t>N
i
The Project Physics Course
Equations 1 Simple Equations
Programmed Instruction
Equations 2 Applications of Simple Equations
Equations 3 Combining Two Relationships
INTRODUCTION
You ore about to use o programmed text.
You should try to use this booklet where there
are no distractions — a quiet classroom or a study
area at home, for instance. Do not hesitate to
seek help if you do not understand some problem.
Programmed texts require your active participa
tion and ore designed to challenge you to some
degree. Their sole purpose is to teach, not to
quiz you.
This book is designed so that you con work
through one program at a time. The first program.
Equations], runs page by page across the top of each
page. Equations 2 porallels it, running through the mid
dle part of each page,andEquations3similarly across
the bottom.
This publication is on« of fh« many instructional materials
developed for the Project Physics Course. These ma
terials include Texts, Handbooks, Teacher Resource Books,
Readers, Programmed Instruction Booklets, Film Loops,
Transparencies, 16mm films and laboratory equipment.
Development of the course has profited from the help of
many colleagues listed in the text units.
Directors of Project Physics
Gerald Holton, Department of Physics, Harvard University
F. James Rutherford, Chairman, Department of Science
Education, New York University
Fletcher G. Watson, Harvard Graduate School of Education
Copyright© 1974, Project Physic*
Copyright© 1971, Project Physic*
All Rights Reserved
ISBN: 00308964UX
012 008 98765
Project Physics is a registered trodemork
A Component of the
Protect Physics Course
Distributed by
Holt, Rmehort and Winston, Inc.
New York — Toronto
Cover An by Andrew AKIgren
Equations 1 SIMPLE EQUATIONS
In physics and many other subjects it is useful to be
able to handle simple algebraic equations. The purpose of
this first program is to help you review the basic operations
necessary to solve equations.
Equations 2 APPLICATIONS OF SIMPLE EQUATIONS
In this program you will gain practice in applying the
basic algebraic operations you learned in Equations 1 to
equations actually used in physics.
Equations 3 Combining Two Relationships
You saw in Equations 2 that relationships among physical quantities may be repre
sented by equations and with the help of operations developed in Equations 1 you gained
experience in solving these equations for some particular quantity or variable.
Often in physics we find a particular quantity in more than one relationship. For
example, the quantity F occurs in each of the two equations F = ma and W = Fd.
In this program we are going to see how two relationships having some quantity in
common may be combined into a new relationship. For example, we can combine the two
equations above and find an expression for W in terms of d, m, and a.
Often in this program more than one frame is used to solve a certain problem. We
have followed the practice of repeating the problem in a box at the top of each of the
frames concerned.
INSTRUCTIONS
1. Frames: Each frame contoins a question. Answer the question by writing in the blank space next to the frame.
Frames are numbered 1, 2, 3, . . .
2. Answer Blocks: To find an answer to a frame, turn the page. Answer blocks are numbered Al, A2, A3, . . .
This booklet is designed so that you con compare your answer with the given answer by folding
bock the page, like this:
1 
1
r^
"73^
i
—
3. Always write your answer before you look at the given answer.
4. If you get the right answers to the sample questions, you do not have to complete the program.
INSTRUCTIONS: Same as for Equations 1, above.
INSTRUCTIONS: Some OS for Equations 1, above.
Introductory Frame A
Answer Space
Adding and subtracting. Study a, b, and c; then write
down the results for d, e, and f.
a. 6b^ 4b= 106
b. 4  10 = 6
c. p + 2q  5p = 2q  5p + p = 2q  4p
d. 6k * 2k =
e. 4t  lOf =
f . X + 2y + X =
1
Answer Space
In physics we often represent physical quantities by symbols
and then represent relationships among physical quantities by
equations.
For example, suppose a car which is moving with an initial
speed v/ is given an acceleration of magnitude o and thus acquires
a final speed of v/ in time f . The equation which describes the
relationship among these four quantities is
— V; = at
(i) How many symbols representing physical quantities appear in
the above equation?
(ii) List the symbols.
Turn page to begin Equations 3.
