PHYSICS SEMESTER ONE
UNIT 2
UNIT 2: KINEMATICS IN ONE DIMENSION
Relationship to Labs: This unit is supported by Lab 2: One dimensional motion (RWSL)
MOTION IN ONE DIMENSION
In physics as in many other courses, many concepts are based on earlier concepts. One dimensional
(1D) motion is the most basic form of motion. Analysis of motion in two or three dimensions often
involves separating the motion into 1D components, and applying the 1D motion relationships.
Terms: General Concepts
Kinematics is the study of motion of objects without concern for the forces that cause the motion.
A frame of reference is a choice of coordinate system that defines a staring point for measuring any
quantity.
In physics, we often treat items as being point-like with infinitesimally small size. This is known as the
particle model. This greatly simplifies the initial analysis. More detailed models can be constructed
based on the simple model.
A scalar is a quantity that can be defined entirely by its size. We will denote variables that are scalars
with italics, A, b , etc.
Anne is driving at a speed of v  125 km h .
Bob flew a distance of d  4.0  104 m in his trip around the world.
A vector is a quantity having both magnitude (size) and direction. In this course, we will denote vectors
in bold, normal print with an tiny arrow “ ” above, A, b , etc. For motion
The dolphin’s displacement is d  150 m North of its original position.
Chi threw the ball with a velocity of v  15 m s at an angle of 45° above horizontal.
The direction of 1D motion can be expressed as a sign on a scalar (positive direction or negative
direction), so we will often use scalar notation for 1D vectors.
Subscripts denote different values of the same type of variable. An initial value is denoted with the
subscript “i”, “1” or “0”. Final values are indicated by subscripts “f”, “2” or no subscript. Letters
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PHYSICS SEMESTER ONE
UNIT 2
distinguish between properties of different items or different directions. vx is the velocity in the x direction. viE is the East component of the initial velocity.
The magnitude of a vector is expressed as the scalar form or as the absolute value of the vector.
The magnitude of A can be expressed as A or A .
Terms: Specific Physical Quantities
As this is a physics course, the following descriptions of the specific physical quantities used in this
section are more detailed than the general concept terms described above.
The position, x, of an object is its location relative a reference location and direction (usually origin and
positive x-direction).
The concepts of velocity and acceleration require a bit more We learn about
Displacement, Δr or d ( x or d for one dimension), is the change in position. In 1D
d  x  x f  xi ,
Distance is a scalar quantity (no direction) equal to the magnitude of the displacement. When there is a
sum of several displacements, the total distance is the sum of the magnitudes of the individual
displacements (sum of individual distances).
Example
A bird on a telephone line moves from initial position xi  6.0 m to final position
x f  1.5 m . The displacement is
x  x f  xi
 1.5 m  6.0 m
 4.5 m
The bird moves 4.5 m in the negative direction or, because we are dealing with only one
dimension, we can say that the displacement is –4.5 m. The distance traveled is 4.5 m.
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PHYSICS SEMESTER ONE
UNIT 2
The bird then moves a displacement of 4.5 m in the positive direction, back to its
original position. The displacement from the original position is now zero because the
initial and final positions are the same. The total distance traveled is 9.0 m, 4.5 m in the
positive direction and 4.5 m in the negative direction.
The average velocity v ( vx in one dimension) of a particle is the particle’s displacement divided by the
time interval t during which that displacement occurs.
vx 
x x f  xi

t
t f  ti
Like displacement, velocity is a vector. We will assume that the time interval is always positive, so the
direction of the average velocity is the same as that of the displacement. When position is plotted
versus time on a position-time graph, the average velocity between A and B is the slope of the line
segment joining two points on the curve representing the position.
Position-Time Graph
40

