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Full text of "Velocity and Acceleration - Project Physics Programmed Instruction"

Velocity and Acceleration 




Digitized by the Internet Archive 

in 2010 with funding from 

F. James Rutherford 



http://www.archive.org/details/velocityacceleraOOfjam 



Acknowledgment 

This document is only one of many instructional materials 
being developed by Harvard Project Physics. Like all existing 
Project materials--additional text units, laboratory experi- 
ments, teachers guide, and the rest--it is now an experimental, 
intermediate stage. This text is based on earlier versions 
used in cooperating schools, and its development has profited 
from the help of many friends and colleagues, both within Pro- 
ject Physics and outisde that group. Successive revisions in 
this text are planned in the light of further experience and 
use. In the final experimental edition scheduled for 1967 ,a 
detailed acknowledgment will appear of those contributions that 
were found to be of greatest use and permanence in the develop- 
ment of the new course. 

The work of Harvard Project Physics has been financially 
supported by: The Carnegie Corporation of New York, The 
National Science Foundation, The Sloan Foundation, and The 
United States Office of Education. 



Copyright (6) F. James Rutherford, 1965. All rights reserved. 
This material, or any part thereof, may not be reproduced or 
quoted in any form without prior written approval of the Pro- 
ject Director, HARVARD PROJECT PHYSICS. 



The material you are about to study is a programed text-- 
often called a "program" for short. It is designed to 
help you learn some useful techniques and important con- 
cepts which are essential for you to know in the course 
you are presently taking. 

Although you will be asked questions at every step in the 
program, there is no penalty for errors. In fact, you will 
often be asked to guess at answers you probably don't know. 
Occasionally hints will follow a question. When you can 
answer the original question, you may skip the rest of the 
hints; if you have trouble with the problem, keep working 
out the hints. There are no time limits for a program 
either. You should work at your own pace. And if you 
have any questions about the material, see your teacher 
for assistance. Remember, A PROGRAM IS NOT A TEST. 

When you are asked a question in the program, try to answer 
it or puzzle it out on your own. Every question is answered 
in the program immediately after the answer space. (Some- 
times below the question, or below and to the right, or on 
the page following the question.) If your answer does not 
coincide with the answer given, try to see why the program's 
answer is more reasonable. If you do not see why, consult 
with your teacher. 

Since you do not want to see the answer before you have 
tried the question yourself, take a sheet of paper and 
place it below the material you are studying. When you 
have finished your answer, slide the sheet down to find 
the correct answer. You will find this "answer shield" 
a handy place to make calculations. In general, you 
should have additional scrap paper for doing arithmetic. 
For most of these programs, a ruler is also needed. 



SAMPLE FRAMES 



1 + 3 = 4 = 22 

1+3 + 5= 9 = 32 

1+3+5+7= 16 =42 

1 + 3 + 5 + 7 + 9= = 

(Fill in the blanks.) 



ANSWER: 25, 52 



The sum of the first 8 odd numbers is , or 



ANSWER: 64, 82 



You have seen a film about mass, acceleration, velocity, 
force, and related concepts. You have made some investi- 
gation of motion in the laboratory. You have read about 
early approaches to these phenomena and about present ways 
of analyzing them. Now we are going to ask you to trans- 
late these phenomena into symbols and graphs. Physicists 
have found this translation helpful in dealing with more 
complex problems involving these phenomena. 



Consider a block sliding on a horizontal board, assumedly 
with very little friction. The diagram gives a side view 
of the situation; you are to fill in the missing symbolic 
and verbal information. When you see a blank, fill it in, 
then look further down the page, or on the next page, for 
the correct answer so that you can check your own progress 
Cover up material below the dotted line until you have 
written your own answer. 

These two diagrams simulate an experiment such as you saw 
in the film and/or did yourself in the laboratory: 



INITIAL 
POSITION, 

s . 





FINAL 
POSITION, 





The diagram is repeated below. Fill in the missing symbols 
and then check your version with the one on the previous 
page. 








Give an algebraic expression for the elapsed time 



Give an algebraic expression for the distance traveled 



(Cover up the answers below until you have answered yourself.) 



Elapsed time: t^ - t.. (Remember: t. means "initial 
Distance traveled: s^ - s.. time," s. means "initial posi- 
tion," t^ is "final time," etc.) 



Which of the problems below can be answered on the basis of 
information given so far? (Give the answer where possible . 



(1) Give an expression for the speed at a point halfway 



between s. and s^: 
1 f 



(2) Give an expression for the average speed: 



(3) Is a force acting on the block? 



(4) Is the block moving at a constant speed? 



(1) You can't answer this one. The speed at a point is a 
complex concept which we have not covered as yet; you may 
understand it, but we have no way of deriving an expression 
for it from the information given. 

(2) The average speed is the distance covered divided by 
the time required to cover it: 

^f - ^i . 

(3) You can't tell from the information given. 

(4) You can't tell from the information given. 



For the last two answers: to determine whether a force is 
applied, one has to know whether or not the block has ac- 
celerated. There is no information about that: the block 
may have speeded up and/or slowed down between initial and 
final positions, or its speed may have been constant. 

What further information would you need to have in order to 
determine whether the block accelerated? 



You would have to know whether the average speed was 
changing; in other words, you would have to know at least 
one more value for both s and t. 



Consider for a moment the symbolic value (s, - s.). Rather 
than use two symbols with a minus sign, it is easier to have 
a symbol which means "the change in." The symbol used to 
represent change is the Greek letter "delta," which looks 
like this: A. Thus the change in position (s^ - s.) is given 
as follows: As. 

How would you symbolize the elapsed time? . 



(Don't look until you write.) 



Elapsed time = At 



Now write an expression for the average speed of the block 
using delta notation: . 



Average speed = — 



On the following page is a diagram representing a view of 
two blocks seen from directly above. They are going in 
different directions, and the two paths of the blocks are 
marked off in units of measurement chosen arbitrarily for 
this problem. Examine the diagram, and then give answers 
to the questions which follow. 



LOOKING VERTICALLY DOWN 
ON TWO MOVING BLOCKS 



BLOCK A 




J : 





BLOCK B 



How do the speeds of the two blocks compare? 



Technically speaking, you can't answer until we specify 
whether we mean instantaneous speed at some point or the 
average speed from initial to final states. Assuming we 
mean average speeds, they are about equal (they are meant 
to be equal, but our drawing is only so accurate). 

How do the velocities compare? 



Assuming that we mean average velocities, you might be 
tempted to say that they are equal also, which is not 
true. Velocity is a vector quantity, involving both speed 
and direction; the speeds of the blocks are equal, but the 
velocities are not equal because 



The velocities are not equal because the blocks are moving 
in different directions . 

We should like to be a little more precise, however, about 
the difference between the motions of A and B. It would be 
helpful to have some common frame of reference to measure 
their displacements. You have already learned how to use 
graphs to plot the coordinates of vectors; we can use the 
same method for the displacements of A and B. We consider 
each of them to be the vector sum of vertical and horizontal 
vectors drawn on suitably chosen coordinate axes. On the 
next page the displacements of A and B are plotted. 



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8 



You can draw vertical lines to the horizontal axis from the 

points s. , s , and s^ on the vectors A and B. Now consider 

^ 1 m f 

the initial, middle, and final positions as they are pro- 
jected onto the horizontal axis. (The horizontal projection 
of a vector is called its "horizontal component," and the 
process of finding the components of a vector is called the 
"resolution of vectors.") Now, how does block B's displace- 
ment compare with A's when the displacements are projected 
on the horizontal axis; or, more precisely, how does the 
horizontal component of displacement B compare with the hor- 
izontal component of displacement A. 



