PHYS.1301: Intro.

to Physics -- Lecture Slides Spring 2022 Page 1

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Chapter 2:
One-Dimensional Kinematics
Position, Distance, and Displacement.
Speed, Velocity, and Acceleration.
Motion with Constant Acceleration.
Freely Falling Objects and the
Acceleration due to Gravity.

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Kinematics 2022 I.R.B.
• In this chapter, and in Chapter 3, we will deal
with a particular branch of physics called
kinematics.
kinematics Kinematics is the study and
description of motion, without concern with
the causes to changes in the motion. In
Chapter 4, we will broaden our study to
include the causes to changes in motion and
thus enter the study of dynamics.
dynamics Together,
kinematics and dynamics form the branch of
physics known as mechanics.
mechanics

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Objects as Point Particles 2022 I.R.B.
• To simplify our discussion of objects, in this
chapter and many future chapters, we will
reduce the objects we deal with (such as
cars, people, rabbits, turtles, etc.) down to
single points located at the center of the
object (center of mass). For example:

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1-D Coordinate System 2022 I.R.B.
• A one dimensional coordinate system consists of a
straight line, with an origin (the zero position) and
an indication of direction. We usually take the
straight line to be the x-axis,
-axis for horizontal motion,
and the y-axis,
-axis for vertical motion.
• In 1-D, there are only 2 possible directions, relative
to the origin (x: right and left, or y: up and down).
To represent these directions, we will assign one
as plus (+) (i.e. the positive direction)
direction making the
other minus (-) (i.e. the negative direction).
direction
• The location of the origin and choice of the positive
direction are arbitrary, but once determined, they
must be used consistently throughout the analysis.
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Position 2022 I.R.B.
• The position of a body locates that body
relative to some fixed reference point
(origin)
origin in a coordinate system. In 1-D, our
coordinate system is a line with a marked
origin (0) and an indication of direction,
represented as positive, to one side of the
origin, and negative, to the other side.
The red dot is at position x = +6 m.

x
-5 0 (origin) +5 (m)
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Displacement 2022 I.R.B.
• The displacement vector points from an
object’s initial position, xi, to its final position,
xf, with its magnitude being the shortest
distance between the two. Mathematically,
the displacement, x (say “delta x”), is:
   Eqn. (2.1)
x  x f  xi
SI unit: m
Only the starting point (xi) and finishing point
( xf ) are important - the path the body takes
is irrelevant. In 1-D, the direction for x is
represented by + or –.
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Distance 2022 I.R.B.
• The distance traveled by an object is
defined as the total length of the path taken.
Thus, the distance traveled is path
dependant – it matters how you get from the
starting point to the finishing point.
• A direction is not assigned to distance (i.e.
no + or –). Distance is always a positive
scalar.
• The SI unit of distance is: meter (m).

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Average Velocity 2022 I.R.B.
• Average velocity,
velocity v, is the displacement
divided by the elapsed time:
  
 x x f  xi displacement Eqn. (2.2)
v  
t t f  ti elapsed time SI unit: m/s

The direction of v is represented by either +

or – in 1-D, relative to the coordinate system
in use. Because t is always positive, the
direction (sign) of v is the same as x.

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Velocity and Speed 2022 I.R.B.
• Average velocity in 1-D is defined when
magnitude (size), direction (+ or -) and unit
are specified. It is a vector.
• Average speed of an object is a defined as
the total distance traveled by the object
divided by the elapsed time:
total distance
average speed 
t
Average speed is always a positive scalar.

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Instantaneous Velocity 2022 I.R.B.
• Instantaneous velocity is a vector,
mathematically defined as:

 x Eqn. (2.3)
v  lim
t  0 t SI unit: m/s
which is interpreted to mean to “calculate the
average velocity, x/t, over shorter and
shorter time intervals approaching zero in the
limit (limt 0), and centered around the
instant in time of interest.” In 1-D, the sign (+
or -) of the instantaneous velocity represents
its direction.
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Average vs. 2022 I.R.B.
Instantaneous
• Instantaneous speed is defined as the
magnitude (size) of the instantaneous
velocity, and is a positive scalar.
• Instantaneous quantities give more specific
information than average quantities because
you are narrowing in on an instant in time,
time
t. With average quantities, you only know
information relating what has occurred on
interval t,
average over the entire time interval,
you do not know what has happened at each
instant in time.
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Graphical Interpretations of v 2022 I.R.B.
• On an x vs. t graph (see figure below):
– The average velocity over a time interval
between two points is equal to the slope of the
line joining the two points (green line).
– The instantaneous velocity at an instant in time
is equal to the slope of the tangent line at the
instant in time of interest (red line).
x Slope of tangent line = instantaneous
velocity at instant of time t2
x3

Slope = average velocity between

x1
times t1 and t3
t
t1 t2 t3
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Example #1: Ch. 2 2022 I.R.B.
• You start off in your car, travelling due east at a
constant 100 km/h for 75 minutes. You then
think that you may not have enough gas to reach
the next station east, and stop the car. After 15
minutes of considering your options, you decide
to turn around and head west, travelling at
constant 120 km/h, for 10 minutes, before
reaching the gas station you had recently passed.
– a) Find your average velocity for the trip, so far.
– b) Find your average speed for the trip, so far.

