Kevin at 13:43, 8 September 2013

2013-09-08T13:43:53Z

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''Kreshnik Angoni'' 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/1DMotion/Module%201D%20Motion%20Worksheet.docx Module 1D Motion Worksheet.docx] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/1DMotion/Module%201D%20Motion%20Worksheet.pdf Module 1D Motion Worksheet.pdf] 

 
Other Resources:
*Haliday & Resnick, Fundamentals of Physics 2.1-2.6;2.10
*[http://cnx.org/content/m42103/latest/?collection=col11406/1.7 Openstax] 

Kinematics is the study of motion, without paying attention to the reason ''why'' it happens (that will come later = force) Simply put, it uses one set of physical quantities (position: <math> x,\ y,\ z\ </math> coordinates, velocity and acceleration) to describe motion without the forces that cause it. 
[http://www.youtube.com/watch?v=JP8A70bq0uc&list=PLEAD6A21917D22918&index=12&feature=plpp_video A good overview video: also follow the video sequence for more detail]

==ONE DIMENSIONAL KINEMATICS==

===Introduction===

In this course we will be looking at two basic types of motion: '''translation''', and '''rotation''' . Note that in both translation and rotation there is no deformation; the object shape and dimensions remain the same.

'''Translation:''' All points in the object move in the same way. For each point of the object, the final position can be found by using the same displacement vector (fig.1). Basically the object is moving in '''straight lines''' 

[[image: Basic_Concepts_KIN_1.png|thumb|none|upright=3.0|
The same displacement vector <math>\overrightarrow{D_1}</math> shifts each object point from position (0) to position (1). The same displacement vector <math>\overrightarrow{D_2}</math> shifts each object point from position (1) to position (2).]] 

'''Rotation:''' The object rotates i.e. it changes its orientation in space. Each point on the object point has a different displacement vector. Even if a rotation is followed by a translation the displacement vectors are different (fig 2.b). 

[[image: Basic Concepts_KIN_2.png|thumb|none|upright=4.5|Figure 2.a: Only rotation. Figure 2.b: Rotation followed by a translation or vice-versa.]] 

Often, in real life, the two types of movement combine together. For instance, an asteroid tumbling through space: rotating as it moves forward, or a football being thrown with spin.
 
However, we often study them separately, taking a complicated situation and making it easier to understand by building a physical model for each motion.
 In this chapter we will deal only with movement along a fixed direction in space (one dimension, 1-D translation).

As mentioned above, in a translation, all object points have the same displacement vector. This simplifies greatly our work; “We study the movement of just one point in the object and the results apply over the object as a whole”. So, one uses '''a particle motion''' as a model for the object motion. 

===Displacement Versus the Traveled Distance===

Consider an object (plane, car..) moving along a straight line (1D). We model this motion by a particle in motion along an axis. To define the position of this particle, we need only one axis. So, we select:

# A Frame origin O
# A Positive direction
# A Length unit "m"
# A Time unit "sec"
# Define t=0 at the initial location (the initial conditions of the motion)

Figure 4 presents the 1-D motion of an object. The particle in the model starts its motion at time <math>t</math> = 0sec and at position <math>x_i</math> = +1m. It moves 4m right, turns and stops at <math>x_f</math> = -3m. The motion lasts for 2sec. 
We define the particle '''displacement''' as <math>\vec{\Delta x} = \vec{x_f} - \vec{x_i}</math> where <math>x_f</math> is the '''final''' location and <math>x_i</math> is the '''initial''' position. 
That is to say the displacement depends '''only''' on where you begin and where you end. It is independent of the ''path'' 

Note that we are dealing with vectors here! 
Note that if <math>\vec{\Delta x} > 0</math> the particle is shifted along in the positive x direction and if <math>\vec{\Delta x} < 0</math> it is shifted in the opposite direction (going in the negative x direction). Remember: The displacement is not the same as the '''travelled distance''' which is always positive. In the case of figure 4, the travelled distance is 4 + 8 = 12m while the displacement is <math>\vec{\Delta x} = (-3) - (1)\mathbf{\hat{ \imath } } = -4\ m\ \mathbf{\hat{ \imath } }</math>. 

[[image: Basic Concepts_KIN_4.png|thumb|none|upright=4.5|1-D motion of an object.]] 

===Velocity Versus Speed===

[http://www.youtube.com/watch?v=r_nqioGm4O0&feature=youtu.be Video Explanation] 

How fast is a particle moving? In everyday vocabulary one uses the '''speed''' (positive scalar) to answer this question in the case of cars, planes etc. If one does not have specific information about the way the particle is moving in particular portions of its path, one has to refer to the '''average speed''' along the path:

<math>speed_{average} = \frac{\ distance\ travelled}{\Delta t} </math>

The average speed along the path is always a positive scalar. 

