• The acceleration is not constant. The
magnitude is constant but the direction
changes.