Chapter 3.

Motion in Two and Three Dimensions

2-1 Displacement, Velocity, and Acceleration

2-2 Special Case 1: Projectile Motion
2-3 Special Case 2: Circular Motion

Phys. 1421.Spring 2015- Chapter 3 1

 SAILBOATS DO NOT TRAVEL IN
STRAIGHT LINES TO THEIR
DESTINATIONS, BUT INSTEAD
MUST “TACK” BACK AND FORTH
ACROSS THE WIND. THIS BOAT
MUST SAIL EAST, THEN SOUTH,
AND THEN EAST AGAIN, IN ITS
JOURNEY TO A SOUTHEASTERN
PORT

How can we calculate the

boat’s displacement and its
average velocity?

The motion of a sailboat tacking into the wind cannot be fully described using the equations we
presented in Chapter 2. Instead, to describe its motion, we must extend the idea of motion in one
dimension discussed in Chapter 2 to two and three dimensions. To do this, we must revisit the
concept of vectors and look at how they can be used to analyze and describe motion in more than
one dimension.

The material in this chapter presumes you are familiar with the material that introduces vectors in
Sections 6 and 7 of Chapter 1. You are encouraged to review these sections before proceeding in this
chapter. 2
Phys. 1421.Spring 2015- Chapter 3
In Chapter 2, the concepts of displacement, velocity, and acceleration were used to
describe the motion of an object moving in a straight line. Now we use the concept
of vectors to extend these characteristics of motion in two and three dimensions.

 Position & Displacement Vectors.

The position vector of a particle is a vector drawn from the origin of a coordinate
system to the location of the particle. For a particle in the x, y plane at the point
with coordinates (x, y), the position vector 𝒓 is

Example 1: The position vector of a particle that

is located in the x,y plane at (3,2) is

Phys. 1421.Spring 2015- Chapter 3 3

 Position & Displacement Vectors.
Consider a particle moving in the x,y plane. At initial time ( 𝒕𝒊 ), the particle was at the
point (𝒙𝒊 , 𝒚𝒊 ) and on a later time (final time) (𝒕𝒊 ), the position was (𝒙𝒇 , 𝒚𝒇 ). Then the
particle’s change in position (Displacement) is

but
(𝒙𝒊 , 𝒚𝒊 )

then

A particle moving in the xy plane .The

displacement of the particle as it
moves from A to B in the time interval
∆𝑡 = 𝑡𝑓 − 𝑡𝑖 is equal to the vector
∆𝑟Ԧ = 𝑟Ԧ𝑓 − 𝑟Ԧ𝑖
Phys. 1421.Spring 2015- Chapter 3 4
 Velocity Vectors.
Recall that average velocity is defined as displacement divided by the elapsed
time. The result of the displacement vector divided by the elapsed time interval
∆𝑡 = 𝑡𝑓 − 𝑡𝑖 is the average-velocity vector

(𝒙𝒊 , 𝒚𝒊 )

 and the instantaneous-velocity vector

and the magnitude of the velocity is

Phys. 1421.Spring 2015- Chapter 3 5

 Acceleration Vectors.
The average-acceleration vector is the ratio of the change in the instantaneous
velocity vector ∆𝑣,
Ԧ , to the elapsed time interval ∆𝑡:

 and the instantaneous-acceleration

vector

and the magnitude of the velocity is

Phys. 1421.Spring 2015- Chapter 3 6

Example 2: (Ex. 3-1 Textbook)
A Sailboat has coordinates (x1, y1) = (130 m, 205 m) at t1 = 60.0 s. Two minutes later, at time t2,
it has coordinates (x2, y2) = (110 m, 218 m).
(a) Find the average velocity for this two-minute interval. Express 𝑣Ԧ𝑎𝑣 in terms of its
rectangular components.
(b) Find the magnitude and direction of this average velocity.
(c) For 𝑡 ≥ 20.0 s, the position of a second sailboat as a function of time is 𝑥 𝑡 = 𝑏1 + 𝑏2𝑡
and 𝑦 𝑡 = 𝑐1 + 𝑐2 𝑡, where 𝑏1 = 100 m, 𝑏2 = 0.500 m/ s, 𝑐1 = 200 m, and 𝑐2 = 360
m/ s. Find the instantaneous velocity as a function of time t, for 𝑡 ≥ 20.0 s.

Phys. 1421.Spring 2015- Chapter 3 7

Example 3: (Problem 46 page 87, Textbook)
A particle’s position coordinates (x,y) are (2.0 m, 3.0 m) at t=0; (6.0 m, 7.0 m) at t= 2.0 s; and
(13 m, 14 m) at t=5.0 s.
(a) Find the magnitude of the average velocity from t=0 to t=2.0 s
(b) Find the magnitude of the average velocity from t=0 to t=5.0 s

Phys. 1421.Spring 2015- Chapter 3 8

Example 4: (Problem 51 page 87, Textbook)
A particle has position vector , where 𝒓 in meter and t in second.
Find the instantaneous-velocity and instantaneous- acceleration vectors as a function of time.

Phys. 1421.Spring 2015- Chapter 3 9

TWO – DIMENTIONAL MOTION WITH CONSTANT
ACCELERATION
 The particle’s motion in two- dimensions can be visualized as a resultant of a motion
a long the x-axis and a motion a long y- axis.