ANSWER: (In your own words.) Block B's displacement is 
less than A's when the displacements are projected on the 
horizontal axis; or, more precisely, the horizontal com- 
ponent of displacement B is less than the horizontal com- 
ponent of displacement A. 



Now, what can we say about the components of the velocities 
A and B? (Remember: At is the same for each of the posi- 
tions s., s , and s^.) 
1 m f 



HINT: Since v = As/At, and At remains constant for each 

interval, v is directly proportional to As (symbolically: 

V =^ As). Therefore, we can be confident that the horizontal 

of velocity A is (greater/less) than the 

of velocity B. 



ANSWER: The horizontal component of velocity A is greater 
than the horizontal component of velocity B. 



Remember that the speeds are equal, though the velocities 
are not. Now consider the vertical components of dis- 
placements A and B: 

(1) How do these components compare? 

(2) How about the vertical components of velocity 
vectors based on the same information about A and B? 



ANSWERS: (1) Displacement A's vertical component is less 
than displacement B's. 

(2) We should confidently expect the vertical 
component of velocity A to be less than the vertical 
component of velocity B. 



We can check our ideas about the velocity components by 
plotting the velocity "sum" vectors on a graph and re- 
solving their components, just as we did with the displace- 
ment vectors. The velocity vectors will have the same di- 
rections as the displacement vectors; their magnitudes will 
be the magnitudes of the displacement vectors divided by 
the time elapsed during each interval (As/At) . On the next 
page is a graph showing the velocity vectors. Observe that 
we could have chosen any convenient unit to measure the 
magnitude of v, but since we chose the same scale for v 
as for As, and since At = 1, the velocity vectors look pre- 
cisely the same as the displacement vectors, although they 
mean quite different things. 





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11 



Now consider the symbols used above. The time elapsed be- 
tween initial and middle is the same as that between middle 
and final; give three symbolic expressions for the time 
intervals between exposures : 



t-t. t^-t At 
mi f m 



Now give the simplest way of expressing the change in dis- 
tance between exposures for block A: 



As (As = s - s . = s^ - s ) 
m 1 f m 



Finally, v = 



- As 
^ = At 



12 



Consider the two events pictured below. Two blocks are 
photographed simultaneously at intervals of one second. 
One block moves at a constant speed; the other accelerates 
Which is which? (Answer only if you can on the basis of 
information provided here.) 



s . 
1 
, ^ 

I 
I 



s 
m 
I 







s . 

1 

1 T 




^m 






^f 










B 




1 1 
1 1 
1 1 




1 1 














-J i L. 




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. 



A represents a block moving at constant speed; B represents 
a block which accelerates. 

What can you deduce further about situations A or B? 



HINT: What about the force acting in A? in B? 



ANSWER: Since the speed in A is constant, there is no force 
acting. (Assuming, of course, that there is no friction; 
otherwise the force acting on the block A is just equal to 
the opposing force of friction.) In B, since the block ac- 
celerates, there is an unbalanced force acting. 



13 



Now consider for a moment the distance covered in B as com- 
pared with that in A. In B the block covers an additional 
(how many?) unit(s) of distance each second. 



ANSWER: In each second the block in B covers one unit of 
distance more than it did the previous second. This speed- 
ing up, as you know, is called acceleration. 



Suppose the block is thought of as gaining one additional 
unit of distance each second; then it has gained speed, 
has it not? How much average speed has it gained; in other 
words, how much speed does it gain each second? 



HINT: It covers one additional unit of distance in one 
second, every second. Describe in your own words how much 
speed it gains each second. 



ANSWER: It covers one additional unit of distance each 
second, so i_t gains an average speed of one unit per second 
during each second, or one unit per second per second. For 
instance, if the units were feet, then it would gain one 
foot a second, and it would gain this much every second, or 
one foot per second per second. This may sound a little 
complicated, but it should become clearer as you do some 
numerical problems related to it. 



14 



Suppose that the distance an object travels is 4 cm during 
the first second, 6 cm during the second second, and 8 cm 
during the third second. What is the average speed during 
the first second? during the second second? during the third 
second? 



First second: 4 cm per second; 
Second second: 6 cm per second; 
Third second: 8 cm per second. 



Now what is the change in average speed from second to 
second? 



The change in average speed is 2 cm per second each second; 
that is, 2 cm per second is added to the velocity each sec- 
ond. The acceleration, therefore, is 



ANSWER: 2 cm/sec/sec, or 2 cm/sec^ 



NOTE- 2^"^/sQg = 2cin,_l_ ^ ^ ^"^ 
sec sec sec sec' 



Now compute the acceleration of another object from the 
data given below. 

DISTANCE COVERED DURING 
FIRST SECOND SECOND SECOND THIRD SECOND 

7 ft 19 ft 31 ft 



15 



HINT: What is the average speed during each second? What 
is the increase in speed during each second? What, then, 
is the acceleration? 



ANSWER: Since the unit of time is conveniently one second, 
the distance covered is numerically equal to the average 
speed. The increase in speed during each second, therefore, 
is 12 ft/sec. ([19 ft/sec - 7 ft/sec] = [31 ft/ sec - 19 
ft/sec] = 12 ft/sec.) The acceleration, therefore, is 12 
ft/sec/sec, or 12 ft/ sec^ . 



Compute the acceleration of another object. 

TOTAL DISTANCE AT THE END OF 
THREE SECONDS SIX SECONDS NINE SECONDS 

3 m 21 m 54 m 



HINT: Your first step might be to set up a revised table 
like the following: 

DISTANCE COVERED DURING 
FIRST 3 SECONDS SECOND 3 SECONDS THIRD 3 SECONDS 
3m 18 m 33 m 

(21 m [total (54 m - 21 m 

distance] - = 33 m) 

3 m [distance 
covered pre- 
viously] = 
18 m) 

Now, what was the average speed during the first three 
seconds? the second three? the third three? What, then, 
was the acceleration? 



16 



ANSWER: 

Average speed during the first three seconds: 1 m/sec. 
Average speed during the second three seconds: 6 m/sec. 
Average speed during the third three seconds: 11 m/sec. 

The increase in average speed is 5 m/sec each 3 seconds, 

5 / 9 

so the acceleration is zr m/sec^ . 



These observations and results can be presented in tabular 

form, 

the second, and so on.) 



(Note that s, is the first observed position, s„ 



s- Om >^ 



s . 54m 
4 



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► 18m 3sec 



s- 21m 1 



As At 



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sec 



. m 
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i 



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m 
sec 



A (As/At) 

At 

or a 



5_ m 
3 sec^ 



5 m 
3 sec^ 



Some of the symbols above may be new to you, although the 
concepts they represent should be more or less familiar. 
"v" stands for average velocity; it is pronounced "v-bar" 
and is the algebraic equivalent of As/At. (Be sure that 
you don't confuse v with v.) "a" stands for acceleration, 
the change in average velocity during a given time inter- 
val (Av/At) . 