30
x
x(m)
20
10

t
0
-10
-20
0
10
20
30
40
50
t(s)
vx 
x 34 m  10 m 24 m


 1.2 m s in the positive x direction
t
20 s  0 s
20 s
The average speed v is a scalar quantity equal to the total distance traveled divided by the total time
interval required to travel the distance.
average speed=
total distance
total time
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PHYSICS SEMESTER ONE
UNIT 2
Example
If the bird on the telephone wire takes 10 s to travel 4.5 m down the telephone wire and
back to the starting point, the average velocity is 0 m/s while the average speed is
v
9.0 m
 0.9 m s .
10 s
The slope of the curve on the position-time plot on the previous page is not constant. When the velocity
is constantly changing, we define the instantaneous velocity v (vx in one dimension) as the velocity at a
single point on the curve. The instantaneous velocity is found as the limit of the average velocity
equation when the shrinks to zero. Once again, the scalar notation is used for velocity in one direction.
x
t  0 t
v x  lim
In calculus, this limit is the derivative of x with respect to t.
v x  lim
t  0
x dx

t dt
If you are unfamiliar with the concepts of limits, don’t worry, you will learn this in your math course.
You can check out limits on-line by entering “limits mathematics” or “limits calculus” into your favourite
search engine.
We can look at the following time graph. As shown on the previous page, the average velocity is the
slope of the secant (line through two points on a curve) of the position-time curve. In the limit as time
between points, t , approaches 0 s, the slope of the secant line approaches the slope of the tangent to
the curve at a single point. A tangent to a curve is a line that touches the curve at one point in the
region near the point, and has the same slope of the curve at that point.
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PHYSICS SEMESTER ONE
UNIT 2
Position-Time Graph
40
35
x (m)
30
25
slope approaches
tangent of curve at
point A
20
15
10
5
0
0
5
10
15
20
25
t (s)
t shrinks to 0
Example
As a car accelerates, its position varies with time according to the equation
x  3.500t  0.08750t 2
where x is in metres and t is in seconds. This is shown in the above graph.
a) Find the average velocity between t1  0.000 s and i) t2  1.000 s , ii) t3  0.100 s , iii)
t4  0.010 s .
The position at t1 is
x1  3.500t1  0.08750t12
 3.500  0.000   0.0875  0.000 
2
 0.000 m
i)
The position at t2 is
x2  3.500t2  0.08750t22
 3.500 1.000   0.0875 1.000 
2
 3.4125 m
The average velocity between t1 and t2 is
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PHYSICS SEMESTER ONE
UNIT 2
x
t
x x
 2 1
t2  t1
v12 
3.4125  0.000
1.000  0.000
 3.4125 m s

 3.413 m s
ii)
with the correct number of significant figures
The position at t3 is
x3  3.500t3  0.08750t32
 3.500  0.100   0.0875  0.100 
2
 0.349125 m
The average velocity between t1 and t2 is
x
t
x x
 3 1
t3  t1
v13 
0.349125  0.000
0.100  0.000
 3.49125 m s

 3.49 m s
iii)
with the correct number of significant figures
The position at t4 is
x4  3.500t4  0.08750t42
 3.500  0.010   0.0875  0.010 
2
 0.3499125 m
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PHYSICS SEMESTER ONE
UNIT 2
The average velocity between t1 and t2 is
x
t
x x
 4 1
t4  t1
v14 
0.3499125  0.000
0.010  0.000
 3.499125 m s

 3.5 m s
with the correct number of significant figures
In this case, the velocity seems to be approaching 3.5 m/s as t approaches 0.
b)
Using calculus to find the instantaneous velocity at t1, we can show that
vx 
dx
 3.500  2  0.0875t   3.500  0.1750t
dt
Evaluating at t1,
vx  3.5  0.1750  0.0   3.5 m/s
This is the same as the limit of the average velocity as we reduced t to 0.
The instantaneous velocity is the limit of the average velocity as t approaches 0.
Velocity is the rate of change of displacement with respect to time. v  2.0 m/s North means the
displacement will increase by 2.0 m North each second. Similarly, acceleration
a  ax or a in one dimension  is the rate of change of velocity with respect to time.
a  9.8 m/s2 down means the velocity increases by 9.8 m/s in the downward direction each second.
Acceleration can be either a scalar or a vector.
The average acceleration a , a of a particle is defined as the change in velocity v or speed v divided
by the time interval
ax 
v x v xf  v xi

