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

Note: as with velocity, an acceleration vector does not
extend from one point to another, only shows direction and
magnitude
Answer: (1) x:yes, y:yes, a:yes
(2) x:no, y:yes, a:no
(3) x:yes, y:yes, a:yes
(4) x:no, y:yes, a:no
Two-dimensional kinematics is similar to one-dimensional kinematics,
but we must now use full vector notation rather than positive and
negative signs to indicate the direction of motion.
Example 1: the x and y components of an object moving in xy plane are given as 𝑥 = 𝑡 3 − 5𝑡 𝑚
and 𝑦 = 3𝑡 2 + 6 (𝑚) where t is measured in seconds. Determine the position vector, velocity and
acceleration of the object at 𝑡 = 2 s.
Solution:
𝑟 𝑡 = 𝑥 𝑖 + 𝑦𝑗 = (𝑡 3 − 5𝑡 )𝑖 + (3𝑡 2 + 6)𝑗
Evaluate 𝑟 at 𝑡 = 2 𝑠
1) The position vector
𝑟 𝑡 = 2 𝑠 = 23 − 5 × 2 𝑖 + 3 × 22 + 6 𝑗 = −2𝑖 + 18𝑗 𝑚
2) The velocity vector
𝑣 = 𝑣𝑥 𝑖 + 𝑣𝑦 𝑗
𝑑𝑥
𝑑 3
=
𝑡 − 5𝑡 = 3𝑡 2 − 5
𝑑𝑡 𝑑𝑡
𝑑𝑦
𝑑
𝑣𝑦 =
=
3𝑡 2 + 6 = 6𝑡
𝑑𝑡 𝑑𝑡
𝑣𝑥 =
The velocity at any time are:
𝑣(𝑡) = 𝑣𝑥 𝑖 + 𝑣𝑦 𝑗 = 3𝑡 2 − 5 𝑖 + 6𝑡𝑗𝐴𝑡 𝑡 = 2 𝑠
𝑣 𝑡 = 2 𝑠 = 3 × 22 − 5 𝑖 + 6 × 2 𝑗 = 7𝑖 + 12𝑗 𝑚/𝑠
Also the acceleration is:
𝑎 = 𝑎𝑥 𝑖 + 𝑎𝑦 𝑗
The components of the acceleration at any time are:
𝑑𝑣𝑥
𝑑
𝑎𝑥 =
=
3𝑡 2 − 5 = 6𝑡
𝑑𝑡
𝑑𝑡
𝑑𝑣𝑦
𝑑
𝑎𝑦 =
=
6𝑡 = 6
𝑑𝑡
𝑑𝑡
𝑎(𝑡) = 𝑎𝑥 𝑖 + 𝑎𝑦 𝑗 = 6𝑡 𝑖 + 6𝑗
𝑨𝒕 𝒕 = 𝟐 𝒔
𝑎 𝑡 = 2 𝑠 = 6 × 2 𝑖 + 6 𝑗 = 12𝑖 + 6𝑗 𝑚/𝑠 2
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.
Parabolic
• Path traced by an object accelerating only in the vertical direction while moving at constant horizontal
velocity.


The most important experimental fact about projectile motion in two dimensions is that the
horizontal and vertical motions are completely independent of each other. This means that motion in
one direction has no effect on motion in the other direction.
In general, the equations of constant acceleration developed in previous lectures follow separately
for both the x-direction and the y-direction. An important difference is that the initial velocity now
has two components.
Projectile Motion

A projectile is
o
A particle moving in the vertical 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. (4-19)
Eq. (4-20)
Projectile Motion

Therefore we can decompose two-dimensional motion into
2 one-dimensional problems
Figure 4-10
Answer: Yes. The y-velocity is negative, so the ball is now falling.
Projectile Motion

Horizontal motion:
o
No acceleration, so velocity is constant (recall Eq. 2-15):
V
Eq. (4-21)