17 



Below is a table giving the positions of an object (s,, s_ , 
etc.) and the times between positions (At). The differences 
between positions (As) , average speeds (v) , changes in speed 
(Av) , and acceleration (a) are to be filled in. 



s, -3 cm -^ 



Ma) 



s„ 4cm < 



s^ 11cm -s 



s . 18cm -^ 
4 



(b) 



(c) 



As 



At 



As/At A (As/At) 
or v or Av 



A (As/At) 

At 

or a 



•1 sec (d) ^ 



Mg) 



1 sec (e) 






■1 sec (f) 



(h) 



(i) 



(J) 



(a) 7cm 

(b) 7cm 

(c) 7cm 



(d) 7 

(e) 7 

(f) 7 



cm 



sec 












(g) 





(i) 





cm 










sec 












(h) 





(J) 





cm 











sec 



Now consider the data above as they describe the motion of 
an object; what can you say about this motion? 



HINT: Is the speed constant? Is there any acceleration? 
Is there any force acting on the object? 



18 



ANSWER: The speed is constant; there is no acceleration; 
therefore, there is no net force acting on the object. 



We have been observing an object which moved in a straight 
line : 




There is no objection to tilting our frame of reference so 
that the vectors are horizontal: 



A&. 



The vector as now points to the right. Draw different 
vectors s, and s„ s^ 
points to the left. 



vectors s, and s„ so that their difference (s_ - s, ) 



19 



Two cases are possible: 



2i ^1 , 

T r 



fr 



ft = and >f 

I • 

JiS_J L AS 



A similar situation occurs with the velocity vector. In 
Straight-line motion, the average velocity can remain con- 
stant from interval to interval: 



^1 ^1 



or 
->■ -»• 

l2 ll. 



or it can change. But since the direction is fixed, only 
the magnitude (the speed) of the velocity vector can change 
Draw the velocity vectors for the two cases where the speed 
increases : 



20 



ANSWERS 



^1 ^1 



■^ 



< and > 



Av Av 

> 



Since there are only two possible directions in which these 
vectors can point, we can simplify computations considerably 
if we assign plus signs to vectors pointing in one direction, 
and minus signs to vectors pointing in the opposite direc- 
tion. By convention, vectors pointing to the right (or up, 
if you use the vertical axis) are plus, and vectors pointing 
to the left (or down) are minus. Our computations are sim- 
plified because we can now use ordinary arithmetic on the 
magnitudes of our vectors. In our horizontal reference sys- 
tem, +2 cm is a displacement of 2 cm to the right, and -2 cm 
is a displacement of ____. 



ANSWER: -2 cm is a displacement of 2_ cm to the right 



Furthermore, +7 cm/sec is a velocity of 
and -4 cm/sec^ is an acceleration of 



ANSWERS: +7 cm/sec is a velocity of 7 cm/sec to the right , 
and -4 cm/sec^ is an acceleration of 4 cm/sec^ to the left . 



21 



A speed of 7 cm/sec may be a velocity of either +7 cm/sec 
or -7 cm/sec. Consider a body with an initial speed of 
7 cm/sec. Draw a vector diagram for a change in velocity 
of -4 cm/sec. (Label the vector which is v.) 



Again, two cases are possible: 





7 cm/sec 
11 cm/sec 


and 


7 


cm/sec ^ 




3_ 


crn/sec 


*- 


Av 



In what follows, we shall always be dealing with vector 
displacement, vector velocity, and vector acceleration. 
To simplify the writing we shall omit the arrows over 
vector quantities. We shall also omit the plus sign; a 
number with no sign if front of it, if it is a vector 
quantity, is a vector pointing to the (right/left) . 



ANSWER: A vector quantity with no sign in front of it points 
to the right . 



22 



Now consider ar. object which moves acccrdir.g re z'r.e data 
recorded ir. z'r.e r.exz la^'zle . ^Fill ir. z'r.e blanks.) 



As/At A (As/At) 

is At or V or Av 



t{Ls/Lt) 



(a) 1 ~:r.— {d) 



(g) 



(b) 1 -1--— (e) 



S3 70ft 



(c) 1 --.T.— {f) 



(h) (i) 



s , 






(a) -15 fz 



(b) -20ft 



(c) -22f- 



(d) -15 

(e) -20 

(f) -22 



min 



ft 
mm 



(g) -2ffz- (i) -2^Tzri 



•Ft -Ft 

th) -2^ (J) -2^: 
nun nun' 



What car. vcu sav abcur the motion cf this cb-ect? 



:-:_:;. : Ir. addition to z'r.e ir.f crrr.aricr. asked fcr previously, 
vrhar car. -.-cu sav about: z'r.e r.arure cf rhe acceleraricr.? 



23 



ANSWER: The speed changes; from what can be observed in 
the table, the change is uniform, that is, the acceleration 
is constant. Furthermore, the acceleration is such that 
the speed increases. The minus signs in the As column in- 
dicate that the object moves in a negative direction; As 
is a vector , and so has both magnitude and direction. Since 
velocity is also a vector quantity, it too has a direction 
(which depends on As) , and so has a negative sign in this 
case. The change in velocity (Av) , however, depends, in 
this case, not only on the direction of v, but also on 
whether the difference between the average velocities is 
an increase or a decrease. Acceleration depends directly 
on Av, and so has the same sign. 



Now consider the motion of an object as described in the 
table below: 



As 



At 



As/At 
or V 



A (As/At) 
or Av 



A (As/At) 



or a 



s. Oft 



(a) 



5 sec (d) 



s^ 35ft^ 



'(g) 



(i) 



(b) 



5 sec (e) "* 



s^ 60fti 



Uh) 



(i) 



y (c) 



•5 sec (f) 



s, 75ft 
4 



24 



ANSWER: 



(a) 35ft 

(b) 25ft 

(c) 15ft 



(d) 7 

(e) 5 

(f) 3 



ft 



(i,j) 



sec 

ft 


(g) 


_2ft 
sec 


2 ft 
5 sec^ 


sec 
ft 


(h) 


sec 


°'' .ft 
-•^sec2 



sec 



Describe the motion of this object: 



HINT: Is it speeding up or slowing down? 



ANSWER: The speed is changing, and the change seems to be 
uniform — that is, the acceleration is constant. In this 
case, however, the object is slowing down, so you might 
want to call this motion "deceleration." If an accelerating 
force acts in the same direction as the object's motion (re- 
gardless of whether that motion has a positive or a negative 
direction), then we say the object is accelerating; if the 
accelerating force acts in the opposite direction to that of 
the object's motion, then we usually say the motion is de- 
celerating. To the physicist, however, both speeding up and 
slowing down are cases of acceleration, a vector quantity 
(with direction as well as magnitude) . 



25 



Now consider this final problem: 



As/At A(As/At) 



A (As/At) 



s^ 150ft 



_ _ ^^ 

As At or V or Av or a 



(a) 5 min (d) 



52 120ft (g) (i) 

(b) 5 min (e) 

53 95ft (h) (1) 



(c) 5 min (f) 



s. 75ft 
4 



(a) -30ft (d) -6^ ^^'^^ 

mm 



(b) -25ft (e) -5^^ 



, . , f t 1 ft 

(g) 1— : — T —■ — 2 
^ min 5 min^ 



mm or 

(h) 1^^ .2-^ 



(c) -20ft (f) -4^ 



min mm' 



mm 



In this problem, the direction of v is (positive/negative) 
[choose one] . This means that the object is moving hori- 
zontally from (right to left/left to right) . 

The direction of the acceleration vector is (positive/ 
negative); in other words, the accelerating force is acting 
in a direction that is (the same as/ opposite to) that of 
the body's velocity. Therefore, the body is speeding up/ 
slowing down) . 



26 



The direction of v is negative . In other words, the object 
is moving horizontally from right to left . 

