MEASUREMENT
AND MOTION IN
ONE DIMENSION
Topic
Chapter 1:
Physics and
Measurements
Chapter 2:
Motion in One
Dimension
Learning outcomes
 State and effectively use the SI
units of length, mass, and time
 Apply dimensional analysis to show
the consistency in relations and
functions
Key Terms
 Length
 Mass
 Time
 Dimensional analysis
 Estimate the order of magnitude
and use significant figures to
determine the accuracy in
calculation
 Significant figures
 Describe position and differentiate
between instantaneous and average
values for speed, velocity and
acceleration
 Position
 Analyse particle under constant
velocity and constant acceleration
 Analyse and solve problems based
on linear motion and free fall
 Average speed
 Instantaneous
velocity
 Average acceleration
 Function of time
Study time
You should be able to complete this section by the end of
Week 1.
Readings/ Resources
Chapters 1 and 2 of your textbook: R. A. Serway and J. W. Jewett (Jr) 2010,
Physics for Scientists and Engineers with modern Physics, 8th ed., Thomson
Learning Academic Resource Centre, USA.
Week 1: Measurement and Motion in One Dimension
1.2
1.0
Introduction
Classical Physics (PH102) deals with classical mechanics and thermodynamics.
Other branches of physics such as optics and electromagnetism would be
introduced in PH101 and quantum physics would be dealt later at 200 level
courses.
This chapter deals with kinematics of classical mechanics. It explains objects
position, velocity and acceleration as a function of time. Kinematic equations for
motion of a particle under constant acceleration are developed.
You would notice that the materials are introductory in nature; however, it is
important that the foundation of the physics is clearly spelt out. You would be
guide through the physics and measurement and motion in one dimension in this
section.
1.1
Standards of Length, Mass,
and Time
International standard for the fundamental quantities of science
use the SI units. The units for length, mass and time are meter,
kilogram, and second respectively. You may check your textbook
on other fundamental units of measurement. It is important to
understand the definition of the SI fundamental units of length,
mass, and time.
More often students do not appreciate the feel for numbers. It is recommended
that you go through Table 1.1 and feel the approximate values of lengths given.
Note that the length is stated in meters. For example length of a standard car may
be around 4 m and not 2 m. Similarly, Tables 1.2 and 1.3 give the approximate
values of mass and time. Note the mass of earth from table 1.2 and the time one
day in seconds. Note that length, mass, and time are fundamental units, whereas
other quantities such as area, volume, and speed are derived units. Read through
Table 1.4 and try to remember some of the common prefixes and abbreviations.
1.2
Dimensional Analysis
The dimension (physical nature of the quantity) for length, mass, and time are L,
M, and T respectively. Note that the dimensions are written in upper case letters.
Dimensional analysis is a powerful and useful way to check the consistency of a
formula. You should note that dimensions can be treated as algebraic quantise.
Note L+L = L, but L  L = L2. It is important to realise that constants have no
significance in dimensional analysis, hence the accuracy of the equation can not
Week 1: Measurement and Motion in One Dimension
1.3
be checked using dimensional analysis. The next section deals with conversion of
units, you should be comfortable with converting non- SI into m k s units
1.4
Conversion of Units
Students generally have the tendency to drop the units through their calculation
and hence difficult to establish the errors once they realise the answer is incorrect.
So unless you feel confident with your units, it is recommended that you follow
through with the associated units. As an example, consider the following:
On an interstate highway in a rural region of Wyoming, a car is travelling
at a speed of 38.0 m/s. Is the driver exceeding the speed limit of 75.0 mi/h?
Convert meters in the speed to miles.
 38.0 m/s  
1 mi 
2
  2.36  10 mi/s
 1609 m 
Convert seconds to hours
 2.36  10
2
 60 s  60 min 
mi/s  