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Example #2: Ch. 2, Prob. #68 2022 I.R.B.
• A bus makes a trip according to the position–time
graph shown in the drawing.
– a) What is the average velocity (magnitude and
direction) of the bus during each of the segments A, B,
and C?
– b) What are the average
velocity, and speed, for
the entire trip?
Answer in km/h.

Fig. Problem 68, Page 52

PHYSICS, 11th Edition,
Cutnell and Johnson,
FROM: PHYSICS, 11th Edition, Cutnell and Johnson, John Wiley & Sons, Inc., 2018 John Wiley & Sons, Inc., 2018

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Acceleration 2022 I.R.B.
• Average acceleration,
acceleration a, is a vector defined
as change in velocity over the change in
time:  v  v
 v f i Eqn. (2.4)
a 
t t f  ti SI unit: m/s2
Instantaneous acceleration,
acceleration at an instant
in time, is calculated by taking shorter and
shorter time intervals (limt 0), centered
around the time of interest:

 v Eqn. (2.5)
a  lim
t  0 t SI unit: m/s2
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Notes About Acceleration 2022 I.R.B.
• Acceleration is the rate of change of velocity
and has units of meter per second per
second: (m/s)/s or m/s2.
• Again, for 1-D motion, we will use a + or - to
represent the direction of a.
• When velocity and acceleration have the
same direction (same sign), speed
increases. When velocity and acceleration
are oppositely directed (opposite signs),
speed decreases in time and the object is
said to be decelerating.
decelerating

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Graphical Interpretations of a 2022 I.R.B.
• On a v vs. t graph (see figure below):
– The average acceleration over a time interval
between two points is equal to the slope of the
line joining the two points (green line).
– The instantaneous acceleration at an instant in
time is equal to the slope of the tangent line at
the instant in time of interest (red line).
v Slope of tangent line = instantaneous
acceleration at instant of time t2

Slope = average acceleration

between times t1 and t3
t
t1 t2 t3
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Example #3: Ch. 2, Prob. #16 2022 I.R.B.
• Over a time interval of 2.16 years, the velocity of
a planet orbiting a distant star reverses direction,
changing from +20.9 km/s to -18.5 km/s, where
the + and - signs represent opposite directions.
Find
– a) the total change in the planet's velocity (in m/s) and
– b) its average acceleration (in m/s2) during this
interval. Include the correct algebraic sign with your
answers to convey the directions of the velocity and
the acceleration.

FROM: PHYSICS, 11th Edition, Cutnell and Johnson, John Wiley & Sons, Inc., 2018

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Example #4: Ch. 2, Prob. #23 2022 I.R.B.
• Two motorcycles are traveling due east with
different velocities. However, four seconds later,
they have the same velocity. During this four-
second interval, cycle A has an average
acceleration of 2.0 m/s2 due east, while cycle B
has an average acceleration of 4.0 m/s2 due east.
By how much did the speeds differ at the
beginning of the four-second interval, and which
motorcycle was moving faster?

FROM: PHYSICS, 11th Edition, Cutnell and Johnson, John Wiley & Sons, Inc., 2018

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1-D Motion with 2022 I.R.B.

Constant Acceleration
• Deriving the Kinematic Equations for
Motion in 1-D with Constant Acceleration:
Acceleration
• In 1-D, it is customary to neglect the arrow
notation for the vectors in the following
equations and simply convey the directions
with plus and minus signs.
• Remember that each term, except t, is a
vector and must have a direction associated
with it!
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The Kinematic Eqn’s 2022 I.R.B.
• Constant acceleration means that the
velocity increases or decreases at the same
rate throughout the motion.
• So, for a = constant:
v f  vi
aa a from Eqn. (2.4)
t

v f  vi  a( t ) Eqn. (2.6)

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Kinematic Eqn’s Cont’ 2022 I.R.B.
• Because a = constant, and thus v changes
uniformly, we can write: v  1 v  v
2  i f 
x  xi
Eqn. (2.2) gives: v  x  f
t t
Equating the above two expressions gives:
1
x f  xi  ( vi  v f )( t ) Eqn. (2.7)
2
Using vf = vi +a(t) in the above we have:
x f  xi  vi ( t )  1 a( t ) 2 Eqn. (2.8)
2
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Kinematic Eqn’s Cont’ 2022 I.R.B.
• Solving Eqn. (2.6) for t gives:
v f  vi
t 
a
which, when combined with Eqn. (2.8), gives:
2
 v f  vi  1  v f  vi 
x f  xi  vi    2a a 
 a   

Finally, after some algebra, we have:

v 2f  vi2  2a( x f  xi ) Eqn. (2.9)

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The Kinematic 2022 I.R.B.