t
t f  ti
The average acceleration is the slope between two points on a velocity-time curve. This is the same
relationship we had between the average velocity, and the slope between two points on the positiontime curve.
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PHYSICS SEMESTER ONE
UNIT 2
Velocity-Time Graph
40
35
v (m/s)
30
vx
25
slope = ax
20
t
15
10
5
0
0
5
10
15
20
25
t (s)
As with the instantaneous velocity, we can also define the instantaneous acceleration a, a as the limit
of the change in velocity as the time interval approaches zero.
a x  lim
t  0
v x dv x

t
dt
We will limit ourselves to accelerations with constant magnitudes in this course. In this case, both the
average and the instantaneous accelerations are equal to the constant acceleration. We will look at
acceleration where the magnitude is constant but the direction changes when we cover circular motion.
The sign of the acceleration may differ from the sign of the velocity. When the object’s velocity and
acceleration are in the same direction, the object is speeding up. When the object’s velocity and
acceleration are in opposite directions, the object is slowing down. The different cases are shown in the
following table.
velocity
acceleration
motion
+
+
speeding up
+
–
slowing down
+
0
constant velocity
–
+
slowing down
–
–
speeding up
–
0
constant velocity
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PHYSICS SEMESTER ONE
UNIT 2
ONE-DIMENSIONAL MOTION WITH CONSTANT ACCELERATION
Kinematics Equations
In our 1D motion analysis, we will work with relationship between five variables: displacement d, initial
velocity vxi, final velocity vxf, constant acceleration ax and time t. These relationships are found in the
following kinematics equations. There are four common kinematics equations. Each equation relates
four of the five variables. If values of three variables are known, the fourth can be determined by one of
the kinematics equations (and a bit of algebra).
Important Notes:
-
Your text may use an alternate notation for the variables, xi may be xo , x f may be
simply x , etc. Become familiar with any formula sheets provided for exams (check out
sheet before the exam) to avoid confusion due to any discrepancies.
-
The following derivations are provided in your text, probably in more detail. These notes
are not meant to replace the text, but to provide a quick summary of the text material. It
is strongly recommended that you read the applicable sections of the text.
We will define the kinematics equations for constant acceleration motion in the x-
direction.
These equations apply to other directions; just replace the x subscripts with the appropriate subscript.
Under constant acceleration, the average and instantaneous acceleration are equal so we will use the
symbol ax. We will also set ti  0 so t  t f  t . Under these definitions, the average velocity in the
x-direction is
vx 
x f  xi
t
or, with a bit of algebra
x f  xi  vxt
(0)
If the velocity is constant (no acceleration), vx  v so
x f  xi  vt
Noting that the displacement is the difference between the two x-positions
d  x f  xi   xi  vt   xi
or
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d  vt
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PHYSICS SEMESTER ONE
UNIT 2
Although, not strictly one of the kinematics equations promised early, equation (0) is a useful equation
when there is no acceleration.
If there is constant acceleration, the average velocity is
v  vx 
vxi  vxf
2
so we can write
 vxi  vxf
d 
2


t

(1)
This is our first kinematics equation relating four of the variables.
Rearranging the formula for the average acceleration gives the final velocity in terms of initial velocity,
time and constant acceleration in the x-direction.
v xf  v xi  a x t
(2)
Under constant acceleration, the average velocity is equal to the average of the initial and final
velocities
vx 
v xi  v xf
2
We can insert this into equation (0) above to find


x f  xi  12 vxi  vxf t
We can now replace vxf using equation (1) to get
x f  xi  12  vxi t   vxi  ax t  t 
 xi  vxi t  12 a x t 2
The displacement is then


d  x f  xi  xi  vxit  12 axt 2  xi
or
d  vxit  12 axt 2
(3)
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PHYSICS SEMESTER ONE
UNIT 2
Under the constant acceleration conditions, the formula for the average acceleration can be rearranged
as
t
vxf  vxi
ax
We can insert this into equation 1, to give
 vxf  vxi 

 ax 
2
2 
 vxf
 vxi
1

 2
 ax 


d
1
2
 vxi  vxf  
Which we can rearrange as
2
2
vxf
 vxi
 2ax d
(4)
The four kinematics equations and the variables involved are then
 vxi  vxf 
d 
t
2