Answer A
91^
6t
f. 2y
A1
(i) four
(ii) yf
Introductory Frame B
Multiplying and dividing. Study a, b, and c; then write
the results for d, e, and f.
Answer Space
a. (1) (+3) = ^3
b. (5) (a) = +5o = 5o
6a . 6a 6a
c. — = a (Note: — =
6 6 6
d. (1) {3b) =
e. (4) (2p) =
6a
6
First operation: \( we add the same quantity to both sides of an
equation, both sides will still be equal to each other.
Example: o  t = 4c
ab + b = 4c *b
a = 4c + b
Note that by odding Jb to both sides, we get a alone on one side of
the equation, that is, we have solved the equation for o.
Solve the following equation for vi in the same way:
Vf — V = at
If m = 2p and p = 3, what is the value of m?
Answer B
3b
8p
A2
Vr  V; = af
V/ — V; ♦ V;  of ♦ V
v/ = Of ♦ V.
A1
m « 2p
m  (2) (3)
m ■ 6
Introductory Frame C
Operations with parentheses. Study a, b, c, and d; then
write results for e, f, and g.
Answer Space
a. 3(o  2c) = 3o ► 6c
b. 3(o 2c) = 3o 6c
c. 3(o  2c) = 3a
6. {p * q) = p  q
e. 5{k  2m) =
f. (f5) =
g. 4 {7a . b) =
6c
Second operation: If we subtract the same quantity from both sides
of an equation, both sides will still be equal to each other.
Example:
fc + 4 = a
fc+44=o4
fc = o  4
Note that by subtracting 4 from both sides, we get b alone on one
side of the equation, that is, we have solved the equation for b.
Solve the following equation for v:
v + at = Vr
If V, = 4vj
V, represents?
ind
2, what is the number which
Answer C
5fc 
10m
t +
5
80
4b
A3
V ♦ of = Vf
V. •> at — at  v/ — of
V. = vt — at
A2
V, = 4V2
but V2 = 2
so V, = (4)(2)
V, = 8
Introductory Frame D 1 Answer Space
More dividing! Study a, b, and c; then write results ,
for d, e, and f. .
a.^'''^ = 2a36 d.^:2i^ 1 d.
3 2 1
3 2 1
c. = 3o + — f. = 1 ^•
2 2 _3 ,
4 1
Third operation: If we divide both sides of an equation by the same 1
quantity, both sides will still be equal to each other. •
Example: c = 4o 
4 4 1
i" \
Note that by dividing both sides by 4, we have solved for a. 
If we return to the physics relationships we have been using i
and if the initial speed v^ is equal to 0, we have this relationship 
between Vf, o, and t: 
vt = at 1
Solve for VI as in the example above. I
3 j
If V, = 4v2 and V2 = 4fc, what is the value of Vi in
terms of k? ■
Answer D
d.
f.
2o * b
7a * b
A4
Vf = at
IL 21
t ' t
t
or
''^
A3
V, r 4V2
but V, r 4tc
so V, = 16ik
1 j
The equal sign (=) in an equation means that the symbols on I
one side represent the same quantity as the symbols on the other 1
side. For example, 1
2*4 = 3.3 (3) (4) = (2) (6) j
o + 4 = i>+3 3m = 2n 1
2, 3, 4, and 6 are called numerals, a, b, m, and n are called '
variables and may be replaced by numerals. '
In the equation (4) (3) = fc + 2, what numeral will replace b so '
that both sides of the equal sign represent the same quantity? '
5 1
Fourfh operation: If we multiply both sides of an equation by the 
same quantity, both sides will still be equal to each other. 
Example: — = b '
4 1
{4)^={4)(b) 1
a = 4b j
Note by this operation we have solved for a. i
Solve the following equation for v in the same manner: i
'^1 j
4 j
In the previous frame we began with two equations v^ = 4v2 
and V2 = 4k, and we obtained a new equation v, = ]6k. 
Note that the new equation does not contain vj. We have 
eliminated the quantity vg which is common to both of these equations. 1
How was vj eliminated? By taking the value of Vj given in the 
second equation and substituting it for Vj in the first equation. 