The direction of the acceleration vector is positive ; this 
means that the accelerating force is acting in a direction 
that is opposite to that of the body's velocity. Therefore, 
the body is slowing down. 



PART II 



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44 



ANSWER: (3) At 



Symbolically, if v„ is the velocity at the beginning of 
the time interval At, and a is the constant acceleration 
during At, the final velocity, v^, at the end of the in- 
terval is: 

^f = ^0 ■" • 



v^= v_ + aAt 



The area under a v against t curve represents 



The area under an a against t curve represents 



The area under a v against t curve represents distance cov - 
ered . 

The area under an a against t curve represents change in 
average speed . 



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■ M ' 


1 , ' . 






1 ' 


'III 






1 1 1 


1 


1 i 


1 : 1 




1 




1 i 1 


' ' h 


' 1 


' i ' ' 


III' 




i i 1 




\ 


1 t 1 


1 


1 1 


; ; 1 






1 . i 1 


1 ' i 




■ ! ■ ! 




1 1 1 




1 1 ' 1 




' 


1 1 




1 1 


J i 


1 1 ' 







' ■ ■ r-r- "^ - !■■'■'; • i ' 1 ■ TT" "7 


• tr " 1 i i I . . "■' ' ■ ; 1 1 1 1 1 ! 1 j 46 




1 i 1 1 i 1 i , , 1 : ! ! ! , . . . , 




"^ 1 1 ■ ' , 1 1 . 1 


1 ■ ' Mill ■ : ; ' ' i ' ' ' ' • 


i ; 1 1 1 ! 1 ! . ; ! , i ; i i ■ i l i 


- J ' ' 1 1 ! ' ' •■*:*■• * ; I ; j j ! • • ■ 


1 ._^__ ^L I i ! ! , ! ' i 1 1 , , . i ' ; ' , , , 1 




I ' ' ■ 1 • 1 ' ' J ' ' 1 


1 ' ' ' - ' ' ' 1 1 ■ ■ 


Orid point is qj v©n en tth<i grai h ba] owj . i ! 




M i 1 i i 1 1 ' 1 i 1 1 ' i , 1 . . ; . , , , 1 


• J 1 t 1 ] _ 1 1 1 1 ! 1 ' ' ' ' .....,,,..». 


1.1 


! Plot two J inds one dott* d ^ tJe other ccntiniious . . 




L -I k « >« « 1 1 « 


(1^ i.et the dotted ].ine £ how "(rpr^stMit speed. i ' 


' 


( 1 i\ 1 .IiOt thf cor feiniifiUfi li t\(^ sViOW a AaciTennirxc . . ! . 




"^ 1 i -< . rt » . ■ ' 


"accele raticiH of 2 ft/secl. ; i;; '■■' 


1 1 1 II ' j : i ; ' 








1 1 ' ■ ' 


it til', 


1 1 1 i ' ' ■ ' 


...,,,, 1, 1 ill;,,' 


, 1 ; ' 


1 M \ • \ '••••'. \ 


I 11 • , 


' i ' 






•11 ' . h i 1 




;. 1 1 1 1 1 1 1 i ! ' ' i 11 1 


i ti • i 1 > ' ' '■':''■! 1 M 


^- ' ^^ 1 1 i i ■ 1 1 i ' , 


'V' 1 1 ! ' i ! ' i 


1 ^. i i 1 ; 1 , 1 ; 1 , 1 ! 1 ! i 


1 ' ffl- ) 1 ! 1 ■ : I 1 ■ ' 1 ' ■ M i 






■ , 1 ! 1 ■ . i i ' i 1 1 


1 1 1 ; 1 1 i 


' ' ' ' • ' ' M 




111 "^ ■ 1 1 ' 1 "T ' j 


' ' ' t 1 , ' , 


' ' ' , ' ' ' • , . , , ,. . , i i 


' 1 M 


..4_ , ■,.,._ , , . i i i 


I , ; 1 1 ' i 


' : ;'. ^ ' 






It (sec) 1 




"j 


1 


1 r 1 I 1 ' 


1 T 1,11 ' 










_^_ T 


[ ' ^ 


l_ J 


1 


] 


1 










' 1 I 


"■'"'■"■ . .„ ^. , , , , . . _. . ( 






' 








... „ . _, . 1 . ii i. i 


-+- ^ - - - ; .14 _L 


. . • . ,,.,. 1 , , , ; 1 t ' ( 




t 1 ' ' ^ I 


t 1 


v^ 4 _ T '. .... : ; . : . . : . ' ' ' . . . . ! .^ - 


I i '-!':. 





"T 


I 1 1 1 


■" " 


1 1 ■ 






[ 




_ 




■T'|-T-i yt 






■ r 




I'll 


















1 ' ' M 




■ 47 


1 i 


Ml! 




" 1 ! 










1 




Mil • 


1 




1 1 


' ' M 




! 1 


; i 




1 j 


' 1 


1 


' 


1 1 


, 




! 1 


: : ! 1 




1 


1 






i 1 




MM II 






' 


1 


1 i 1 1 






1 








' 1 


1 M 1 


MM 


j 




. 




1-1 






1 




i 


















III' 






1 




1 ' ' 


I'll 




[Ml 


1 






i ; ■ 


' 1 ! 1 






1 


i 1 


' ! ' 




1 1 M 




1 




. , ,. ,. 1=' 1 


. i 




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1 ' 


1 1 1 ' 


M ! I 










-'I 






■ i i i 








I'M 








1 


1 i 






1 


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j 




1 i 


i t ' 


1 




i 




















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1 


1 I 






' 




( ' 


1 , 










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1 ! I 








\ ' 1 1 ' ' 


a' 


■ ' 


















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i i 






1 ' 1 ■ ; 




1 
















1 


M 






1 . ! 


: ; f 1 i 1 


1 1 


1 














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■ M ' 1 


1 1 


1 


1 i 


1 


I 










' 










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1 1 


•' i 1 




1 


1 




,—p t 


"VMcn^a 


sTni r""D I 


^tt'tS 




' ! 1 


II ! 1 . 




1 1 ' 


0' 








/ 


^ 


















1 AjSl 1"^ 








C 


I 




1 


j 






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I 


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( 


1 1 


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i "A,'t! ill 










V 








i 


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■ 1 


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1 1 


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ill. ' ' '' \ 






1 ^ ^ 2 










; 






! I 








1 i • I ■ 1 




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1 






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1 










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' 


: ' i ' ; U 1 




'\ 






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0U2 MM 




1 1 ' ! , ' M i 


III 


' ' \ 


nBGAT 


IVE ^l 


^CElir ] 


RATIO 


N OF Z Lt/S 


eic 1 ' 




. , . ; , i ' , ,V 




\ 


















1 1 




V (1 


le sic 


;>r>e O 


z—thje 


line is -Z 


. ) 




' , • i ■ 


; : 1 . 




\ 












1 


i i 1 , • 


' ! , 




\ ' 


' ' 1 i 








, ' 






, ' 




\ 


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, , 






V 










1 ^ 1 , 




i ' • , C 




1 ' 












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1 1 






1 


2 


^ 


i 


5' , 




1 ' ' j i M 


1 1 1 


' 1 ■ ■ ■ 














1 i . . 


' ' 


, M i 










(sec) 




1 1 


j 1 












t 




1 ! 








; ! ■ 1 1 ■ . 














1 1 ' , ! 




1 


t \ . \ , 


1 






1 1 






1 i ■ 1 1 ' 


' M > M 


' 1 1 




■ 1 ' ' 






1 1 








ill 1 1 ' 






., . 