*Example: In the upper mentioned example of 1-D motion, one would find: average speed = 12m / 2sec = 6m/sec. 

In physics, one uses the velocity (''vector'') to describe the way a particle is moving. For 1D motion along Ox one starts by defining the '''average velocity''': 
<center> <math>
\vec{\upsilon_{Av}} = \frac{\vec{\Delta x}}{\Delta t}
</math>
</center>

Note that the velocity may be '''negative''' or '''positive'''. 

*Example: In the 1-D motion example, Vav = -4m / 2sec = -2m/sec, which is different from the average speed (6m/sec). 

'''Remember''': Although the speed and velocity have the same units (m/s in SI) they are very different. Speed comes from distance travelled, a scalar quantity, velocity comes from the displacement, where you begin to where you end, a vector quantity. 

'''Note''': The displacement and the velocity are both '''positive''' when the particle moves along <math>+Ox</math> direction (along the positive x axis).
When the particle moves along opposite direction <math>(-Ox)</math>, they are both '''negative'''. 

Very often one uses the graph <math>x = x(t)</math> to present the history of a particle motion. The figures 5 & 6 present two such graphs. The '''average velocity''' is calculated easily from the slope at these graphs. In the graph of figure 5, it does not depend on initial moment or the length of time interval. The motion follows all time with the same velocity. In the graph of fig. 6, we see that average velocity increases for smaller time interval (<math>\Delta x</math> ~ equal, <math>\Delta t</math> smaller). So, <math>\upsilon_{Av}</math> does not offer good information for all situations. 
For instance, imagine I throw a ball to you, there will be a displacement for the ball and a change in time. Now imagine I throw the ball, and an eagle swoops down, grabs the ball, flies around with it before dropping it for you to catch. The displacement is the same as the first case, but the time is much longer. This average velocity will not be very useful! 

[[image: Basic Concepts_KIN_5&6.png|thumb|none|upright=5.0|Figure 5: The particle moves at a constant velocity. Figure 6: The particle moves at a changing velocity.]] 

===Instantaneous Velocity===

Imagine you are taking a bus trip to school. You could calculate the average velocity of the trip, which would be the displacement vector divided by the change in time for the trip. But the velocity itself will be changing over this trip, the bus will be stopping, starting, speeding etc. The bus driver can know the speed (and velocity since the bus has a direction) by looking at the odometer. You can ''calculate'' this velocity at any time by dividing up the trip into many small time intervals and looking at the displacement across these time intervals. When these small time intervals get really really small (approaching zero) the average velocity approaches the instantaneous velocity: essentially the velocity at a point in time. Although it is measured using points which straddle the time point in question. 
Look at figure 7 which shows the position as a function of time. You can calculate the average velocity across different time intervals, represented as the slope of the straight lines. The figure show what happens when the time interval <math>\Delta t </math> is made smaller and smaller. What happens to the slope?

[[image: Basic Concepts_KIN_7.png|thumb|none|upright=2.5|For the smallest time interval, the slope (average velocity) fits the curve more accurately.]] 

As you decrease the time interval, the slope starts approximating the slope of the tangent line of the function at point P. Taking this observation further, you may see that the best estimation for the velocity close to point P is the limiting value of the average velocity when the time interval goes to zero. This is the definition of '''instantaneous velocity''' at point P:

<center> <math>
{\upsilon_x^P} = lim_{\Delta t \rightarrow 0} {\Delta x \over \Delta t} \mid_{t_P} = {dx \over dt} \mid^P
</math>
</center> 
[[image: Staycalmandcarryon.jpg|thumb|left|upright=1.0]]

 
'''Don't Worry''': This equation will make perfect sense to you when you cover this in your calculus courses. What you are doing is taking the ''derivative'' of the function. All you have to understand at this point is the concept of making the time interval very small makes the slope approximate the tangent slope at the time point. 
[http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-2/ Ready for this? Check out this video]
 

==How to get instantaneous velocity from a position vs. time graph==
[http://www.youtube.com/watch?v=DWB_FbSI6hA&feature=relmfu How to find a slope in 47s] 
As the velocity (derivative of the position function) is equal to the slope of the tangent of the curve at point P, you may find the velocity straight from the x vs. t graph. 