Equations
• And so we have the 4 basic Kinematic
Equations for 1-D Constant a Motion:
Motion
v f  vi  a( t ) Eqn. (2.6)
1
x f  xi  ( vi  v f )( t ) Eqn. (2.7)
2
x f  xi  vi ( t )  1 a( t ) 2 Eqn. (2.8)
2

v 2f  vi2  2a( x f  xi ) Eqn. (2.9)

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Watch Your Signs! 2022 I.R.B.
• In the 4 Equations (2.6 to 2.9), remember
that each term (except time, t) carries with
it a + or - sign (as determined from the
coordinate system assigned) conveying the
direction of the quantity. Once you’ve set up
your coordinate system, be sure to use it
consistently through the problem.
• Also, note that these 4 equations are valid
only for motion under constant
acceleration.

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Solving Kinematic Problems 2022 I.R.B.
• Draw a picture to represent and assign the positive
and negative directions. Stick with these
assignments as you work through the solution.
• Organize the information you know and what you
need to find out.
• Watch for the motions of two objects being
interrelated or sharing a common variable.
• When the motion of an object is divided in
segments, the final results for the first segment
become the initial information for the next segment.
• Two possible answers to kinematics problems are
often possible – think about what this means.

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Example #5: Ch. 2, Prob. #33 2022 I.R.B.
• A car is traveling at 20.0 m/s, and the driver sees
a traffic light turn red. After 0.530 s (the reaction
time), the driver applies the brakes, and the car
decelerates at 7.00 m/s2. What is the stopping
distance of the car, as measured from the point
where the driver first sees the red light?

FROM: PHYSICS, 11th Edition, Cutnell and Johnson, John Wiley & Sons, Inc., 2018

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Example #6: Ch. 2, Prob. #79 2022 I.R.B.
• This is an example of a “catch-up” problem!
• A cheetah is laying motionless in the grass, ready
to hunt its prey. A hare passes by, running at a
constant velocity of 9.0 m/s, due east. Assume
the chase begins at the instant the hare passes the
cheetah.
– a) Starting from rest, what constant acceleration must
the cheetah maintain in order to catch the hare, 3.0 s
after starting the chase?
– b) How far do the animals run during the chase?

FROM: PHYSICS, 11th Edition, Cutnell and Johnson, John Wiley & Sons, Inc., 2018

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Example #7: Ch. 2, Prob. #82 2022 I.R.B.
• Another example of a “catch-up” problem!
• The leader of a bicycle race is traveling with a
constant velocity of +11.0 m/s and is 10.0 m
ahead of the second-place cyclist. The second-
place cyclist has a velocity of +9.50 m/s and an
acceleration of +1.20 m/s2. The + signs indicate
direction.
– a) How much time elapses before he catches the
leader?
– b) How far does he travel before catching the leader?
– c) What is his velocity when he catches the leader?
FROM: PHYSICS, 11th Edition, Cutnell and Johnson, John Wiley & Sons, Inc., 2018

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Freely Falling Bodies 2022 I.R.B.
• A freely falling body is one which moves
under the influence of gravity alone.
• If we neglect air resistance, all objects that
are in free fall near the Earth’s surface
experience a constant acceleration directed
towards the Earth’s center (down,
perpendicular to the Earth’s surface).
• The magnitude of this acceleration is taken
to be g  9.80 m/s2
and is called the acceleration due to gravity.
gravity
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Free Fall and g 2022 I.R.B.
• An object is in free fall immediately after
being released, regardless of whether it is
dropped from rest, thrown up, or thrown
down.
• Your choice of coordinate system will
determine the sign (+ or -) of acceleration:
+y

If up is + If down
then
a = -g OR is + then
a = +g

+y
ground
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Example #8: Ch. 2, Prob. #54 2022 I.R.B.
• A pellet gun is fired straight downward from the
edge of a cliff that is 15 m above the ground. The
pellet strikes the ground with a speed of 27 m/s.
How far above the cliff edge would the pellet
have gone had the gun been fired straight
upward?

FROM: PHYSICS, 11th Edition, Cutnell and Johnson, John Wiley & Sons, Inc., 2018

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Example #9: Ch. 2, Prob. #58 2022 I.R.B.
• Two stones are thrown simultaneously, one
straight upward from the base of a cliff and the
other straight downward from the top of the cliff.
The height of the cliff is 6.00 m. The stones are
thrown with the same initial speed of 9.00 m/s.
Find the location (measured up from the base of
the cliff) of the point where the stones cross
paths.

FROM: PHYSICS, 11th Edition, Cutnell and Johnson, John Wiley & Sons, Inc., 2018

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Example #10: Ch. 2, Prob. #49 2022 I.R.B.
• A hot-air balloon is rising upward with a constant
speed of 2.50 m/s. When the balloon is 3.00 m
above the ground, the balloonist accidentally
drops a compass over the side of the balloon.
– a) How much time elapses before the compass hits the
ground?
– b) What will be the velocity of the compass just
before it hits the ground?

FROM: PHYSICS, 11th Edition, Cutnell and Johnson, John Wiley & Sons, Inc., 2018

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