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Unformatted text preview: Motion 81 Description of motion In order to ﬁnd the laws governing the various changes that take place in 8—1 Description of motion
bodies as time goes on, we must be able to describe the changes and have some way
to record them. The simplest change to observe in a body is the apparent change 8—2 Speed
in its position With time, Which we call motion. Let us consider some solid object 8—3 Speed as a derivative
with a permanent mark, which we shall call a point, that we can observe. We
shall discuss the motion of the little marker, which might be the radiator cap of an
automobile or the center of a falling ball, and shall try to describe the fact that it 8—5 Acceleration
moves and how it moves. These examples may sound trivial, but many subtleties enter into the descrip
tion of change. Some changes are more diﬂicult to describe than the motion of
a point on a solid object, for example the speed of drift of a cloud that is drifting
very slowly, but rapidly forming or evaporating, or the change of a woman’s
mind. We do not know a simple way to analyze a change of mind, but since the
cloud can be represented or described by many molecules, perhaps we can describe
the motion of the cloud in principle by describing the motion of all its individual
molecules. Likewise, perhaps even the changes in the mind may have a parallel
in changes of the atoms inside the brain, but we have no such knowledge yet. At any rate, that is why we begin with the motion of points; perhaps we should
think of them as atoms, but it is probably better to be more rough in the begin Table 8'1
ning and simply to think of some kind of small objects—small, that is, compared 1 (min) 3 (ft)
with the distance moved. For instance, in describing the motion of a car that is .._____.___.__
going a hundred miles, we do not have to distinguish between the front and the
back of the car. To be sure, there are slight differences, but for rough purposes we
say “the car,” and likewise it does not matter that our points are not absolute
points; for our present purposes it is not necessary to be extremely precise. Also,
while we take a ﬁrst look at this subject we are going to forget about the three
dimensions of the world. We shall just concentrate on moving in one direction,
as in a car on one road. We shall return to three dimensions after we see how to
describe motion in one dimension. Now, you may say, “This is all some kind of
trivia,” and indeed it is. How can we describe such a onedimensional motion—
let us say, of a car? Nothing could be simpler. Among many possible ways, one
would be the following. To determine the position of the car at different times,
we measure its distance from the starting point and record all the observations.
In Table 8—1, s represents the distance of the car, in feet, from the starting point,
and t represents the time in minutes. The ﬁrst line in the table represents zero
distance and zero time—the car has not started yet. After one minute it has started
and has gone 1200 feet. Then in two minutes, it goes farther—notice that it picked
up more distance in the second minute—it has accelerated; but something hap
pened between 3 and 4 and even more so at 5—it stopped at a light perhaps? Then
it speeds up again and goes 13,000 feet by the end of 6 minutes, 18,000 feet at the
end of 7 minutes, and 23,500 feet in 8 minutes; at 9 minutes it has advanced to
only 24,000 feet, because in the last minute it was stopped by a cop. That is one way to describe the motion. Another way is by means of a graph.
If we plot the time horizontally and the distance vertically, we obtain a curve some
thing like that shown in Fig. 8—1. As the time increases, the distance increases, im: m amms‘ a '0
at ﬁrst very slowly and then more rapidly, and very slowly again for a little while
at 4 minutes; then it increases again for a few minutes and ﬁnally, at 9 minutes, Fig. 3—1 , Graph of distance versus
appears to have stopped increasing. These observations can be made from the time for the car. 8—1 84 Distance as an integral 1200 9500
9600
13000
18000
23500
24000 \Dmﬂa‘tMBUJNHO §§§§ DISTANCE TRAVELED IN FEET
an Table 82 1 (sec) 3 (ft) O
16
64 144
256
400
576 ONUIwat‘O 400 0‘
O
O N
O
0 I00 DISTANCE FALLEN IN FEET l 2 3 4 5
“VS IN SECONDS Fig. 8—2. Graph of distance versus
time for a falling body. graph, without a table. Obviously, for a complete description one would have to
know where the car is at \the halfminute marks, too, but we suppose that the graph
means something, that the car has some position at all the intermediate times. The motion of a car is complicated. For another example we take something
that moves in a simpler manner, folldwing more simple laws: a falling ball.
Table 8—2 gives the time in seconds and the distance in feet for a falling body.