vxf  vxi  axt
with variables d , vxi , vxf , and t
no a x
vxi , vxf , a x , and t
no d
d  vxi t  12 a xt 2
d , vxi , a x , and t
no v xf
2
2
vxf
 vxi
 2a x d
vxi , vxf , a x , and d
no t
Each of these equations has four of the five different variables ( d , vxi , vxf , ax , and t ). These four
kinematics equations will be provided for you on the formula sheet for the exams. Note that all of the
kinematics equations contain vxi . If you need to calculate a desired variable besides vxi and you
don’t have vxi , then you have to calculate vxi first (even when it is not required) before you can
calculate the desired variable. This type of problem will show up so be ready to calculate vxi first.
Finally, if there is no acceleration, these expressions simplify to
vxf  vxi  v
and
d  vt .
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PHYSICS SEMESTER ONE
UNIT 2
1D MOTION PROBLEMS
In 1D motion problems, you are usually given three of the five variables, and are asked to find one or
both of the two missing ones.
Editor’s note: The names Anne and Bob show up often in this course, especially Bob. They are
convenient names because they lead to alphabetical subscripts “A” and “B”. Bob is used especially often
because it sounds the same if you say it backwards. It also helps that it is short and easy to spell.
Problem solving steps are noted in red italics. You should follow the general problem solving form
shown here to get full marks on your assignments and exams.
Explanatory comments are in blue.
The symbol “~” indicates approximate values for estimates.
We will assume all motion is in the x-direction. You can leave out the direction subscripts when you
solve 1D motion for assignments or exams, but it helps to get in the practice of doing so for when we do
2D motion problems.
Example 1
Moving with a constant acceleration Bob went from 12 m/s to 36 m/s over a distance of 120 m.
a)
How long did it take Bob to cover the 64 m distance?
b)
What was Bob’s acceleration?
a)
Define variables or make a sketch and define variables.
vxi  12 m/s
vxf  36 m/s
d  120 m
t ?
State equation and solve.
We need an equation that includes the three variables we are given and the fourth variable
we wish to find.
 vxi  vxf 
d 
t
2


This equation has all four variables, but we need to isolate the time variable.
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PHYSICS SEMESTER ONE
UNIT 2
t
2d
vxi  vxf
Substituting variables
t
2 120 m 
12 m/s  36 m/s

240 m
48 m/s
 5.0 s
Check
Note that the units are correct
m
s
 m
 s.
m/s
m
The value seems reasonable too. The average speed is ~24 m/s and it would take ~5 s to
travel 120 m at the average speed. This only works with works with constant
acceleration, but that describes all of our kinematics problems.
State
It takes Bob 5.0 s to cover the 120 m.
b)
Define variables or make a sketch and define variables. We really only need to define the
acceleration here, the other terms are already defined.
vxi  12 m/s
vxf  36 m/s
d  120 m
ax  ?
It is often best to use the original data whenever possible. The original data is free of
round-
off and other errors.
State equation and solve.
We need an equation that includes the three variables we are given and the fourth variable
we wish to find.
2
2
vxf
 vxi
 2ax d
This equation has all four variables, but we need to isolate the time variable.
ax 
2
2
vxf
 vxi
2d
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PHYSICS SEMESTER ONE
UNIT 2
Substituting variables
2
2
36 m/s   12 m/s 

ax 
2 120 m 

1152 m 2 / s 2
240 m
 4.8 m / s 2
Check
Note that the units are correct.
m2 /s2 m 2 1
m
 2
 2
m
s m s
The value seems reasonable too. The speed changes by ~25 m/s in ~5 s which is
around 5 m/s2.
State
Bob accelerates at a rate of 4.8 m/s2.
Example 2
Anne accelerates at a rate of –3.5 m/s2 to a complete stop in 6.0 s.
a)
Why did I circle the “to a complete stop” part?
b)
Which of the following should I calculate first?
i) How far did Anne travel?
ii) How fast is Anne moving after 6.0 s?
a)
Some variables are not stated numerically. As you work through the problems in the text
and assignments, you will become more familiar with phrases that define different variables.
A list of common expressions is provided at the end of this example.
“To a complete stop” indicates that the final velocity is zero (see table on page 17).
b)
The known variables are ax , vxf , and t . None of the original four kinematics equations
contain these three variables and d. Stated another way, all of the original four kinematics
equations contain the variable vxi . As noted earlier, if you need to calculated variable
besides vxi and you don’t have vxi , then you have to calculate vxi first.
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PHYSICS SEMESTER ONE
UNIT 2
ii)
Define variables or make a sketch and define variables.
ax  3.5 m/s2
vxf  0
t  6.0 s
vxi  ?
State equation and solve.
v xf  v xi  a x t
vxi  vxf  a x t