  85.0 mi/h
 1 min  1 h 
The driver is exceeding the speed limit.
1.5
Significant Figures
Significant figures tell us about the uncertainty in a measurement. They are rules
used to determine the final significant figures in answer. Whether dividing or
multiplying the answer should have the same number of significant figures as the
number of figures in a quantity with the lease number of significant figures.
However, when adding or subtracting, we consider the decimal places. Again, the
final answer should have the same number of decimal places as the least number
of decimal places in any quantity. As an example, consider the following:
Given a room with a length of 12.71 m and width of 3.46 m, determine the
area of the room.
12.71 m  3.46 m = 43.976 m2 . However, the rule of thumb is that the
number of significant figures in the final answer should be equal to the lowest
number of significant figure in the measured quantity. The lowest number of
significant figures in this example is in 3.46 m, which is 3 significant figures.
Hence, the final answer is 44.0 m2.
Week 1: Measurement and Motion in One Dimension
1.4
1.6
Position, Velocity and Speed
The location of a particle with respect to a reference point,
generally the origin of the coordinates describes its position.
However, the displacement of a particle is defined as its position
in some time interval. As a particle moves from its initial
position to a final position its displacement is given by
x

≡ x f  xi
[1.1]
Note: we use  to denote change, and x as the symbol for position, where
subscripts f and i are final and initial respectively.
The distance is the total length of a path followed by a particle. For example,
Timoci walked 100 m to a close by discount shop and returned in 400s. He
travelled a distance of 2 x100 m = 200 m. However, the displacement is zero
because the starting and the finishing points are same.
The average velocity vx,avg of a particle (a vector quantity) is defined as the
particles displacement x divided by the time interval t during which that
displacement occurs:
vx , avg

≡
x
t
[1.2]
Hence, in the above example Timoci’s average velocity was zero since his
displacement was zero. The average speed (a scalar quantity) is defined as the
total distance travelled divided by the total time taken to travel the distance.
v avg
≡
d
t
[1.3]
In the above example, Timoci’s average speed was 200/400 = 0.5 m/s.
1.7
Instantaneous Velocity and
Speed
The instantaneous velocity vx equals the limiting value of the ratio x/t as t
approaches zero.
vx ≡ lim
 t 0
x
t
Week 1: Measurement and Motion in One Dimension
[1.4]
1.5
In calculus notation, it is expressed as:
v x ≡ lim
 t 0
x
t
≡

dx
dt
[1.5]
The instantaneous speed of a particle is defined as the magnitude of its
instantaneous velocity. Now, consider the following example:
Consider the following one-dimensional motions: (A) a ball thrown
directly upward rises to a highest point and falls back into the throwers
hand; (B) a race car starts from rest and speeds up to 100 m/s; and (C) a
spacecraft drifts through space at constant velocity. Are there any points
in the motion of these objects at which the instantaneous velocity has the
same value as the average velocity over the entire motion? If so, identify
the point(s).
(A) The average velocity is zero, because the displacement is zero. There is
one point at which instantaneous velocity is zero; at the top of the motion.
(B) The car’s average velocity canno0t be evaluated unambiguously with the
information given, but it must have some value between 0 and 100 m/s.
because the car will have every instantaneous velocity between 0 and 100 m/s
at some time during the interval, there must be some instant at which the
instantaneous velocity is equal to the average velocity over the entire motion.
(C) Because the spacecraft has constant velocity, its average speed and
instantaneous velocity are equal at all times.
1.8
The Particle Under Constant
Velocity
The analysis model deals with the behavior of some physical entity or its
interaction with the environment. The particle under constant velocity model can
be applied to any situation in which an entity that can be modelled as a particle is
moving with constant velocity. If the velocity is constant then
xf
 xi  vxt
Week 1: Measurement and Motion in One Dimension
[1.6]
1.6
The slope of the straight line under a position-time graph is the value of the
constant velocity.
A scientist is studying the biomechanics of the human body. She
determines the velocity of an experimental subject while he runs along a
straight line at a constant rate. The scientist starts the stopwatch at the
moment the runner passes a given point and stops it after the runner has
passed another point 20 m away. The time interval indicated on the
stopwatch is 4.0 s.
a) What is the runner’s velocity?
The problem states the subject is running at a constant rate, hence:
vx 
x x f  xi 20.0  0


 5.0 m/s
t
t
4.0
b) If the runner continues his motion after the stopwatch is stopped, what
is his position after 10 s has passed?
x f  xi  vxt  0   5.010   50 m
So far we have seen particle under constant velocity, how about if the velocity is
not constant. We begin by examining the acceleration of a particle.
1.9
Acceleration
The average acceleration ax, avg of the particle is defined as the change in velocity
vx divided by the time interval t during which that change occurs:

≡
ax , avg
vxf  vxi
 vx
≡
t
t f  ti
[1.7]
Hence, the instantaneous acceleration is defined as the limit of the average
acceleration as t approaches zero.
a x,
≡
 vx
t 0  t
lim

≡
dv x
dt
[1.8]
The instantaneous acceleration equals the derivative of the velocity with respect
to time.
Week 1: Measurement and Motion in One Dimension
1.7
When an object’s velocity and acceleration are in the same direction, the object is
speeding up.
When the velocity and acceleration are in the opposite direction, the object is
slowing down. Note that negative acceleration does not necessarily mean that the
object is slowing down.
Reading Exercise
Read carefully the Conceptual Example 2.5 on page 31 of the textbook and find
out how velocity and then acceleration curves are derived from the position time
curve.
Your text book gives clear and concise diagrams to show motion, in the next
section we would look at that.
1.10
The Particle Under Constant
Acceleration
This section develops the kinematic equations for motion under constant
acceleration. You are very strongly advised to use the symbols for displacement,
position or distance in X direction as x. You are not to use any other symbols
apart from this, some student from their previous encounter/learning use s and u
to denote distance and velocity respectively.. This is not acceptable anymore. The
table below is adopted from your textbook. It is imperative that you learn the
equations given in this table.
We will now consider an example, to understand how these equations can be
utilized. It is however suggested strongly that you work-through other example
questions in the textbook to reinforce knowledge on these equations further.
Week 1: Measurement and Motion in One Dimension
1.8
A jet lands on an aircraft carrier at a speed of 63 m/s.
a) What is its acceleration (assumed constant) if it stops in 2.0 s due to an
arresting cable that snags the jet and brings it to a stop?
Define the x-axis as the direction of motion of the jet. Then:
ax 
vxf  vxi
t

0  63
 32 m/s2
2.0
b) If the jet touches down at position xi = 0, what is its final position?
x f  xi 
1.11
1
1
vxi  vxf  t  0   63  0  2.0   63 m

2
2
Freely Falling Objects
A freely falling object is any object moving under the influence of gravity alone,
regardless of the initial motion. Note that objects thrown upward or downward
and those released from rest are all falling freely once they are released.
Freely falling object experiences acceleration downward, regardless of its initial
motion.
Note that g is a positive quantity, however for vertical motion (the y direction),
the acceleration is downward and has a magnitude of 9.80 m/s2. Therefore, we
always choose ay = -g = - 9.80 m/s2. The negative sign indicates that the
acceleration in the y direction is downward. We conveniently choose upward as
positive y direction.
This concludes Week 1. We have provided you with some solved problems at the
end of every week. These answers are to be used as guide only. There may be
more than one way to solve the same problem. In solving problems, be consistent
with units and lay out the working clearly for the reader.
Week 1: Measurement and Motion in One Dimension
1.9
Solved Problems
The radius of a solid sphere is measured to be (6.50  0.20) cm, and its mass is
measured to be (1.85  0.02) kg. Determine the density of the sphere in
kilograms per cubic meter and the uncertainty in the density.
r   6.50  0.20 cm   6.50  0.20  102 m
m   1.85  0.02 kg

m
   r3
4
3
** Remember while dividing add the percentage uncertainty
also,
   m 3 r



m
r
In other words, the percentages of uncertainty are cumulative. Therefore,
  0.02 3 0.20


 0.103
 1.85
6.50

1.85
    6.5 10
2
4
3
m

3
 1.61 103 kg m
3
and
     1.61 0.17  103 kg m 3  1.6  0.2  103 kg m 3
A truck covers 40.0 m in 8.50 s while smoothly slowing down to final speed
2.80 m/s. (a) Find its original speed. (b) Find its acceleration.
(a)
xf  xi 


1
vi  vf t becomes 40 m  1  vi  2.80 m s  8.50 s which yields
2
2
vi  6.61 m s .
(b)
a
vf  vi
t

2.80 m s 6.61 m s
 0.448 m s2
8.50 s
Week 1: Measurement and Motion in One Dimension
1.10