Chapter 4
Motion in Two and Three
Dimensions
Position and Displacement

A position vector locates a particle in space
o
Extends from a reference point (origin) to the particle
Eq. (4-1)
Example
o Position vector (-3m, 2m, 5m)
Figure 4-1
Position and Displacement

Change in position vector is a displacement
Eq. (4-2)

We can rewrite this as:
Eq. (4-3)

Or express it in terms of changes in each coordinate:
Eq. (4-4)
Average Velocity and Instantaneous Velocity

Average velocity is
o

A displacement divided by its time interval
We can write this in component form:
Example
Eq. (4-8)
Eq. (4-9)
o A particle moves through displacement (12 m)i + (3.0 m)k in 2.0 s:
Average Velocity and Instantaneous Velocity

Instantaneous velocity is
o
o

The velocity of a particle at a single point in time
The limit of avg. velocity
as the time interval shrinks to 0
Eq. (4-10)
Visualize displacement and instantaneous velocity:
Figure 4-3
Figure 4-4
Average Velocity and Instantaneous Velocity

In unit-vector form, we write:

Which can also be written:
Eq. (4-11)
Eq. (4-12)

Note: a velocity vector does not extend from one point to
another, only shows direction and magnitude
Average Acceleration and Instantaneous Acceleration

Average acceleration is
o
A change in velocity divided by its time interval
Eq. (4-15)

Instantaneous acceleration is again the limit t → 0:
Eq. (4-16)

We can write Eq. 4-16 in unit-vector form:
Average Acceleration and Instantaneous Acceleration

We can rewrite as:
Eq. (4-17)
Eq. (4-18)

To get the components of acceleration, we differentiate the
components of velocity with respect to time
Figure 4-6
Average Acceleration and Instantaneous
Acceleration
© 2014 John Wiley & Sons, Inc. All rights reserved.
Average Acceleration and Instantaneous
Acceleration
© 2014 John Wiley & Sons, Inc. All rights reserved.
Average Acceleration and Instantaneous
Acceleration
HW#:
At t = 0 the velocity of a particle is measured to be 5 m/s
along the positive x–axis and its acceleration is
a = -3î + 4ĵ (m/s2) , Find
• The velocity vector as a function of time ?
{answer v = (5 – 3t)î + 4t ĵ}
© 2014 John Wiley & Sons, Inc. All rights reserved.
Projectile Motion
Characteristics of Projectile Motion:
Horizontal
• Motion of a ball rolling freely along a level surface.
• Horizontal velocity is always constant.
Vertical
• Motion of a freely falling object.
• Force due to gravity.
• Vertical component of velocity changes with time.
Projectile Motion

A projectile is
o
A particle moving in the x-y plane
o
With some initial velocity
o
Whose acceleration is always free-fall acceleration (g)

The motion of a projectile is projectile motion

Launched with an initial velocity v0
Eq. (1)
Eq. (2)
Projectile Motion

Horizontal motion:
o
No acceleration, so velocity is constant (ax=0):
Eq. (3)

Vertical motion:
o
Acceleration is always g (ay= g):
Eq. (4)
Eq. (5)
Eq. (6)
Projectile Motion
Horizontal range and the maximum height of a projectile:
To find the maximum height we use the equation
H = v2osin2θ / 2g
© 2014 John Wiley & Sons, Inc. All rights reserved.
Projectile Motion
Therefore, R is a maximum when 𝜽 = 𝟒𝟓°
Projectile Motion
EXAMPLE :
A ball is thrown with an initial velocity of 20 m/s at an angle
of 30 above the ground. Find
•
The maximum height the ball can reach?
•
The range of the ball travel?
•
The time taken to hit the ground again?
© 2014 John Wiley & Sons, Inc. All rights reserved.
Projectile Motion
Solve:
• H = v2osin2θ / 2g
= (20 x sin 30)2 / (2 x 9.8 ) = 5.1 m
• R = v2osin2θ / g
= (400 x sin 60) / 9.8 = 35.3 m .
• vy = vosinθ – gt
vy (at maximum height) = 0
t = vosinθ / g = 20 x sinθ / 9.8
t = 1.02 s
(time needed to achieve the maximum height)
Ttotall = 2 x t = 2.04 s
© 2014 John Wiley & Sons, Inc. All rights reserved.
Projectile Motion
EXAMPLE:
A ball is thrown horizontally with an initial velocity of 25 m/s from
the top of a 60 m high building .
•
How much time will the ball take to hit the ground ?
•
How far does the ball travel horizontally ?
•
What is the velocity of the ball as it hits the ground ?
© 2014 John Wiley & Sons, Inc. All rights reserved.
Projectile Motion
solve
•
y = y0 + (v0sinѳ)t – ½ gt2
Ѳ= 0
y = -1/2gt2
t = 3.5s
•
•
x = (v0cosѳ)t
x = 87.5 m
vx= v0cosѲ =25 m/s
vy= v0sinѲ – gt = -34.3 m/s
| v |= 42.44 m/s
© 2014 John Wiley & Sons, Inc. All rights reserved.
Projectile Motion
Example :
:a
projectile is fired to achieve a maximum range of 120 m .what should
its speed be ?
Rmax =V20/g
V0=37 m/s
Example 4
: a projectile is fired with an angle Θ above the horizontal . take 15s to
reach its range of 110 m . What is its speed at the highest point ?
x = (v0cosѲ)t
v0cosѲ = v0x = x/t = 9.34 m/s
© 2014 John Wiley & Sons, Inc. All rights reserved.
Projectile Motion
HW #:
A particle is projected with an initial velocity vi = ( 10î + 10ĵ ) m/s
The angle of projection is : ( answer = 45֯ )
© 2014 John Wiley & Sons, Inc. All rights reserved.
Uniform Circular Motion