The motion a long the x-axis can be described by the three kinematic equation

The motion a long the y-axis can be described by the three kinematic equation

Phys. 1421.Spring 2015- Chapter 3 10

TWO – DIMENTIONAL MOTION WITH CONSTANT
ACCELERATION
Example 5:
A particle starts from the origin at 𝑡 = 0 with an initial velocity having an 𝑥 − component of
20 m/s and a 𝑦 − component of −15 m/s. The particle moves in the 𝑥𝑦 plane with an 𝑥
component of acceleration only, given by 𝑎𝑥 = 4.0 m/s2.
(A) Determine the components of the velocity vector at any time and the total velocity vector
at any time
(B) Calculate the velocity and speed of the particle at 𝑡 = 5.0 s
(C) Determine the 𝑥 and 𝑦 coordinates of the particle at any time 𝑡 and the position vector at
this time.

Phys. 1421.Spring 2015- Chapter 3 11

 Projectile Motion
The Projectile motion is a special case of two-dimensional Motion in which the
acceleration in the 𝑥-axis is zero (𝑎𝑥 = 0) and the acceleration in y-axis is -g (𝑎𝑥 = −g)

The parabolic path of a

projectile that leaves the
origin with a velocity 𝒗𝒊 . The
velocity vector 𝒗 changes
with time in both magnitude
and direction. This change is
the result of acceleration in
the negative y direction.
The x component of velocity
remains constant in time
because there is no
acceleration along the
horizontal direction. The y
component of velocity is zero
at the peak of the path.

Phys. 1421.Spring 2015- Chapter 3 12

 Horizontal Range and Maximum Height of a Projectile
Assume a projectile is launched from the origin at 𝑡𝑖 = 0 with a positive 𝑣𝑦𝑖 component,
as shown in the figure. Two points are especially interesting to analyze: the peak point
(A), which has Cartesian coordinates (R/2, h), and the point (B), which has coordinates
(R, 0). The distance R is called the horizontal range of the projectile, and the distance h
is its maximum height.

Let us find h and R in terms of 𝑣𝑖 , 𝜃, and g.

Phys. 1421.Spring 2015- Chapter 3 13

 Horizontal Range and Maximum Height of a Projectile

A projectile launched from the origin with an initial speed

of 50 m/s at various angles of projection. Note that
complementary values of i result in the same value of R
(range of the projectile).

Phys. 1421.Spring 2015- Chapter 3 14

Example 6: (Ex .6 page 72, Textbook)
A delighted physics graduate throws her cap into the air with an initial velocity of 24.5 m/s at
36.9° above the horizontal. The cap is later caught by another student. Find
(a) the total time the cap is in the air, and
(b) the total horizontal distance traveled. (Ignore effects of air resistance.)

Phys. 1421.Spring 2015- Chapter 3 15

Example 7: (Problem 84, page 89, Textbook)
A projectile is fired into the air from the top of a 200-m cliff above a valley as shown in the
figure. Its initial velocity is 60 m/s at 60° above the horizontal. Where does the projectile land?
(Ignore any effects due to air resistance.)

Phys. 1421.Spring 2015- Chapter 3 16

(a) A car moving along a circular path at constant speed experiences uniform
circular motion. (b) As a particle moves from (A) to (B), its velocity vector changes
from 𝒗𝒊 to 𝒗𝒇 . (c) The construction for determining the direction of the change in velocity
∆𝒗, which is toward the center of the circle for small ∆𝒓.

- What is the acceleration (centripetal acceleration )

- What is the total time of traveling a complete circle (Periodic Time)

Phys. 1421.Spring 2015- Chapter 3 17

The acceleration vector in uniform circular motion is always perpendicular to the path
and always points toward the center of the circle. An acceleration of this nature is called
a centripetal or radial acceleration (centripetal means center-seeking), and its magnitude
is 2
𝑣
𝑎𝑐 =
𝑟
where 𝑟 is the radius of the circle. The subscript on the acceleration symbol reminds us
that the acceleration is centripetal

Phys. 1421.Spring 2015- Chapter 3 18

In many situations it is convenient to describe the motion of a particle moving with
constant speed in a circle of radius 𝑟 in terms of the period 𝑇, which is defined as the
time required for one complete revolution. In the time interval 𝑇 the particle moves a
distance of 2𝜋𝑟, which is equal to the circumference of the particle’s circular path.
Therefore, because its speed is equal to the circumference of the circular path divided
by the period, or 𝑣 = 2𝜋𝑟/𝑇, it follows that
2𝜋𝑟
𝑇=
𝑣
Phys. 1421.Spring 2015- Chapter 3 19
The motion of a particle along an arbitrary curved path lying in the 𝑥𝑦 plane. If the
velocity vector 𝒗 (always tangent to the path) changes in direction and magnitude, the
components of the acceleration 𝒂 are a tangential component 𝑎𝑡 and a radial
component 𝑎𝑟 .
𝒂 = 𝒂𝒕 + 𝒂𝒓

Phys. 1421.Spring 2015- Chapter 3 20

Example 8: A car exhibits a constant acceleration of 0.300 m/s2 parallel to the roadway. The
car passes over a rise in the roadway such that the top of the rise is shaped like a circle of radius
500 m. At the moment the car is at the top of the rise, its velocity vector is horizontal and has a
magnitude of 6.00 m/s. What is the direction of the total acceleration vector for the car at this
instant?

Phys. 1421.Spring 2015- Chapter 3 21

Example 9: The shown figure represents the total acceleration of a particle moving clockwise
in a circle of radius 2.50 m at a certain instant of time. At this instant, find
(a) the radial (centripetal) ) acceleration,
(b) the speed of the particle,
(c) its tangential acceleration.

Phys. 1421.Spring 2015- Chapter 3 22