Let us consider two other equations: F = ma and a = =. i
What quantity is common to both of these equations? i
A1
since the left side (3) (4)
is equal to 12, and the
right sid« will equal 12
if b is replaced by 10.
A5
(a) (t) . f (t)
or v/ = at
A4
H
2 r
We con write a new equation by adding the same quantity to '
both sides of a given equation. For example, '
if a = b
then + 2 = 6+2 1
and, in general, a * k = b + k 
Change the equation '
 4fc = c 1
into a new equation by adding 6b to both sides. 1
1
6 1
We can apply any or all of these operations to manipulate symbols 1
of an equation to express a given relationship in a more useful form. 1
For example: the equation F = ma describes a relationship among J
the amount of net force F applied to on object, the mass m of the object, 1
and the amount of acceleration acquired by the object. We can show 1
how this acceleration depends upon the mass and the force by solving 1
the equation F = mo for 0. 1
Divide both sides by m, ^ = ^ 1
''mm
or = ^ 1
m 1
Now solve F = mo for m in a similar way. 
5 1
In the two equations, F = ma and a  — , we may substitute 
re 1
the value of in the second equation for in the first equation, and '
by doing so get a new expression for F. '
What is this new expression for F? '
A3
X  y = c + rf
X  Y * y = c * d
but y ♦ y =
so
X = c * d
A7
7nR
T = (^)(r)
'T = IttR
A6
This is the varioble which is
common to the two original
equations.
4 j
Note that in answer frame A 3 by adding y to both sides of the 1
equation we get x alone on the left hand side. Later we shall see I
an advantage in getting a variable by itself on one side of an equation. 1 [
k  20 = 6
so that k will be alone on the left hand side? 1
8 j
Now solve the new equation vT = 2nR for T by dividing both 1
sides by v. I
7 1
Firxjl speed v/, amount of acceleration a, and time tare three 1
choracteristics of the motion of an object. For an object starting 1
from rest, these ore related in two equations: 1
Vf = at and ^ ' J ^4* 1
Combine these two relationships into a new equation by elimi
nating Vf. This will give us an expression for d in terms of o and /. '
A4
Add 20 to both sides
Thus k 70 6
i  20 > 20  6 ♦ 20
k  26
A8
vT = 2r7f?
vT 2,7/?
T =
2jtR
kJ
Substitute the value of v# m the
first equation (that is, of) for v/
in the second equation
rf  ^(ol) (f)
5 1
Subtracting the same quantity from both sides of an equation '
also gives o new equation. For example, '
if a = b ]
then o  4 = fc  4 .
and, in general, a  c = b  c t
Moke a new equation by subtracting 2 from both sides of j
the equation
X + 2 = y ,
9 i
Another equation which describes circular motion relates speed 
V, radius of path R, and acceleration toward the center of revolution l
Oc 'S v2 i
''^ R 1
Solve this equation for the variable R. You can follow the same '
steps used in the previous two frames. j
8 1
Given the two equations o = ^ and v = ^ — , find a in i
terms of R, T, and n. 1
A5
X ♦ 2 = y
X ♦ 22 = y  2
X = y  2
Note that by subtracting 2 from
both Sides we got x by itself on
the left hond side of the equation.
A9
°^  R
Multiply both sides by R
c^R  ^
or OfR  v2
Divide both sides by o^:
A8
but
» = ^ = '''i'
^  7r,R
' T
o  0f)' (JL)
_ 4,r2/?2
"  UR
„  4rr2^
6 j
We can maice a new equation by multiplying both sides of 
an equation by the some quantity. Here are three examples to study. 
(i) If o = i, then 5c = 5t j
(ii) If p = (q + r), then 2p = 2(q + r) j
or, 2p = 2q + 2r 
(iii) If^ = 4, then (2)© = (2) (4) 
2 2 j
or, m = 8 1
Make a new equation by multiplying both sides of the equation '
^o = i + 2 1
by 3.
10 [
Suppose we wish to solve the equation 
V2 !
"<= R 1
for V. 1
First solve for v2 by performing the necessary operation. 