A , , 




1 ■ ■Imagine an 


OD^ec 


t "wni 


en us 


at r« 


2st a 


L tQ 


— V ariQ s^ 


— U . 




' ; 




















•^nnnoFe vol 


knov; 


■h^HA+" 


its 


noQ "i -f- 


Lon a 




= ,lsed is s 


-, ~ Icrn. 
























i ' ■ ■ ' ' 




















■ 




































, 1 . . ; ' ■ 






' 1 














III ' 




' 1 ' ' ■ 1 i ' i ' : ■ 


; ' i 


















■ 1 ■ 1 ' 1 1 : 1 ' ' I 


'1 


! 1 
















1 'i 


i ' 


1 1 


1 


















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' 1 1 














' , ' 




1 i 


' M ■ 


i 1 


I 










1 1 . 




1 1 : 




1 I 


1 










1 ' ! , ; ' 


^ 






































M •■■' , ' 




















ill, . , ' 


\ 






1 1 1 


i 1 




1 




1 


Jill ■ '^ ' 










! 1 1 


, 


, 


■ ■ 


1 










1 








' 1 ■ 


' 


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' /" 










' 


1 ' voni) X- 








c 


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1 ' 1 ' 




1 ' 












, 


; i 1 


' 1 1 


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M i ' 


1 1 


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1 ' ' ' 








1 ' ' 






Ml 


1 1 


1 1 1 
















I 1 1 1 






" 1 ' ' •■ ' 








' 1 1 






i \ ■ 1 1 


1 , 1 




1 . 






1 


' 1 i i 




' 1 


1 


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1 ■ 






' 






1 


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1 1 , . , 


1 1 i 


' 














Ml! 


j 


















Ill 1 






r 














1 ' M 




■ , 










1 


1 




j ; 1 III 2 




1 


















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/■ nj-. j- 


\. 


' ■ j 


■ 


M , 1 


- . — I 1 ■ ' i ■ 








t 


vsec 


) 




' 1 ' 




1 < 


















1 1 1 










■ c 


hl^jt 


h&::al: 


aye:;i 


Sfa 


1 


1 ' i 


1 1 


it an ycu, or 


the 


haaL£ 


„al 


imation. 


gi\e___^ 


1 1 
~^~*~ 1" 


' ' ' ^ ' 'i_ 




■ .■ ■ ! 


— . — 















M ; :e ny oa — the- 


tqi,liC 


wfingT 


















' 


















! M ' ' 1 ' faV '3 


vetac 


e Idne 


ed 


• ill 


, , 


; 
















1 ' i ! 


1 1 t 1 


1 1 ! 1 


MM 






■ ' ' ' r\n^ '. 


• 


j^ . 
















r T»r "« 




rotic 


n 








1 1 1 




1 i 






I 1 ■ 










'111 


1 1 


' 1 


' : ' ■ {n) P 


pf?ed 


at" t" 


= ic 


ec 










1 






■ ' "i 1 
















1 ' , /■ J \ x. 


, 


y, . 


J_ 


-^ ,C] 




1^ 1 








-" ; : : ■ (a); t 


he gi 


apn it 


EQItl t 


^ 


t!C t 


1 ^ 






; 




Mi! ' ' 1 














M 


1 




I .--i— 


1 i i r-- t • • • ' 1 
















1 






i 1 1 ' 1 ' r ■■ 


1 


! 1 j 




1 






1 






1 


1 ! i i 1 i 1 ; 1 


i , 






! 1 








._L 


M 1 1 1 M 1 ■ 





1 -TT-" — 1 : , ■ , ■ , y -r ■r-rn , . 


" ' , ' ' ; i i ' ' 1 48 


i I ' i ' ' ' i 1 ■ ' . 


1 i , . ) 11! ;,.;.. : ! 1 1 


.- , 1 , ' . ! II : 1 , ' ■ , 1 , . , 1 t 


1 j.i:> J . / . . Ifm , cm i ; i , 


^1 J^verac 6 speed = As/At. = , = 1 i M 


, ~T t • "^ Is eo sec ; ; i ; 


[,,;;, , 1 , f 1 , , 


; ";:;■■' • ' ^ ■ i i I : i i ■ : . ■ ' ........... 


b) C'ne Cc.nnot detejiiiihe the ciccel<;raticn*_j_:__4- 




'1 


1 ' 1 ; 1 1 1 


' , 1 ' ■ 1 


' c) nor 11. s speed ai, i; ;=* l^e<i, , | , , , ! | ' i ! i , ] \ ' • 


I " 1 ■ ' 1 ' ' ' ' i i ' ' 1 1 1 ' I ' 


1 1 i Ml 1 1 ! 1 ' 1 ' ' ' 11 


I , ' ' ' ; i t ' ' ' 


i .. ^ .. , , , , 1 I ^ M ' ; 


'4 ' ' 1 ' 


i ; 1 : , 1 1 ' I 


1 1 i i i ; 1 ! 1 1 , ' ! ! ' i : 1 ■ 


1 '1 ! ■ I ' 1 . 1 ' • ■ 1 ' ' 1 


17 1 . . . ■ ' ' • ' i 1 


4: -Tip -T-!- " .. xin i X 


I ! ' 1 _|_ ' 1- i 






• . ' 


IT 1 1 j 




I'fTit ' ' ' jC^"' 


{CU) ■ J ^"FH* ' 1 ' ■ 1 1 ■ ■ j : ■ 


■* ■( ft i . 1 ■ '^ "3^ '' . j 1 ; ' , . ■ . 




1 1 L ^r ', 1 J^ "L/M i ; 1 1 i .i . ! . . 


• ■ ■ I t>ri 1 JrT Jr 1 1 1 1 1 ' : 


' ' ' ' ' ' Tb Z"^' J> ' 1 ' 1 ! ' ' ' : 


1 ' ^ ^<^ 1 1 1 , . , , 


I 1 ■ i ,1^'^'^ ^y'^ 1 1 I ' ' ' '1 


' ' A '''^ _,y ' ' 1 ' ' 1 


j j i .^'"^^/^ """"^^^ - ) M ' i ■ . ; i 1 j 




i - Mi' 


1 1 ' ! 1 ■ 


.,_..,. . , . I , ., i \ \ ■\ , tta.f»\ , . i ■ 1 , . 11 . 


1 f ' 1. sec) 


■"'■'■ i j 1 1 1 ■ ■ ■ 




\ '. \ 111 




- a; '• . n© ci apn '(louia De ar y or a numoer ct iir es; two 


1 1 ' 1 


~p "^ 'c f :th<: orie£ ahovn are possible, one not. 


1 ..' i»l»l.*T t^ L i Ul_ V- 


I'xscuiS tn< pnyj ical meaning oJ tnes e qrcpns. (Wn: en 


"*" 


_i_ i lie .i£ .impc ssib] e?l . 


-|- -|-i- ^ -^ * 




1 




M Ml 


_^ ^ MM' [iX 




t 1 


' J 1 , j 




1 


-1 -H-i -? 4 - ... 1 


-f H J -J - 




11'' 


1 1 


1 t 




., ,^ 


-\ ■ 1 , . , , , . 1 . 




1 


^ , . T 




: ~ ,-_ " - -. -f a . . . ^ I":.: 


- - ! . , , . , 






' "*" 






• • ni ni. 2 ' Keraenixmr «>' sxop* =» v< iocxt y.' 1 


"T ' I ■ • • ■ • ■ • ■ ■ "^ .... , P . 