 0  3.5 m/s 2
  6.0 s 
 21 m/s
Check
Note that the units are correct.
m
s
2
s 
m
s
The value seems reasonable too. The acceleration is negative so we expect an initial
positive velocity. The acceleration is ~4 m/s2 for 6 s so the initial speed should be ~24 m/s
faster than the final speed.
State
Anne’s initial velocity was 21 m/s.
i)
This is one of those cases where we have to use a calculated value vxi to find another value.
In this case, the answer wasn’t rounded to get the proper number of significant figures. If
the answer had been rounded, use the pre-rounding value to reduce round-off error.
Define variables or make a sketch and define variables.
ax  3.5 m/s2
vxf  0
t  6.0 s
vxi  21 m/s
d ?
State equation and solve.
We can use any of the kinematics equations. There is one we haven’t used so we will use it.
d  vxi t  12 axt 2


  21 m/s  6.0 s   12 3.5 m/s 2  6.0 s 
2
 63 m
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PHYSICS SEMESTER ONE
UNIT 2
Check
m
m 2
s
s mmm
2
s
s
Note that the units are correct.
The value seems reasonable too. The average speed is ~10 m/s for 6 s so the distance
traveled is should be ~60 m.
State
Anne traveled 63 m.
More Notes on Problem Solving
You don’t always need to carry the units through the calculation. If all of the units are
consistent (all lengths in m, all masses in kg…) and your formula is dimensionally correct,
then your answer should be in the expected units. Recognizing when you can leave units out
comes with practice. You will only need to include the units for a line or two when you
solve for the unknown variable algebraically.
A value of 0 does not need units. 0 m = 0 km.
Variable Phrases
variable
initial velocity
phrase
“from rest”, “from the stop light”
value
vxi  0
“dropped”
“left the stop light/sign”
final velocity
“to a stop”, “to a complete stop”
vxf  0
“up to the stop light/sign”
displacement
“round trip”
d 0
“complete cycle”
“return to”
acceleration
“constant velocity”
ax  0
“no acceleration”
“freefall”
ay = –g = –9.8 m/s2
“acceleration due to gravity (on
+y is assumed to be up
Earth’s surface)”
“g” or “G”
a = g = 9.8 m/s2
“multiple g or G”, (Bob pulled 0.2 G on
a = ng = n  9.8 m/s2
the merry-go-round)
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PHYSICS SEMESTER ONE
UNIT 2
Be careful to read the question. In physics, we usually assume people obey the law in our problems.
There are exceptions like
“Carla barrelled through the stop light at 85 km/h and accelerated up to 160 km/h over a
distance of 120 m. Find Carla’s acceleration?”
where a value is stated for the variable. In these cases, use the stated value.
There will be a set of problems with each lecture that you are expected to try. Solutions, and sometimes
hints, are provided via links.
Problems
1. Carla barrelled through the stop light at 85 km/h and accelerated up to 160 km/h over a
distance of 120 m. Find Carla’s acceleration?
2.
This question requires you to use the derivative definitions of velocity and acceleration. The
distance (in m) from a squirrel to a chestnut tree is
x  2.0t  0.10t 2  0.025t 4
where t is the time in seconds.
Find the distance, velocity and acceleration of the squirrel at t  2.0 s .
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PHYSICS SEMESTER ONE
UNIT 2
Working with Freely Falling Objects
A freely falling object is any object moving freely under the influence of gravity alone, regardless of its
initial motion. The free-fall acceleration g is equal to 9.80 m/s2 near the surface of the earth. We will
assume this value is constant. The convention is to define “up” as the positive vertical y-direction so
acceleration due to gravity is a = –g = –9.80 m/s2.
This is not quite true. The actual value depends on elevation and the density of the local rocks. The
website http://gdr.nrcan.gc.ca/gravity/index_e.php provides access to thousands of values across
Canada.
3.
Compare the final velocities of two balls thrown from the top of a 65 m high building. Ball
1 is thrown vertically upwards at a velocity of vyi1 = 12 m/s. Ball 2 is thrown vertically
downwards at a velocity of vyi2 = –12 m/s (negative because up is the positive direction).
Ignore air resistance.
4.
After receiving the baton, relay racer Bob accelerated from an initial velocity of 4.0 m/s with
a constant acceleration of 1.0 m/s2 over a distance of 42 m. Calculate the length of time that
Bob was accelerating.
5.