9 j
Let us look at a situation which requires an additional step. .
If V = ot and F = ma, find an expression for F in terms of m,
V, and t. In other words, combine the equations to eliminate o.
To do this we can ,
(i) solve one equation for a, and
(ii) substitute this value for o in the remaining equation. .
To solve the first equation v = of for a, what operation would j
you perform? 
A6
^O  fc * i
1^ = 3(1). 2)
Note thot  = 1
and 3(f> +
2) = 3fc * 6
so
a 2b* 6
AlO
"^  R
Multiply both sides by R:
Oc/?= v2
,2 = OcR
A9
Divide both sides by t
1
1
7 j
Make c  fc = 3 
into a new equation by multiplying both sides by —1. .
11 1
If v2 = acR,
find an expression for v, that is, solve this equation for v. 1
■
10
OUR PROBLEM: If v = o/ and F = ma, find F in terms
of m, V, and t.
F
When we divide both sides of v = of by t, we get
V at
f " t
V
or a = r
Now substitute this value for a in the second equotio
= ma.
n, '
A7
c 6 = 3
He W= (1)(3)
Note that (l)(c) = c
and (l)(fc) = *b
so c ♦ b  3
b 
3
All
y2 = OcR
Take the square root of both sides:
= \ a^R
If you are not familiar with squore
roots, osic your teacher for help.
AlO
F = mo
but o = ?
so F
.(P
F = ^
8 1
Dividing both sides of an equation by the same quantity gives 
a new equation.
Make a new equation by dividing both sides of the equation j
3fc= 12 1
by 3. [
12 1
In the equation 
2d = afi 1
d represents the distance an object moves from rest in time f when '
given a constant acceleration o. '
To solve this equation for /, solve first for t^ and for t. '
11 1
In the study of electricity you will become familiar with 1
power P, voltage V, current /, and resistance R. 1
Suppose we are given the relationships represented by 
the equations P = VI and / = = and we wish to find P in 
terms of / and R. 1
What two steps would you perform? 
A8
2b = 12
3fc =
3
12
3
b =
4
A12
2<y  at2
Divide both sides by a:
2d at2
^ f^
f2 
2^
Take the square root of both sides
, = V^
All
(1) Solve ' = ^ for V.
(2) Substitute this value for
V in P = V/.
(Alternatively, it is possible to
solve P = V/ for V or>d to sub
stitute this volue in / = i .
R
This would not be as direct o
method, however.)
1
1
1
9 [
Make a new equation by dividing both sides of the equation 
2b = c  4a
by 2. 1
13 j
Consider again the equation: '
Vf . V; + of 1
If we wish to find an expression for t, we must get af alone j
on the right hand side of the equation. How can we do this? j
.
1
2
OUR PROBLEM: If P = V/ and / = ^ find P in
terms of / and R.
Perform the two steps listed in answer All.
A9
25 = c  40
2b _ c  4a
2 " 2
fc _ c  4o
2
or
i = 7  2o
A13
Subtract v from both sides
of the equation.
A12
If
/
V
■ R
then
IR
 V
or
V
^ IR
Subsl
■ tut
e this
VO
ue
for
V in
P =
V/
thus p r f/ff; (1)
or P . /2f?
10 [
The two sides of an equation will still be equal to each 
other if we do any of the following: 
(i) add the same quantity to both sides 1
(ii) subtract the same quantity from both sides 1
(iii) multiply both sides by the same quantity 1
(iv) divide both sides by the some quantity. 1
These four operations which you have learned will be used •
many times in this program. Use one of these operations to change •
the equation ,
c + i = 3 1
so that c is the only symbol on one side of the equation; we call '
this "solving the equation for c." j
14 1
Subtracting v from both sides of the equation 
vf = V,. + at 1
gives us: ,
v^_ V. = V. + at  V, 1
Vf  Vj = of j
or at = Vf  Vj 1
Now soive this new equation for t. ,
13 1
We are going to use the following equations to introduce one '
further step in combining two relationships: '
if 5v * 7t = k, 1
ond V * t = m, j
find an expression for v in terms of k and m. 1
First of all, what quantity should we eliminate? 1
A10
To get c alone on the left
side of the equotion, sub
tract b from both sides.
c i= 3
C ♦ fc t r 3  i
C = 3  b
You hove thus solved th«
equation c * b = 3 for c.