Vertical motion:
o
Acceleration is always -g (recall Eqs. 2-15, 2-11, 2-16):
Eq. (4-22)
Eq. (4-23)
Eq. (4-24)
Projectile Motion
Horizontal Range and Maximum Height of a Projectile:
To find the maximum height h we use the equation
𝒗𝟐 𝐬𝐢𝐧𝟐 𝜽𝒐
𝑯=
‫أقصى أرتفاع للمقذوفة‬
𝟐𝒈
Maximum Range of the Projectile:
The maximum value of R from Equation is
𝒗𝒐 𝟐 𝐬𝐢𝐧𝟐 𝜽𝒐
𝑹=
𝒈
‫المسافة التي تقطعها المقذوفة افقيا‬
𝑹𝒎𝒂𝒙
𝒗𝒐 𝟐
=
𝒈
Projectile Motion
Therefore, R is a maximum when 𝜽 = 𝟒𝟓°
Projectile Motion
Example 1: A long-jumper leaves the ground at an angle of 20° to the horizontal and at a speed of 11
1. How far does he jump?
2. The maximum height reached?
Solution:
1. 𝑹 =
𝒗𝒐 𝟐 𝒔𝒊𝒏𝟐 𝜽𝒐
𝒈
2. 𝑯 =
𝒗𝟐 𝒔𝒊𝒏𝟐 𝜽𝒐
𝟐𝒈
=
=
(𝟏𝟏)𝟐 𝒔𝒊𝒏𝟐(𝟐𝟎° )
𝟗.𝟖
(𝟏𝟏)𝟐 (𝐬𝐢𝐧𝟐𝟎° )𝟐
𝟐(𝟗.𝟖)
= 𝟕. 𝟗𝟒 𝒎
= 𝟎. 𝟕𝟐𝟐 𝐦
𝑚
𝑠
.
Projectile motion
EXAMPLE 2 :
A ball is thrown horizontally with an initial velocity of 25 m/s from the top of a
60 m high building .
a) how much time will the ball take to hit the ground ?
b) how far does the ball travel horizontally ?
c) what is the velocity of the ball as it hits the ground ?
© 2014 John Wiley & Sons, Inc. All rights reserved.
a)
Y = Y0 +(V0sinѳ)t -1/2gt2
Ѳ= 0
Y= -1/2gt2
t= 3.5s
b)
X=(v0cosѳ)t
X=87.5 m
c)
vx=v0cosѲ =25 m/s
vy=v0sinѲ – gt = -34.3 m/s
‫ ן‬v ‫ =ן‬42.44 m/s
© 2014 John Wiley & Sons, Inc. All rights reserved.
Projectile motion
Example 3 : 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 # 4
A ball is thrown up with initial velocity of 15 m/s at an angle of (30) from the top
of a 50 m high building . It lands on the ground at a point P .
1- Calculate the maximum height the ball can reach ?
2- how long does it take the ball to hit the ground at p ?
3- how far from the building edge does the ball hit the ground ?
4- what is the speed of the ball just before hitting the ground ?
© 2014 John Wiley & Sons, Inc. All rights reserved.
Uniform Circular Motion

A particle is in uniform circular motion if
o
It travels around a circle or circular arc
o
At a constant speed

Since the velocity changes, the particle is accelerating!

Velocity and acceleration have:
o
Constant magnitude
o
Changing direction
Figure 4-16
Uniform Circular Motion
Uniform Constant speed, or, constant magnitude of velocity.
Circular motion along a circle: Changing direction of velocity.
Centripetal Acceleration: centripetal means center-seeking.
. ‫ يعني مركز التماس‬: ‫تسارع الجاذبية‬
Uniform Circular Motion

Acceleration is called centripetal acceleration
o
Means “center seeking”
o
Directed radially inward
Eq. (4-34)

The period of revolution is:
o
The time it takes for the particle go around the closed path
exactly once
𝟐𝝅𝒓
𝝉=
𝒗
1
𝜏
To find the frequency: 𝒇 = =
1
2𝜋𝑟
𝑣
=
𝒗
𝟐𝝅𝒓
Eq. (4-35)
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
1 , 3 , 6 , 13 , 56 , 62
© 2014 John Wiley & Sons, Inc. All rights reserved.