Officer Anne was examining the 38 m long tread marks left by a lawn mower to determine
the lawn mower exceeded the 2.0 m/s2 limit for the magnitude of lawn mower accelerations.
Witnesses claimed that the lawn
mower skidded for 6.0 s before coming to a stop. Did the
lawn mower exceed the acceleration limit?
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18
PHYSICS SEMESTER ONE
UNIT 2
POSITION-TIME, VELOCITY-TIME AND ACCELERATION-TIME GRAPHS
The following graphs show the curves that would result if acceleration, velocity and position are plotted
versus time under the noted conditions. The value of the velocity-time curve at any time equals the
slope of the position-time curve at that time, and the value of the acceleration-time curve at any time is
equal to the slope of the velocity-time curve.
a) ax = constant > 0, vxi = 0, xi = 0
ax
vx
increasing slope
constant slope
constant
t
0
x
0
t
0
t
In this set of graphs, acceleration is constant and positive. That means that the slope of the velocitytime curve is positive with a value equal to the acceleration. The velocity has an initial value of zero, as
defined by the initial velocity.
The velocity is the slope of the position-time curve. With the velocity increasing, the slope of the
position-time curve also increases. The position has an initial value of zero, as defined by the initial
position.
Starting from the position-time curve, we see that the slope is increasing so the velocity must be
increasing. The constant positive slope of the velocity-time curve indicates that the acceleration is
constant and positive.
b) ax = constant < 0, vxi > 0, xi > 0
ax
vx
0
x
0
t
0
t
t
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PHYSICS SEMESTER ONE
UNIT 2
Here, acceleration is constant and negative. That means that the slope of the velocity-time curve is
positive with a value equal to the acceleration. The velocity has an initial positive value so it must go
through zero at some point.
With the velocity initially positive, the slope of the position-time curve is initially positive. The positiontime curve levels off and becomes negative as the velocity goes through zero into the negative values.
c) ax = constant > 0, vxi = 0, xi < 0
ax
vx
x
0
t
0
0
t
t
This set of graphs is almost identical to the set of curves in part a). The only difference is that the
position value is initially negative so the position-time graph has a negative x-intercept.
d) ax = 0, vxi < 0, xi > 0
ax
0
vx
t
x
0
t
0
t
We can work from the position-time curve to the acceleration curve for this set. The slope of the
position time curve is constant and negative. This indicates that the velocity is also constant and
negative. The acceleration is zero because the velocity-time curve is zero (velocity is constant).
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PHYSICS SEMESTER ONE
UNIT 2
Kinematics Equations and Position-Time, Velocity-Time, and Acceleration-Time Curves
Under zero acceleration conditions, the change in position (displacement) due to a constant velocity vxi
over a time interval t is given by the equation
x  vxi t
Graphically, this is the area between the curve in the velocity-time graph and the time axis.
x  vxi t
vx
Rectangle with
length t and
vxi
height vxi
vxi
0
t
t
Note that this applies to both positive and negative displacements.
vx
t
0
x  vxi t  0
t
vxi
vxi
Using conventional algebra, calculating the area under the velocity-time curve is limited to the simple
cases of zero acceleration (rectangle), constant acceleration (triangle) and a few other recognizable
curves.
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PHYSICS SEMESTER ONE
UNIT 2
We usually need tools like calculus to determine slopes or areas for curves. There are however special
cases, when velocity and acceleration are constant or have constant slope, where we can determine the
area under a curve using geometric relations. You can even work with slightly more complicated curves
by breaking them down into smaller sections, and finding the areas or slopes for those sections. It is
also hoped that this will help as a practical application to consider when you are introduced to these
concepts in calculus.
Examples
velocity – time graph
30
B
velocity (m/s)
25
C
20
3
D
15
2
1
10
4
5
A
0
0
10
20
30
40
50
60
70
80
time (s)
The displacement from A to B is the area of a triangle between the curve and time axis.
A1  12 bh