A14
at = '*t ~ ^■
Divide both sides by a:
at f I
a
a
f
=
^f

^i
a
A13
1
1
 .J.. . 
11 1
Change the equation ■
V, . v^.= 6h 1
so that V is the only symbol on one side of the equation; 1
that is, solve the equation for v . 
15 1
Here is another equation similar to the one which we just i
solved: i
v^ = v;2 + 2ad 1
Use similar steps to solve the equation for d. '
;
14
OUR PROBLEM: If 5v + 2f = /^ and v + f = m
find an expression for v in terms of k and m.
sf
St
To eliminate / we first solve one equation for t. Which equotio
lould we select so that solving for / will require the lest number of
eps?
T 1
All
= 6h 
^1
^'Z
= 6h 
^'i
A15
Vf2 = y.2 ♦ 2ad
Subtract Vy2 from both sides:
Vf2 _ V.2 r V.2 ♦ 2ad
v^  v.2 = 2o</
Divide both sides by 2o:
2od
' 2o
= </
2o
V/2 _ V
z
2o
y^y
2
2o
A14
The second equation,
y * t s m
To solve for t we subtract y
from both sides.
To solve for t in the first
equation, Sv ♦ 2t = k, we must
first subtroct Sv from both
sides and tf>«n divide both
sidis by 2.
1
1
1
— ' 1
12 1
To solve the equation .
xy2 1
for y, we may first change the equation so that y appears alone
on the left side. How can we do this?
16 j
1^ P,V, P2V2 1
n, T, nz T2 1
write an equation to show how T2 depends on the other variables. 
We want to write an eauotion in the form T? = .To qet 
T
2 on top, begin by multiplying both sides of the equotion
^2. 1
•
15
OUR PROBLEM: If 5v + 2f = t and v + t = m, find an
expression for v in terms of k and m.
Solve the second equation for t.
A12
Subtract x from both sides.
A16
n,T.
n,T,
A15
02X2
P2V27;
P,V,T2 PjV.
Subtract v from both sides
V ♦ r = m
* = m  V
1
1
13 •
when X is subtracted from both sides of the equation 1
xy = 3 I
we get xyx = 3x !
or xxy=3x •
so y = 3  x 1
Now solve this new equation for y by multiplying both 
sides by (1). 
17 1
In A16, note that T2 is multiplied by PiV, and divided by 
n,7,. To solve for Tj we must multiply both sides of the equation 
by n 1 and T, (or by O) T,) and then divide both sides byPi and V, 1
(orbyP,V,). 1
First multiply by n^Ty. \
1
16
1 
OUR PROBLEM: If 5v + 2f = fc and v + / = m, find an
expression for v in terms of k and m.
Substitute this value for t in the first equation.
A13
y = 3  jt
(1) (y) = (_1) (3 _x)
y = 2 * X
or y = X _ 3
A17
(„,T,) = ■!^^(n,T,)
"2
T2P,V, =
A16
5v ♦ 2f = t
but f = m — V
SO 5v ♦ 2{iii  v) * k
1
1
14 j
Use the two steps described in frames 12 and 13 to *
solve the equation j
for Sj. 1
18 1
Now divide both sides of the equation in A17 by Pi Vi to j
give an expression for Tz. I
,
17
OUR PROBLEM: If 5v + 2f = it and v + f = m, find an
expression for v in terms of k and m.
Finally solve the equation in answer frame A16 for v.
A14
S2  S, = 3
Subtract S2 from both sides.
S2  s,  S2 ^ 3  S2
s, = 3  $2
Multiply both sides by (1).