_ij ;'!'■■' '■ '!::::::::::: 




1 1 T 1 


: ... 1 ; 1 t , . , 



~ ^P fill 1 1 ' i T" 


4: X XT : 49 


. . 1 . . 


II 1 








1 


1 


, ,.. 1 , ,2 ..,.. ,. . 




1 






" 




/ 


£ y ^ " 


■ i / 1^^'^ 


L, ('f^nil) 1 1 ' 1 ' ; ii^-' . 1, 




1 1 . 1 i ^ i .1 ' . • ^ "^ i.'rt/i 1 


! : 'i4-)H ! : ^ yf\\ 




\ ; i 1 f \y.\ '^ ^ i ' . 1 


■ ] 1 > .■ ^"> ^^ i 


i 1 / .jt^ i J^ '^\' 1 1 


''^M q) 


\ 1 I ^^?--r . 1 1 


."S"^ 1 i 1 i . 


1 ■ ■ j ■ I If T^ — "T""^ ■"" 


' ; . 1 • J 1 1 . : 1 ; ! \ i 


t ■ '■'■'. \ '- \ \ 


l''' ';; 4/rn>-.j-i\ ' 


■ ■ ' ■ ' , ■ ; : : i t. (sec) i 


1 ' ' : 1 ! ! ' I " 




I 


1 


' : 1 


1 1 i 1 1 1 


-iJ-u; ' a) ""'he ol'jf*ct '■•t.iiij.- s;i>f!< c \'(^r\ j-ipic^.y, tl-Ler. ■alovfs; icLcivi. .. 




'11. !, -■^.■'■•l J 1' 1' ii ' f \''' 


' ! : 1 xt 'gees heir oj: tne way :,n aDCiutt ;^d sec)!. 


1 ' ^ ■ , i : 1 : ■ ' t 






. ' ' 1 ' 1 ' : . 1 1 ' 1 j 1 1 1 1 1 1 1 


1 • ■ , 1 1 , > 1 1 1 1 I.I 1 _^ 


1 ' ' ■ i ' ^ ■ 


i4j_^_j J::_4ai ^e-jjta^ aie-i-J Jai^-- : .n^lx£ ^a J^ 






■ ■ ' a cons tant vexotix'-yi wnxci xs jinpossxDxe xr xi.s ye.ocxt-: ci>f [ 


',■.;'! i ! 1 ' : M 1 


■ ■ ' ' f =0 • S . i ' ' I ■ 1 ' ' ! ! ! 


.. i-f^-^v . ..a . w . 1 , 1 j 1 ^ , . . - ^ j 1 




1 1 • ■ ' 1 ■ : • ! i M ! 1 ■ ■ 1 


1 ' 1 ' 1 1 i ' ' ■ ■ ' J ' 


i 1 i r^ i''h-i« <'iirvf> shr^wf; "thf^ nhiect -Snf^f^diniT up CTclCluall^'J l^: 




! j 1 : , , , T , . ■ ' ' ' 1 ' 1 1 1 1 


' 1 ": :$ • cl( 'se tc» a cctnstant acceiexatxon, ' 1 1 ■ ' 


' ■ , ' 1 i ■ i 1 ' 1 , 1 1 1 i ' ; , 


_-^— ' , • , , , ,: \ L- -4- ^ ' ' ' I 


^ 1 i 1 1 Mil 1 _L 1 X -^-i- -^ -M 


' 1 ' ' ' ' 1 _L ' 1 1 1 1 


-^~ — ^- ; ■ ^ 1 ' 1 : 1 i , : .._!_ 




, , , . j . , .,,.,,,,. j ^ , 1 1 ^1 


, 1 i X^ i ■ ■ ^^'^ ^" ' -- X-^ 


_l_ : ■ ' ' ■ ' 1 ' 1 ' _, ' 1 : 1 1 , 1 1 1 


i ' ' i ■ 1 i ■ ' 1 ' ' ■ 1 


" 1 " : 1 : ' Mil Ml n ^ 1 _i_"' ■ -^ 1 , , 1 1 1 


X X^— '''Mil' 


i j 1 . , , . t ! M 1 1 1 


....,,,. ; . . 1 , 1 , 


+ M" j '"^^ " '""'"' ^~ -■ "T X " 




1 ' 1 1 1 1 ' 


j M ' M ' 1 ' ' ^ 1 ,_ 


1 ' i 1 1 ' 1 1 


1 1 ; , ' ■ ' ' 


-^^ ^_ — ^ ' "" ■ ' 1 ■ -■- X 


i 1 i x^" X M X 


M i ! ~^'^ ' ! IL 




X j ! 




; : 1 M ' : i ' ' ' 1 X 




: M 1 tX~ - ' "^^ Ml 


' ' • ' ' ' , ._..! ^4_ ^_ _ 




X xn i M i 1 X X XX 


_i_ _|— I — 1_ -i — 1_(_ 


1 1 1 111 " ■ . . t- 


X , . i , , ' 1 ^ ^ - . 



PART III 



-t-M- 



T^ 



±^^ 



-n- 



ISIR^ 



I I I i 



I'll 
I i I I 



n^ 



rti gy^a: 



«HW^ 



-4—^ 



-^4-1 



i ' I 



:?: 



i^ere 



tcr^ave'^a g'l 



cons fean t 



iximz3=x 



What 



ESai 



X 



-T-t- 



m 



ffl 



n 



riT 



at tfe 



1. therr^t 



i i s ' A 9/ A i t ' 



J_L 



the-^es-i^fej;er^---shcwR-^-— C^n- yD^ir-^3^gaJrafe|eHi^ 



4-H 



-i_L 



u. 






TT 



■H-l- 



EUtSesl 



ob - j e c t p-tii tial 4 y- ar^- 



jdig_Q£X£ 



-*-r 



I I 



for t hie— f i rs-fc ^ eeortdi ? 



S£i2QD^ 



i_L 



-lJ_ 



4-U4 



ryr^snchr^that: 



locR: 



wouli 



i : I i 



■tT 



4-^ 



50 



rest. 



+^T- 



U-U- 



\ I I 



-i-U- 



in 



^TTt 



TT~ 



-r-t- 



-Hr 



51 



First second: -r-r = 1 
At 



Second second: —r = 3 
At 



cm 
sec 

cm 
sec 



./AS. _ ^cm A (As/At) 
^^At' " ^sec' ^° At 



^cm 

a = 2 2 

sec^ 



It seems easy enough, then, to calculate the acceleration 
when we have s and t data for two or more intervals. But 
what happens if we have information about only one interval? 















'1:1 


1 ' 








1 


1 


~" 






I 1 1 < 










[ 








M , 


. i \ \ 


- 52 






Ml. 




' 














1 ' 


M 








; i ; ' 












\ 






' 




1 1 1 


















1 ' 










' 1 
































fi 




.. -i.U. - 


^ _^ ^ 




K-4 A/nt 


.t_ • 


4%— •-*— 


Jv»-*-<w4 












o uppc 




>u OO: 


ervec 


L an c 


DjeCT. 


wmc 


Tr^SXc 


Tuea 


at re 


St ai 


a 






























wa<^ ; 


rr.Al* 


TTAi-fiC 


Hy" ; 


qnr\t 


•H;:^n+- 


fnirna 


^TOT\ 


+• 


n ^r 


t = 


Ir 






















Jjl. 




1 








a . ml 




n -m-r 


y^ rt 






















Sq aJ 


so ;~ 


u > ai 


tor S. 