1
2
 20 s  25 m/s 
 250 m
1
so
dA to B  250 m
Note that this is the same displacement we would find if we use
 vi  v f
d 
 2

t

with vi  0 , v f  25 m/s and t  20 s .
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22
PHYSICS SEMESTER ONE
UNIT 2
The displacement from B to C is the area of a rectangle between the curve and time axis.
A2  bh
  20 s  25 m/s 
2
 500 m
so
dB to C  500 m
The displacement from C to D is the combination of the areas of a rectangle and a triangle
between the curve and time axis.
A3  12 bh

1
2
3
 30 s 10 m/s 
 150 m
A4  bh
  30 s 15 m/s 
4
 450 m
so
dC to D  A3  A4
 150 m  450 m
 600 m
All of this motion is in the same direction (velocity is positive from A to D). The
displacement from A to D is the sum of the displacements for the different sections.
d A to D  d A to B  d B to C  dC to D
 250 m  500 m  600 m
 1350 m
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23
PHYSICS SEMESTER ONE
UNIT 2
If the curve passes through zero, subtract the area below the time axis from the area above the time
axis.
velocity – time graph
15
velocity (m/s)
10
5
0
-5
0
10
20
30
40
50
60
-10
-15
-20
time (s)
The displacement between 0 and 20 s is
A 
1
2
 20 s 10 m/s 
 100 m
The displacement between 20 and 50 s is
A 
1
2
 30 s 15 m/s 
 225 m
The total displacement is then
d  A  A
 100 m   225 m 
 125 m
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PHYSICS SEMESTER ONE
UNIT 2
Often, the velocity-time curve or acceleration-time curve is not made up of linear segments. In this
case, you can estimate the path of linear segments to get an approximate area. Alternatively, you can
use the grid of the graph to count the number of squares of a particular area are between the curve and
the time axis. In this case the area of the indicated rectangle is