At zero seconds the ball starts out at zero feet, and at the end of 1 second it has
fallen 16 feet. At the end of 2 seconds, it has fallen 64 feet, at the end of 3
seconds, 144 feet, and so on; if the tabulated numbers are plotted, we get the
nice parabolic curve shown in Fig. 8—2. The formula for this curve can be written
as s = 1612. (8.1) This formula enables us to calculate the distances at any time. You might say
there ought to be a formula for the ﬁrst graph too. Actually, one may wr1te such
a formula abstractly, as s = f(t), (8.2) meaning that s is some quantity depending on t or, in mathematical phraseology,
s is a function of t. Since we do not know what the functlon is, there is no way we
can write it in deﬁnite algebraic form. We have now seen two examples of motion, adequately described with very
simple ideas, no subtleties. However, there are subtleties—«several of them. In
the ﬁrst place, what do we mean by time and space? It turns out that these deep
philosophical questions have to be analyzed very carefully in physics, and this
is not so easy to do. The theory of relativity shows that our ideas of space and
time are not as simple as one might think at ﬁrst sight. However, for our present
purposes, for the accuracy that we need at ﬁrst, we need not be very careful about
deﬁning things precisely. Perhaps you say, “That’s a terrible thing—I learned that
in science we have to deﬁne everything precisely.” We cannot deﬁne anything
precisely! If we attempt to, we get into that paralysis of thought that comes to
philosophers, who s1t opposite each other, one saying to the other, “You don’t
know what you are talking about!” The second one says, “What do you mean
by know ? What do you mean by talking ? What do you mean by you .7,” and so on.
In order to be able to talk constructively, we just have to agree that we are talking
about roughly the same thing. You know as much about time as we need for the
present, but remember that there are some subtleties that have to be discussed;
we shall discuss them later. Another subtlety involved, and already mentioned, is that it should be possible
to imagine that the moving point we are observing is always located somewhere.
(Of course when we are looking at it, there it is, but maybe when we look away
it isn’t there.) It turns out that in the motion of atoms, that idea alsd is false—
we cannot ﬁnd a marker on an atom and watch it move. That subtlety we shall
have to get around in quantum mechanics. But we are ﬁrst going to learn what the
problems are before introducing the complications, and then we shall be in a better
position to make corrections, in the light of the more recent knowledge of the
subject. We shall, therefore, take a simple point of view about time and space.
We know what these concepts are in a rough way, and those who have driven a
car know what speed means. 8—2 Speed Even though we know roughly what “speed” means, there are still some
rather deep subtleties; consider that the learned Greeks were never able to adequately
describe problems involving velocity. The subtlety comes when we try to compre
hend exactly what is meant by “speed.” The Greeks got very confused about this,
and a new branch of mathematics had to be discovered beyond the geometry and
algebra of the Greeks, Arabs, and Babylonians. As an illustration of the difﬁ
culty, try to solve this problem by sheer algebra: A balloon is being inﬂated so 8—2 that the volume of the balloon is increasing at the rate of 100 cm3 per second;
at what speed is the radius increasing when the volume is 1000 cm3? The Greeks
were somewhat confused by such problems, being helped, of course, by some very
confusing Greeks. To show that there were difﬁculties in reasoning about speed
at the time, Zeno produced a large number of paradoxes, of which we shall men
tion one to illustrate his point that there are obvious difﬁculties in thinking about
motion. “Listen,” he says, “to the following argument: Achilles runs 10 times as
fast as a tortoise, nevertheless he can never catch the tortoise. For, suppose that
they start in a race where the tortoise is 100 meters ahead of Achilles; then when
Achilles has run the 100 meters to the place where the tortoise was, the tortoise has
proceeded 10 meters, having run onetenth as fast. Now, Achilles has to run
another 10 meters to catch up with the tortoise, but on arriving at the end of that
run, he ﬁnds that the tortoise is still 1 meter ahead of him; running another meter,
he ﬁnds the tortoise 10 centimeters ahead, and so on, ad inﬁnitum. Therefore, at
any moment the tortoise is always ahead of Achilles and Achilles can never catch
up with the tortoise.” What is wrong with that? It is that a ﬁnite amount of time
can be divided into an inﬁnite number of pieces, just as a length of line can be
divided into an inﬁnite number of pieces by dividing repeatedly by two. And so,
although there are an inﬁnite number of steps (in the argument) to the point at
which Achilles reaches the tortoise, it doesn’t mean that there is an inﬁnite amount
of time. We can see from this example that there are indeed some subtleties in
reasoning about speed. In order to get to the subtleties in a clearer fashion, we remind you of a joke
which you surely must have heard. At the point where the lady in the car is caught
by a cop, the cop comes up to her and says, “Lady, you were going 60 miles an
hour!” She says, “That’s impossible, sir, I was travelling for only seven minutes.
It is ridiculous—how can I go 60 miles an hour when I wasn’t going an hour?”
How would you answer her if you were the cop? Of course, if you were really the
cop, then no subtleties are involved; it is very simple: you say, “Tell that to the
judge!” But let us suppose that we do not have that escape and we make a more
honest, intellectual attack on the problem, and try to explain to this lady what
we mean by the idea that she was going 60 miles an hour. Just what do we mean?