The graph in figure 8 presents a motion along the x axis:

[[image: Basic Concepts_KIN_8.png|thumb|none|upright=3.5|
Point A: <math>\upsilon_x > 0</math>; Motion along <math>+Ox</math> 
Point B: <math>\upsilon_x > 0</math> (larger); Faster motion along <math>+Ox</math> 
Point C: <math>\upsilon_x = 0</math>; Instantaneous rest, turn back point 
Point D: <math>\upsilon_x < 0</math>; Motion along <math>-Ox</math> 
Point E: <math>\upsilon_x < 0</math> (smaller); Slower motion along <math>-Ox</math> 
Point F: <math>\upsilon_x = 0</math>; Long rest before turning back along <math>+Ox</math>]] 
Not sure how this is working? Take a ruler and follow the curve line with it. You are finding the tangent at each point. 
[http://www.youtube.com/watch?v=C08OcN3z2EA Every curve becomes a straight line if you are super close to it] 
===Acceleration===

In the popular vocabulary the word “acceleration” means "speed increase". In physics, it means simply a '''rate of change of velocity vector''' (magnitude, direction or magnitude & direction simultaneously). Knowing the relation <math>x = x(t)</math> for the motion of a particle, you can build the graph of instantaneous velocity versus time. If it is a straight line, the average acceleration is a constant and is sufficient for the velocity change description. However, such graphs may have different curved parts and one uses '''instantaneous acceleration'''.

[[image: Basic Concepts_KIN_9.png|thumb|none|upright=2.5|]]

So, if the average acceleration is defined as

'''average acceleration = change of velocity / time interval''' or
<center> <math>
\vec{a_{average}} = \frac{\vec{\Delta v}}{\Delta t} = \frac{\vec{v_{f}}-\vec{v_{i}}}{\Delta t} \ \ \ or \frac{\vec{v_{2}}-\vec{v_{1}}}{\Delta t}
</math>
</center>

Analogous to the instantaneous velocity, the instantaneous acceleration at point D is

<center> <math>
{a_x^D} = lim_{\Delta t \rightarrow 0} {\Delta \upsilon_x \over \Delta t} \mid_{t_D} = {d\upsilon_x \over dt} \mid^D
</math>
</center> 

The acceleration at any point D on the graph <math>\upsilon = \upsilon(t)</math> is equal to the graph '''slope''' at that point. The acceleration may be '''positive''' or '''negative'''. It is important to mention that the acceleration sign alone is not sufficient to understand whether the particle is speeding up or slowing down. To get this information, one must compare the <math>a_x</math> sign to the <math>\upsilon_x</math> sign. The following table describes all four possible situations. 

{| border="3" style="margin-left: 3em;"
|-
! scope="col" |
! scope="col" | <math>\upsilon_x</math> is POSITIVE
! scope="col" | <math>\upsilon_x</math> is NEGATIVE
|-
! scope="row" | <math>a_x</math> is POSITIVE
| Velocity and acceleration have the '''same sign''' so the body is '''speeding up'''. Velocity is '''positive''' which means the motion is along <math>Ox</math>.
| Velocity and acceleration have '''opposite signs''' so the body is '''slowing down'''. Velocity is '''negative''' which means the motion is along <math>-Ox</math>.
|-
! scope="row" | <math>a_x</math> is NEGATIVE
| Velocity and acceleration have '''opposite signs''' so the body is '''slowing down'''. Velocity is '''positive''' which means the motion is along <math>Ox</math>.
| Velocity and acceleration have the '''same sign''' so the body is '''speeding up'''. Velocity is '''negative''' which means the motion is along <math>-Ox</math>.
|}
 

'''Remember''': The direction of motion is always shown by the sign of velocity. 

==How to get instantaneous acceleration from a velocity vs. time graph==
As the acceleration (derivative of the velocity function) is equal to the slope of the tangent of the velocity plot, you may find the acceleration straight from the v vs. t graph. 
To get a better meaning of this, let's consider the <math>\upsilon</math> and <math>a</math> graphs in figure 10:

[[image: Basic Concepts_KIN_10.png|thumb|center|upright=3.0|]]

'''A'''-point: <math>\upsilon < 0</math>; motion along <math>-Ox</math>

Since <math>a > 0</math> (opposite sign), there is "slowing down" 

Between '''A''' & '''B''': <math>a</math> = max, (slope on <math>\upsilon</math>) = max; there is "slowing down" 

'''B'''-point: <math>\upsilon = 0</math>, but <math>a \neq 0</math>; this means '''instantaneous rest'''

Since <math>a > 0</math>, ready to move along <math>+Ox</math> 

Between '''B''' & '''C''': <math>\upsilon > 0</math>, <math>a > 0</math>; motion along <math>+Ox</math> 

'''C'''-point: max acceleration 

Between '''C''' & '''E''': <math>\upsilon > 0</math>; motion along <math>+Ox</math>

Since <math>a > 0</math> but decreasing, slight <math>\upsilon</math> increase 

'''I'''-point: zero acceleration; '''instantaneous constant velocity''' 

Between '''I''' & '''J''': <math>\upsilon > 0</math>; motion along <math>+Ox</math>

Since <math>a < 0</math>, there is "slowing down" 

Between '''G''' & '''H''': greater "slowing down" effect 

Beyond '''G'''-point: the stopping effect decreases until it becomes zero at point '''J''' ('''constant velocity''') 

'''Notes''': In real life the velocity cannot change instantaneously. This means that the acceleration has always a finite value (no infinite). At this level, we will study only motions with constant acceleration. To describe a 1-D motion with constant acceleration only "<math>x</math>, <math>\upsilon</math>, and <math>a</math>" parameters are needed. 