A  5 m/s2  5 s   25 m/s
There are approximate 27 of such squares beneath the curve. The change in velocity is then
v  nA
  27  25 m/s 
 675 m/s
acceleration-time graph
35
2
acceleration (m/s )
30
25
20
15
10
5
0
0
5
10
15
20
25
30
35
40
45
time (s)
The University of Colorado at Boulder has a Physics Education Technology website,
http://phet.colorado.edu/new/index.php, with several excellent simulations for a huge range of physics
concepts. Relevant simulations are noted throughout the course, but feel free to explore the different
simulations on you own.
Check out http://phet.colorado.edu/new/simulations/sims.php?sim=The_Moving_Man for a look at 1D
motion graphs. You can set initial positions, velocities and accelerations using the slide bars or number
window to the left of the charts. When the moving man hits one of the walls, his velocity rapidly drops
to zero due to a large acceleration in the direction opposite the velocity.
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25
PHYSICS SEMESTER ONE
UNIT 2
Problems 6
a) i)
Find the displacement between 0 and 40 s in the following curve.
Find the acceleration for each segment of the trip.
velocity (m/s)
ii)
20
15
10
5
0
-5 0
-10
-15
-20
-25
-30
-35
5
10
15
20
25
30
35
40
45
time (s)
b) i)
Find the change in velocity between 5 and 30 s in the following curve.
ii)
Sketch the velocity-time and position time curves. Assume the initial velocity is
zero and the initial displacement is 1000 m.
15
2
acceleration (m/s )
10
5
0
-5
0
5
10
15
20
25
30
35
40
45
35
40
-10
-15
-20
time (s)
60
40
20
velocity (m/s)
6.
0
-20 0
5
10
15
20
25
30
45
-40
-60
-80
-100
-120
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time (s)
26
PHYSICS SEMESTER ONE
UNIT 2
displacement (m)
1000
500
0
0
5
10 15 20 25 30 35 40 45
-500
-1000
time (s)
Estimate the total displacement between 0 and 45s for the following curve.
30
25
velocity (m/s)
c)
20
15
10
5
0
0
5
10
15
20
25
30
35
40
45
time (s)
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PHYSICS SEMESTER ONE
UNIT 2
Calculus gives us the ability to generate analytic expressions (exact equations) for the areas under much
more complex curves. The following is a brief overview of some of the fundamentals of integral
calculus. It is recognised that most students have not studied integral calculus before so, for calculus
students, this is a brief look at a topic you will study in the near future.
To some degree of accuracy, the area under any curve can be calculated by breaking the region beneath
the curve into n narrow time intervals tn.
area of segment xn  vxn tn
vx
vxn
0
ti
tf t
tn
The total displacement between ti and tf
x n.
x   xn   v xn t n
n
n
tn shrinks. In the limit is tn approaches 0 ( 0), this
approximation becomes exact.
x  lim
t n 0
v
xn
t n
n
In calculus, the limit of this sum is the integral of the function vx(t) with respect to t from ti to tf.
x   vx  t  dt
tf
ti
We can use the exact same argument to show that the change in velocity due to the acceleration ax(t)
from ti to tf is
vx   ax  t  dt
tf
ti
Under constant acceleration, ax(t) = ax = constant, this integral becomes
v x   a x dt  a x t f  ti 
tf
ti
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PHYSICS SEMESTER ONE
UNIT 2
Writing vx  vxf  vxi , ti  0 and t f  t , this becomes good old
vxf  vxi  axt
Noting that a x and vxi are constant and writing vx  vxf  vxi  axt , we can further integrate to get
x f  xi   vx  t  dt
t
0
t
t
0
0
  vxi dt   ax tdt
 vxi t  12 ax t 2
or
d x  vxi t  12 axt 2
When the displacement, velocity and acceleration terms were first introduced, it was noted that the
velocity is the slope of the position-time curve and acceleration is the slope of the velocity-time curve.
Here we find that the displacement between two times is equal to the area between the velocity-time
curve and the time axis, and the change in velocity is the area between the acceleration-time curve and
the time axis.
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29
PHYSICS SEMESTER ONE
UNIT 2
SUMMARY OF TERMS
t, t  t f  ti
time
also time interval, elapsed time
position
initial position, final position
x, xi, xf
displacement
vector change in position
d  x  x f  xi
distance
scalar magnitude of displacement
x , Δx, d , or d
or sum of distances
velocity
vector, change in position per unit time v
average velocity in x-direction
vx 
instantaneous velocity in x-direction
x x f  xi

t t f  ti
x dx

t  0 t
dt
v x  lim
you can treat 1D vectors as scalars
speed
acceleration
scalar, magnitude of velocity
v v
average speed
v
total distance
total time
vector or scalar, change in velocity per a, a
unit time
v x v xf  v xi

t
t f  ti
average acceleration in x-direction
ax 
instantaneous acceleration in x-direction
a x  lim
v x dv x

t  0 t
dt
Kinematics Equations with Constant Acceleration
initial/final velocity, acceleration and time
v f  vi  at
displacement, initial/final velocity and time
d
displacement, initial velocity, acceleration and time
d  vit  12 at 2
displacement, initial/final velocity, and acceleration
v 2f  vi2  2ad
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1
2
 vi  v f  t
30
PHYSICS SEMESTER ONE
UNIT 2
NANSLO Physics Core Units and Laboratory Experiments
by the North American Network of Science Labs Online,
a collaboration between WICHE, CCCS, and BCcampus
is licensed under a Creative Commons Attribution 3.0 Unported License;
based on a work at rwsl.nic.bc.ca.
Funded by a grant from EDUCAUSE through the Next Generation Learning Challenges.
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31