= zcr 


I. i ; 










1 , 














' X 




1 1 


1 1 ' 1 






' 






; ' 














, 1 , 


1 




























1 1 ; 


1 




1 






















1 '' 




1 i ' 




























1 




' 1 


' 




























1 1 




1 
























1 1 


. : 1 1 




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; 1 


1 1 


1 


1 ;T 


1 i M 


; 






1 I 


1 


i 1 












1 


1 


1 1 


1 1 






1 








1 


1 






1 














' 


1 


1 1 


: 1 


i I 


' 1 


1 


1 














' 


1 i 




1 1 










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' ; 


i 




1 




1 


i 






















, ' ' - 


ill' 






' 1 




1 






















' i^r" 


1 ' i 


; . 


■ 1 ■ 


• ' 1 ' 








1 










1 1 ' 


1 i 1 1 


i i : 1 




' i : ; 




1 1 


j 




MM 


' 1 i 


1 i 


1 


i i t 




' ' i 


1 




1 1 






' 1 1 i 












' 1 i 


■ 




.111 




TO 


; 1 1 


i M 




1 


1 






1 




■ ' 






. ' . 




















' 


1 




1 \ 1 












, 










, 


1 1 ' 


. ( 




M ' 1 










1 ■ 




1 . ■ 


1 


, 








MM 


'■ '. ' t 










' i 


: i 




• i ■ 


; 1 ■ 








■ ! I 


Mil 




q 












\ ■ ' ' 


1 ■ : • 




' 


. , : 1 




M ' , 














1 1 


1 1 


! 1 1 




• 


. 1 




MM 














; ! ■ 


■ 1 ' 


' 1 ( 




1 M 1 


; i i : 


' 1 ' 










I 


1 ; 




• j 


' ' 


1 1 




1 


Ml 


M 1 i 


1 ' ' 


MM 














i 1 


; 1 




1 I 




' 1 


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53 



You might be able to puzzle out (or guess) the acceleration, 
or you might very well not; in either case, you would prob- 
ably find it a long and complicated process, involving a 
great deal of untidy algebraic juggling. We'd like to help 
you over some of the rough spots, if you'll bear with us 
for a little. What we are after is an equation, or set of 
equations, for determining (1) constant acceleration over 
a given interval, and (2) an object's position at the end 
of successive intervals. Somewhere along the way we should 
also be happy to pick up some information about the velocity 
at various odd moments. 

We have already developed one equation (containing a as one 
of its terms) for the final velocity at the end of an in- 
terval . 



^f = 



ANSWER: (1) ^f = ^ + aAt 



Now, given v_ and v_, what would you do if you wanted to 
discover the average velocity over the interval? 



ANSWER: Divide the sum of the initial (v ) and final (v^; 
velocities by 2. (Elementary, my dear Watson.) 

- ^0 "*■ ""f 
(2) V = -^ ^. 



54 



Now substitute the right side of equation (1) for v^ in 
equation (2). (You may not be able to see, yet, why we are 
doing all this, but keep in mind that we are trying to de- 
velop a formula which will give us the value of a when we 
know only one value for s and t; in other words, we should 
like to develop an equation containing only the terms s, t, 
and a.) Now, what do you get when you substitute the value 
of v^ from (1) in (2)? 



_ v„ + (v_ + aAt) 2v„ + aAt , 
(3) V = -2 2 L_ Vq + ^aAt, 



ANSWER: 



We're getting warm! Let's try to get our friend As into 
the ring: 



As — , 1 A4. 

_ = V = Vq + 2^At, 



So: AS = 



ANSWER: 



(4) As = vAt = v^At + ^a(At)2 



Now can you figure out the value of a? 



2cm' 



Isec 



55 



HINT: You can clean up the general formula a bit by getting 
rid of one of the terms: 

(4) AS = V^At + |a(At)2; 

1 9 

If t„ = 0, then At = t, so As = v_t + ^-at^ . 



But even more important, what about v^? 



ANSWER: When the object starts from rest, v_ = , so 
As = jat^. (The term with v„ drops out, giving us an equa- 
tion with only the terms s, t, and a.) 



Now you should be able to find a with little difficulty: 



To find a: 7^^^ ^ ^^ 

at^ = 2AS 



2 AS 2.2cm .cm 

a = . V = r: y = 4 

t"^ Isec^ sec 



Now what about the position of the object at t» = 2 , if 
the acceleration remains constant? 



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60 



HINT: Use the distance equation previously developed. 



HINT: The equation is As = v^t + ^at' 

1 9 

But v„ = , so As = yat'^, 
from which you can compute a. 



For (a) : When t = 1 sec; s = 1 cm; then, a = ? 



As = 2"^t^; 2 As = at^; a = — p^ 



2 • 1cm ^cm 

a = T r = 2 2 

Isec^ sec^ 



Then, to find the position of the object at t = 2sec, if 

the acceleration (a = 2 o) remains constant: 

sec^ 

As = Ts-at"^ = 77*2 p'4sec'^ = 4cm. 

2 2 sec^ 



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aiNIT: If /ou kiow ttie vaLLie of a., ioes this infortnatio 






IV - • 














• tell /ou anything abDut tie spsed ab ^=2? ' t -.-. ^ -^-.- . 






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the object accelerates to a spaed of 














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64 



Remember, v- = v„ + aAt 



So for t = 0^1, v^ = 0, a = 2^5772 



' sec' 



, J -,cm , „cm 

And v_ = 2 o'lsec = 2 

f sec^ sec 



(We assume for the moment that the final velocity of the 
interval t = 0->-l is the instantaneous velocity at t = 1.] 

Now, what about the average velocity over the following 
interval? (Remember, a = 0.) 



^2 = ^f^ -^ '^2 = ^f^-' ^^4 

But a = (There is no change in speed.) 

_ — ^cm 

So : v„ = V - = 2 

2 f , sec 

In other words, if there is no acceleration, the object con- 
tinues through the second interval at the velocity which it 
had attained by the end of the first interval. 



Now, if you have not already done so, calculate the position 
of (a) at t = 2 and sketch the graph on the preceeding page. 
(What will the section of the curve between t = 1 and t = 2 
look like?) 



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69 



By "instantaneous velocity at a moment" we mean the average 
velocity during an extremely small time interval containing 
that moment. Measure the graph to find the necessary data 
and fill in the table below. 



^■^?J!i^??.^^'^ ^^cm "^^^ INTERVAL At^^^ As/At = v 

(cm/sec) 



INTERVAL ^^ sec 

0^1 1 -. 1 



0.5 ^ 1 0.5 



0.36 ^ 1 0.64 



0.81 -*■ 1 0.19 0.9 H- 1 0.1 1.9 



Since there is always some variability in measurement, we 
shall not supply the "correct" answers. The column for v, 
however, should show gradually increasing values. The 
limit value, as the time intervals become smaller and 
smaller, appears to be 



The limit value, as the time intervals become smaller and 
smaller, appears to be 2cm/sec. 



What is V for the interval from 1 sec to 1.05 sec? 



70 



At = O.OSsec, and As 0.13cm. Hence 
- 0.13cm 



V = 



OSsec 



= 2.6 



cm 
sec 



PART IV 



71 



10 



11 



12 



This is a stroboscopic photo- 
graph of a falling sphere. 
[Source: PSSC Physics , D.C. 
Heath & Co. , 1960. ] 



The time interval between 
exposures is 1/30 sec. The 
scale on the left is cali- 
brated in centimeters. 