A simple way to distinguish an accelerated motion (<math>a \neq 0</math>) from one with constant velocity (<math>a = 0</math>): Fix an interval of time (say 1sec) and measure the travelled distance for several successive intervals of 1 second. If the distances are equal, there is a motion with '''constant velocity''' (fig 11.a). If the successive distances increase (or decrease) by the same quantity, there is a motion with '''constant acceleration''' (fig 11.b). If the successive distances change differently, there is a change of acceleration. 

[[image: Basic Concepts_KIN_11.png|thumb|none|upright=5.5|]]

==Area under Kinematic Graphs==

You have already seen how to find instantaneous velocity from a graph of x vs. t and acceleration from a graph of v vs. t by
using the slopes ( <math>\frac{dx}{dt};\frac{dv}{dt}</math> ) of tangents on the graph at points of interest. The following “inverse” situations happen
in practice, too: 
a) Given the graph v vs.t find the location of particle x at a given moment t. 
b) Given the graph a vs. t, find the velocity of particle v at a given moment t. 
 
[http://www.youtube.com/watch?v=T9v3yteW3hw&feature=plcp Finding the Area Under a Curve]
::::::<youtube>T9v3yteW3hw</youtube>

== Finding the location of particle at time t_2 from the velocity graph==
Knowing that <math>v = \frac{dx}{dt}</math> (1) is a mathematical definition, while in physics, one deals
with measurable ('''small but finite''') interval values <math>\Delta x, \Delta t</math>, one uses <math>v = \frac{\Delta x}{ \Delta t}</math>, 
(2) for a real life velocity and <math>\Delta x = v \times \Delta t</math>, (distance=speed times time) 
(3) For a motion with constant velocity (graph in fig. 1) one may
calculate the displacement from initial location <math>x_{1} at t_{1} as \Delta x = x_{f}-x_{i}=v\times \Delta t = v\times (t_{f}-t_{i})</math> (3) 
As <math> v\times (t_{f}-t_{i})</math> is the area of rectangle (fig.1), the displacement is equal to the area under the velocity
graph for the considered time interval. (<math> v\times (t_{f}-t_{i})</math> has displacement dimensions [m/s]*[s]=[m]) 
 [[File:Figure12.gif|center]] 

The result “displacement <math> \Delta x = x_{f}-x_{i} </math> equal to area under velocity graph” is valid in all cases but the
way one calculates the area is more subtlet if the velocity is not constant (fig.2). One has to
calculate the area by using a set of ‘’extremely’’ narrow rectangles covering the area
under the graph as much as possible. You can probably tell that this method is more precise when using tiny time intervals <math>\Delta t</math>. 
The figures 3,4 show two possible velocity graphs. As the area under the graph in fig.3 is a positive
value, it meand that <math> \Delta x = x_{f}-x_{i} > 0</math> (the displacement is positive)and <math> x_{f}>x_{i} </math> . This means that the particle is moving along the
positive x axis. In fig. 4, the area under the graph is negative, so we have <math> \Delta x = x_{f}-x_{i} < 0</math>
and <math> x_{f}<x_{i} </math>. This means that the particle is moving along the negative x axis. 
 [[File:Figure34.gif|center]] 

B] Finding the velocity of particle at the “moment t2 “ from the acceleration vs t graph 
Similarly, if one has an a vs t graph, one may calculate the change of particle velocity by the area
under the graph. In the frequent case of motion with constant acceleration (figure 5.a,b), one may easily
calculate the velocity at any moment (F-final) if one knows it at an initial (I) moment of time. 
The definition for acceleration is <math>a = \frac{\Delta v}{\Delta t}</math> (4)
As “a” is constant one can express it by “initial and final values” <math>a = \frac{ v_{f}- v_{i} }{ t_{f}- t_{i} }</math> (5)
So, <math> v_{f}- v_{i} = a\times (t_{f}- t_{i})</math> (6)
The time interval being positive, a-sign shows if the velocity value2 increases (5.a) or decreases (5.b). 
 [[File:Figure5.gif|center]] 

Even when “a” is not constant, the area under the a(t) graph gives the change of velocity but in these
cases the velocity does not change linearly with time (Ex. we will see later at “harmonic oscillations”).

Module 3: Intro to 1D Motion - Revision history

Kevin at 13:43, 8 September 2013