A particle is in uniform circular motion if
•
•
•
•
It travels around a circle path
The tangential velocity of a particle has a constant magnitude and
change direction continuously
The centripetal acceleration has a constant
magnitude and towards at the center of the
circle
The velocity and the acceleration perpendicular
to each other
Figure 4-16
Uniform Circular Motion
•
To Find centripetal acceleration use the equation
ɑ - the centripetal acceleration
ν - the velocity of the object
r - the radius of the circle

The period of revolution is:
The time it takes to complete one revolution
The Frequency
the number of revolution per second
1
𝜏
To find the frequency: 𝒇 = =
1
2𝜋𝑟
𝑣
𝟐𝝅𝒓
𝝉=
𝒗
=
𝒗
𝟐𝝅𝒓
Uniform Circular Motion
Example 1: A particle moves in a circular path 0.4 𝑚 in radius with constant speed. If the particle
makes five revolution in each second of its motion, find:
a) The speed of the particle.
b) Its acceleration.
c) The frequency.
Solution:
𝜏=
a) 𝑣 =
b) 𝑎 =
c) 𝑓 =
2𝜋𝑟
2×3.14×0.4
2.51
=
=
𝜏
5 𝑟𝑒𝑣
0.2
2
2
𝑣
12.6
𝑚
=
= 397 2
𝑟
0.4
𝑠
1
1
=
= 5 𝐻𝑧
𝜏
0.2
= 12.6
1𝑠
= 0.2 𝑠
5 𝑟𝑒𝑣
𝑚
𝑠
Uniform Circular Motion
Example 2: A bicycle completes 4 revolutions around a circular path of radius 10 m in 2 min. calculate
its acceleration.
Solution:
𝑣2
𝑎𝑐 =
𝑣 =?
𝑟
𝑣=
2𝜋𝑟
𝜏
𝑇𝑖𝑚𝑒 𝑖𝑛 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
2 × 60 120
𝜏=
=
=
= 30 𝑠
𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑜𝑓 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛
4
4
2𝜋𝑟 2 × 𝜋 × 10
𝑚
∴𝑣=
=
= 2.1
𝜏
30
𝑠
∴ 𝑎𝑐 =
𝑣2
𝑟
=
2.12
10
= 0.44
𝑚
𝑠2
Uniform Circular Motion
Example 3:The Effect of Radius on Centripetal Acceleration The race track contains turns with radii
of 33 𝑚 and 24 𝑚. Find the centripetal acceleration at each turn for a speed of 34
Solution:
𝒗𝟐
𝒂=
𝒓
342
m
a=
= 35 2
33
s
342
m
a=
= 48 2
24
s
𝑚
𝑠
.
Summary
Position Vector

Displacement
Locates a particle in 3-space
Eq. (4-1)

Change in position vector
Eq. (4-2)
Eq. (4-3)
Eq. (4-4)
Average and Instantaneous
Velocity
Average and Instantaneous
Accel.
Eq. (4-8)
Eq. (4-15)
Eq. (4-10)
Eq. (4-16)
Summary
Projectile Motion

Uniform Circular Motion
Flight of particle subject only
to free-fall acceleration (g)

Magnitude of acceleration:
Eq. (4-34)
Eq. (4-22)

Eq. (4-23)
Time to complete a circle:
Eq. (4-35)
EXERCISE
Exercise
1 , 3 , 6 , 13 , 56 , 62
© 2014 John Wiley & Sons, Inc. All rights reserved.