There are a number of questions 
we might want to ask about the 
picture. To begin with, you 
might ask yourself if you can 
tell, simply by looking at the 
picture, whether the velocity 
is constant or not. 



13 



REMOVE THIS PAGE AND REFER TO 
IT FOR THE NEXT FEW PAGES. 



72 



The velocity is increasing, but what about the acceleration? 
How would you go about determining whether it is constant or 
changing? 



I 



73 



You probably decided that you would have to make some 
measurements and some calculations to determine whether 
the acceleration was constant or not. Here are some 
data to help you get started: 

INTERVAL AS V Av a 



P, ^ 6.3cm 189-^"^ 



l->2 * sec 



How much more information will you need to determine 
whether a is constant? Which intervals would you choose 
to measure? 



(When taking your measurements, you will find it easiest to 
measure between the bottoms of each ball. And you might 
find it helpful to use a caliper to transfer your readings 
to the scale.) 



74 



Your calculations might look something like this: 



INTERVAL As V Av 



P, . 6.3cra 189^"^ 



l->2 sec 

P, _ 11.7cm 351^^ 
6-»-7 sec 



P^ „ 12.9cm 387^"^ 



162^H^ 9.72-^2 
sec sec' 



7->'8 sec 

P,„ ,^ 18.4cm 552^^ 
12->-13 sec 



165^^ 9.90^^2 
sec seC^ 



The acceleration would seem to be nearly constant. (If 
your values for a were within 1 or 2 tenths of a meter/sec^ 
of ours, your measurements from the scale were quite ac- 
curate.) As a check you might want to calculate a for Av 
between ?-,_>.') and P^ , : 



INTERVAL AS V Av 



P, ^ 6.3cm 189^^ 

l->2 sec 

P,^ , -, 18.4cm 552^^^ 

12-^13 sec 



363EE_ 9.9-^2 
sec sec"^ 



75 



The last value for a seems in fairly close agreement with 
the others, but the fact that the value of a varies does 
suggest that there is some error involved in the problem. 
You probably observed how difficult it is to measure As 
on the small scale in the picture, and you may have con- 
siderable doubt about whether a is really constant or not. 

In fact, if you chose to calculate Av over shorter inter- 
vals (between P, ^ and P„ ^, P„ -, and P^ ., etc.) you might 
l->2 2-»-3 2^3 3-»-4 ^ ^ 

have found that your values for a varied so much that the 
acceleration did not seem constant at all. We chose the 
largest intervals possible in an effort to increase the 
accuracy of our algebraic calculation. But there is an 
even more accurate way of finding a, which depends on the 
properties of a graph. What happens if you plot v against 
t? (A graph with suitably chosen units is provided on the 
next page; plot the values of v and try to draw the curve.) 




910 IjlOllf 1112^1213 

1X3 II i: 



2 

30 



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78 



In straight-line motion, changes in motion have been either 
a speeding up or a slowing down, that is, the velocity vec- 
tor has changed its , but not its . 



ANSWER: The velocity vector has changed its magnitude , but 
not its direction. 



Below is a schematic representation of the position of an 
object. A reference frame is provided for distance and di- 
rection measurements. Notice that this is not a graph. 



y 


4 


4 






12 


O 









ra 










4 


• 




0*3 












V 




4 


10 
m 


16 





t, = 12sec 
t„ = 18sec 
t^ = 23sec 



We can easily draw a vector diagram of displacements on this 
sketch. Do so. 



79 



Now construct a vector diagram for the average velocities , 
using the frame of reference and scale provided below. 



V 



m 
sec 



80 



HINT 



Our displacement diagram could have looked like this 




The location of a vector is irrelevant; only 
the direction and magnitude are important. 



ANSWER: 



1 . 



m 



sec 




As. - I = 6m 



12 _ 6^ _ 1 m 
At 6 sec 



ASp^l = 10m 



As 



23 



At 



5 sec 



The directions of these vectors are parallel to the di- 
rections of the displacement vectors. 



What is Av? 



81 





\ 




V 

y 


\ 


V23 




^12 










m 
sec 




^x 



ANSWER : 



m 
sec 




-V 



12 



82 



Suppose this change had taken place in a time interval of 
^second. What was the magnitude of the acceleration? What 
was its direction. (Fill in the scale below.) 



m 



sec 



ANSWER: 



AV ==1.7- 



m 



sec 



Therefore , 



= 3.4 



m 



sec 



The di- 



rection is the same as that of Av from the scale above 





t 1 






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a 

y 


I 










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sec^ 


a 

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and for c.ll times : 
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84 



HINT #1: If v^ is constant, how far does the object travel 
horizontally in 1 second? in 2 seconds? 



ANSWERS: s = v t 

X X 



^i^*^^^^ = 2cm 



2i^*2sec = 4cm 



HINT #2: What is the formula for displacement under con- 
stant acceleration when the initial velocity is 0? 



ANSWER: s = jat^ 



HINT #3: What is the vertical displacement after 1 second? 
2 seconds? 



ANSWERS: s = la t^ 
y 2 y 



1 c-cm . , 5 
2-^ii^2-^^^^'= 2^ 

1 -cm . J 
2*^ii^2-4sec2= locm 







. : 




85 




1 i 


^ 1 1 


i 1 1 i ' 1 ' i ' ' 


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t i ' ' ! 1 III. 








h^TiMAt sprjr TTrtNl HF TTHR [ 


p-RrRT.-RM. ' ' I 




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■ ■ Plot' the resultaht- i n^ tail taheo us velocity at t, {=^ lBec> — ^ 


and t^ (= 2sec) . (lB.t._iiiaUQXJ.3lli. of both vectors be t.^5 ' T : ^ - 


^ II ,11 


_. _. iT^i-fnniV ,._.;;i. ....1 


pouit lU/UJ.^ tj 1 i 1] 


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1 M , , , : M M ; M : , , ^ 



87 



HINT #1: What is the velocity component in the x direction, 

V , for both V, and v„? 
X 12 



"*■ cin 

ANSWER: v is constant for the whole motion and is 2- 



X sec 



HINT #2: What is the velocity component in the y direction, 

V , at the end of 1 second? 2 seconds? 
Y 



ANSWERS: v = a t 

y y 



_ cm , r: cm 
5 o • Isec = 5- 

O £=i 1^ ^ < 



sec^ sec 



cm „ 1 A cm 

1 o'2sec = 10 

sec^ sec 



- , , t - - - "■ 1 i i 1 


1 . 


1 1 ' 1 




ZLX^ XJIIII ZL ~^ 


i . 1 i t 


1 ' • ' t 


~f 88 




! 1 1 1 1 _ 




1 




1 1 


1 , 


1 1 . 


1 . 1 ^ , . 




, 


' ' I 




■ i- 


t— 1 1 i 1 1 


1 1 1 i 




- -j- 

* 


1 ; 


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1 i ; 


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1 


1 I ; 




' ■ ■ 1 








1 : . 1 ; . 




. 1 










1 1 1 i 


; 1 


1 : ' 1 ' 


, 


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' 1 i 




11 -H» '^' J: -1- i h -r 




1 i 1 . 




■^21 1 1 1 1 1 1 ; ; : 




1 












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, 






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Cover photograph by Mark Kaufmann, Time/ Inc. 



This unit is the property of HARVARD PROJECT PHYSICS. 
Upon request it is to be returned to Pierce Hall, 29 Oxford 
Street, Cambridge, Massachusetts 02138.