(_!)(_,,) = {_1)(3S2)
s, = 3  S2
= s, 3
A18
PiV,T2
P,V,
P2V2niT,
(P,V,)n2
P,V,n2
A17
5v ♦ 2(m 
v) = k
Remove porenthesis:
5v ♦ 2m  2v = t
or 3v ♦ 2m = it
Sobtroct 2m from both sides:
3v = i  2m
Divide both sid
BS by 3:
3
15 I
Dividing both sides of the equation 1
3fc = 3a  c j
by 3 will get b alone on the left hand side of the equation. 
Solve the equation for b by performing this operation. 
19 1
Returning again to the equation 1
P,V, _ P2V2 ,
Use the steps we used in the lost three frames to solve 1
this equation for T^. ■
18 1
Let us review the steps we have been studying in combining two '
relationships to obtains the value of a particular variable. 1
(i) Examine the two equations to see which quantity should be •
eliminated — the quantity we ore not interested in 1
for the moment. 1
(ii) Select one of these equations and solve for this quantity. 1
(iii) Substitute the value you obtained for this quantity into the 1
other equation. 1
(iv) Solve the new equation for the variable whose value is 
desired. 1
Check these steps carefully and look back at previous frames
if necessary to see where we applied each step. .
A15
3b 
3o 
c
3fc.
3
3o 
3
c
6 =
3o 
3
• c
or
i> =
a 
c
3
A19
P,V,
P,v,
Multiply both sides by T,
P.V. ^ PzVzT,
"1 f^z^z
Multiply both sides by "2^2'
^'^^"'^" = P2V2T,
Divide both sides P z^ z
Pry^n^Tz
P,V
= T,
2'^2ni
T,
P,V,n,
16 [
It is not necessary of course that the symbol being solved for 
always appear alone on the left side of the equation; the right side 
will do as well. For example, an alternative way to solve the equation
s,  s, = 3 1
for Sj. would be to add s to both sides: .
s^  s, .s, =3.s, 1
S2 = 3 + S, 1
and then subtract 3 from both sides: 1
$2 3=3+ s, 3 1
S2  3 = s, 1
Solve the following equation for x in a similar way: 1
6XY 1
20 1
Examine carefully the equation I
PiV, P2V2 j
1 1 '1 "2 '2 1
and then write down the two operations which you would perform 1
to solve for V2. 1
19 [
Consider these two equations describing uniform acceleration ' /.%
from rest' '
V = at and d = jot^ \
(i) If we desire to find in terms of v and d, what quantity . (ij)
would you eliminate? .
(ii) Which equation would you select to solve for this quan
tity most easily? (iii)
(iii) Solve this equation for the quantity to be eliminated. j
A16
6  X = y
6  X . X  y ■• X
6 = y + X
6y=y»xy
6  y = X
A20
(i) Multiply both sides by 02X2
(ii) Divide both sides by P2
A19
(0 \T}
(ii) I V = of
(ill) Divide both sides by a:
17
Here is another similar equation:
Sit = 5f + 15s
Solve this equation for k.
21
Perform the two operations listed in A20 to solve the equation
P,V, P2V2
n , Ti "zTi
for V2.
20
OUR PROBLEM: If v = of and d = jat^,
find a in terms of v and d.
Substitute the value we have found for t into the second
equation, and do any necessary simplification.
A17
5k . 5f * 15s
5* 5f ♦ 15»
k = t * 2s
A21
P,V,
P,V,
"iT,
02X2
P,V,n2T2
=
P2V2
P,V,n
"1^2
T,
=
Vj
V2
=
P,V,„2T,
PjHiT,
A20
but f
ind
d =
r"'
f =
V
d r
F<" (?
d 
j<".4
d =
20
1
1
•
18 1
We can solve the equation 1
l = 2o + c 1
for b if we multiply both sides of this equation by 5. 
Solve this equation for b. \
22 j
Solve ^'^' = ''2V2 ^^^ p^ 1
n, 7, 0272 •
21
I
OUR PROBLEM: If v = 0/ and d = ^at^
find in terms of v and d.
fo
Finally solve the new equation in the answt
r 0.
'r block A20. I
A18
4=2.
5b
5(2o . c)
lOo ♦ 5c
A22
P,V,
P2V,
Multiply both sides by n^T^:
r^y^  —
02*2
Divide both sides by V,;
P,
''2T,V,
A21
^n
Multiply both sides by
ad . if
Divide both sides by d
v2
19 j
We wish to solve the equation i
3{a * b) = c 1
for b. 1
To solve for a quantity within parentheses, we can first remove 
the parentheses by performing the indicated operation. 
Remove the parentheses in the equation 
3(o . t) = c 1
by multiplying out (o + b) by 3. 1
23 1
An expression for kinetic energy £j^ in terms of mass (m) 1
and speed (v) is 1
Solve this equation for v. ■
22 1
Rewrite this relationship as an expression for v '
instead of a. '
y2 1
That is, solve ° " J? ^°^ ^'
A19
3(a ♦ i) = c
3o ♦ 3fc = c
A23
^k = F"*^^
Multiply both sides by 2:
Divide both sides by m:
m
Take the squore root of both sides
W
A22
v2
Multiply both sides by 7d:
7ad = v2 or v^ = 2ad
Take the square root of both
sides:
v « V2orf
20 1
Now we wish to solve the equation 1
3o + 3fc = c j
for b. First get 36 by itself on the left hand side. 
24 j
Our last equation is one which you will study in connection ,
with Newton's Law of Gravitation: j
/?2 1
Solve this equation for R. ■
23 1
We could use the some two equations 
V = of and d = jot^ 
to find d in terms of v and f. 
RemeiTiber to (i) decide which quantity should be eliminated; 
(ii) select one equation to solve for this quantity; 
(iii) substitute this volue into the other equation; 
and (iv) solve, if necessary, to get the required 
variable alone. l
Now find d in terms of v and t. 1
A20
3o . 35 3 c
Subtract 3o from both sides.
3o • 3fc  3o r c  3o
3i = c  3o
A24
^ Cm )/n2
/?2
Multiply both sides by R^'
FR^ = Gm,m2
Divide both Sides by F:
„2 . Cm ,m2
Take the square root of both sides:
\f^
ma
A23
Eliminote o by solving the
equation v = of for o.
and sub
itituting its value in
the equation d  jOt*:
<^"l(?
d')
d « lyf
21 1
Now divide the equation •
3fc = c  3o
by 3 to solve for b. 1
This is the end of Equations 2. The last program in this series,
Equations 3 Combining Two Relationships, starts at the front
of the book.
24 1
Here is o sequence of three equations '
y^ = Fd (1)
F = ma (2) 1
d^^ (3) [
relating to work W, force F, distance d, acceleration a, and speed v i
(for motion beginning from rest). ■
We want to find the amount of work W required to get a body of ■
moss 171 moving at speed v. That is, we want to find W in terms of m ■
and V. First, combine the equations (1) and (2) to eliminate F, a term ■
we are not presently interested in. •
A21
3fc = c  3o
3i c  3o
c  3o
or i = J  o
A24
Substitute the value of F m
equation (2) mto equotion (1):
W = Fd
but F  ma
so W = (ma) (d)
or Mf B mad
22
Here is another equation with a quantity in parentheses:
5(x  3) = 5
Solve for X following the three steps used in the last three frames.
25
OUR PROBLEM: W = Fd (1)
F = mc (2)
d = il (3)
Find W in terms of m and v.
or
Combine the new equation W = mad with equ
d solve for W in terms of m and v by eliminating
ation (3) above i
d. j
A22
5(x  3) = 5
Remove parentheses by multi
plying (x  3) by 5.
5x  15 = 5
Get 5x alone on one side of
the equation by adding 15 to both
both sides.
5x  15 • 15 = 5* 15
5x= 20
Divide both sides by 5
_5x 20
5 5
1Z\
A25
W = mad
■I '4
W  ma
"  "lo
i"^
23
Suppose you
are asl<
ed
to
3o
solve the equot
+ 4b = a * b
on
for b
Note that the quant
ity
ba
ppeors on both
sides
of the
eq
uati
on. I
To solve for
a quon
ity
wh
ich appears more thor
once
in
an
equa
tion, begin by
changi
ng
the
equation so the
t this
quant
ty
app
ears 1
only
on one side.
Change
3o
+ 4b = a + b
so that b appears only on
the
le
ft side; that is.
subtract b f
om
both
sides.
26 1
One final problem. For on object revolving around a central 
point, the amount of centripetal force F^ ond centripetal acceleration 
Oj, the mass m of the object, its speed v, the radius of revolution R 
and the time of one revolution T, are related by these equations: 
Fc = moc yT  2ttR 0(. = p
Find on expression for F^ in terms of m, T and R. i
A23
2a * 4b = a * b
3a * 4b  b  a * b  b
3o + 35 =
A26
*n'mR
If you got this answer, you
can feel confident of youf
ability to Kandle simple
equations. If you missed
this problem, review frames
24 and 25, and then try again.
Should you still have trouble,
asl( your teacher for help.
24 j
Now solve 1
3o + 3i) = o 1
for b, thot is, get b alone on one side of the equation. •
You hove now reviewed simple equations ond the combining of two relationships.
You should have little difficulty in understanding the development of many relatiorv
ships found in your study of physics.
You may want to review these programs at some time later. Just take some
blank pages and place them over your former answers and record your answers again
and compare with the answer blocks.
A24
3o ♦ 3fc = o
Subtract 3o from both sides.
3o*3fc3o = o3o
3fc = 2o
Then divide both sides by 3.
2b 7a
T ' 1~
b  
2a
25
Here
s
an
eq
jatio.T
with
q on both
si
des of the
equa
tion:
2(p 
q) =
2h . q
We want
to
so
Ive th
is eq
uation for
<J
First remove
the 1
parentheses
ar
d get
q on one s
de of the
eq
uation.
A25
2(p q) =2h ^ q
Remove parentheses.
2p  7q  2h • q
Subtract q from both sides:
7p 3q  3/1
26
In the new equation in A25 on the opposite page, solve for
q by getting 3q by itself on the left hand side and then dividing
both sides by —3.
A26
or
2p
 3q= 3A
1p
 3<j  2p = 3/)  2p
Zq = 2h 2p
3q 3/»  2p
3 3
r
2h 2p
'  3
q = /. . fp
r
q = fp  /.
27
Following the steps described in frames 25 and 26, solve the
following equation for s:
3(r  2s)  r + 3$
A27
Note: As was pointed out before, equations can be solved by iso
lating the symbol being solved for on the right hand side of the
equation. Thus, an alternative way to solve for s after removing
brackets from
3(r  2s) = r + 3s
would be to (i) add 6s to both sides, or (ii) subtract r from both
sides, and then (iii) divide by 9.
3r  6s = r + 3s
(i) 3r= r+9s
(ii) 2r = 9s
(iii)
3(r  2s) 3 r . 3s
Remove brockets.
3r  6s = r ♦ 3$
Sufctract 3s from both sides.
3r  6s  3s = r ♦ 3s  3s
3r  9s = r
Subtract 3r from both sides.
3r  9s  3r = r  3r
9s = 2r
Then divide both sides by 9.
9s 2r
9 " 9
See rx>te at left.
28
We shall conclude this program with a few simple
equations to solve.
Solve 6p  2t = s
for p.
A28
6p  2f = s
Add 2f to both sides.
6p ^ s * 2f
Divide both sides by 6.
P =
s * 2/
s f
p = — + —
6 3
29 1
Solve 1
m  n = 8 1
for n. 1
I
A29
m —  n  S
Subtract m from both sides.
— z n  S — m
Multiply both sides by 1.
— n = — 8 ♦ m
or — n = m  8
Multiply both sides by 2.
n = 2(m  8)
or n  2/n  16
1
30 1
Solve 1
2(3o + fc) = + 7fc 1
for 0. 1
A30
2(3o
* b)  a ^
7b
60
* 2b a * 7b
600
* 2b 7b
5o
^ 2b 7b
5a  7b 2b
5a  5b
a  b
You have now completed Equations 1 and are able to handle
the main algebraic operations. You can practice this skill in the
context of physics equations by going through the program Equations
2. It begins at the front of this book just below this program.
003089641 X