We say, “What we mean, lady, is this: if you kept on going the same way as you
are going now, in the next hour you would go 60 miles.” She could say, “Well,
my foot was oﬁ“ the accelerator and the car was slowing down, so if I kept on going
that way it would not go 60 miles.” Or consider the falling ball and suppose we
want to know its speed at the time three seconds if the ball kept on going the way
it is going. What does that mean—kept on accelerating, going faster? No—kept
on going with the same velocity. But that is what we are trying to deﬁne! For if
the ball keeps on going the way it is going, it will just keep on going the way it is
going. Thus we need to deﬁne the velocity better. What has to be kept the same?
The lady can also argue this way: “If I kept on going the way I’m going for one
more hour, I would run into that wall at the end of the street!” It is not so easy to
say what we mean. Many physicists think that measurement is the only deﬁnition of anything.
Obviously, then, we should use the instrument that measures the speed—the
speedometer—and say, “Look, lady, your speedometer reads 60.” So she says,
“My speedometer is broken and didn’t read at all.” Does that mean the car is
standing still? We believe that there is something to measure before we build
the speedometer. Only then can we say, for example, “The speedometer isn’t
working right,” or “the speedometer is broken.” That would be a meaningless
sentence if the velocity had no meaning independent of the speedometer. So we
have in our minds, obviously, an idea that 15 independent of the speedometer,
and the speedometer is meant only to measure this idea. So let us see if we can get
a better deﬁnition of the idea. We say, “Yes, of course, before you went an hour,
you would hit that wall, but if you went one second, you would go 88 feet; lady,
you were going 88 feet per second, and if you kept on going, the next second it
would be 88 feet, and the wall down there is farther away than that.” She says,
“Yes, but there’s no law against going 88 feet per second! There is only a law 8—3 against going 60 miles an hour.” “But,” we reply, “it’s the same thing.” If it is
the same thing, it should not be necessary to go into this circumlocution about
88 feet per second. In fact, the falling ball could not keep going the same way
even one second because it would be changing speed, and we shall have to deﬁne
speed somehow. Now we seem to be getting on the right track; it goes something like this:
If the lady kept on going for another 1/1000 of an hour, she would go 1 / 1000 of
60 miles. In other words, she does not have to keep on going for the whole hour;
the point is that for a moment she is going at that speed. Now what that means
is that if she went just a little bit more in time, the extra distance she goes would
be the same as that of a car that goes at a steady speed of 60 miles an hour. Per
haps the idea of the 88 feet per second is right; we see how far she went in the last
second, divide by 88 feet, and if it comes out 1 the speed was 60 miles an hour.
In other words, we can ﬁnd the speed in this way: We ask, how far do we go in a
very short time? We divide that distance by the time, and that gives the speed.
But the time should be made as short as possible, the shorter the better, because
some change could take place during that time. If we take the time of a falling
body as an hour, the idea is ridiculous. If we take it as a second, the result is
pretty good for a car, because there is not much change in speed, but not for a
falling body; so in order to get the speed more and more accurately, we should
take a smaller and smaller time interval. What we should do is take a millionth
of a second, and divide that distance by a millionth of a second. The result gives
the distance per second, which is what we mean by the velocity, so we can deﬁne
it that way. That is a successful answer for the lady, or rather, that is the deﬁnition
that we are going to use. The foregoing deﬁnition involves a new idea, an idea that was not available
to the Greeks in a general form. That idea was to take an inﬁnitesimal distance
and the corresponding inﬁnitesimal time, form the ratio, and watch what happens
to that ratio as the time that we use gets smaller and smaller and smaller. In other
words, take a limit of the distance travelled divided by the time required, as the
time taken gets smaller and smaller, ad inﬁnitum. This idea was invented by
Newton and by Leibnitz, independently, and is the beginning of a new branch
of mathematics, called the diferential calculus. Calculus was invented in order to
describe motion, and its ﬁrst application was to the problem of deﬁning what is
meant by going “60 miles an hour.” Let us try to deﬁne velocity a little better. Suppose that in a short time,
a, the car or other body goes a short distance x; then the velocity, v, is deﬁned as v = x/e, an approximation that becomes better and better as the e is taken smaller and
smaller. If a mathematical expression is desired, we can say that the velocity
equals the limit as the e is made to go smaller and smaller in the expression x/e, or . x
v = 11m —~ (8.3) e—.0 6
We cannot do the same thing with the lady in the car, because the table is in
complete. We know only where she was at intervals of one minute; we can get
a rough idea that she was going 5000 ft/ min during the 7th minute, but we do not
know, at exactly the moment 7 minutes, whether she had been speeding up and the
speed was 4900 ft/min at the beginning of the 6th minute, and is now 5100 ft/min,
or something else, because we do not have the exact details in between. So only
if the table were completed with an inﬁnite number of entries could we really
calculate the velocity from such a table. On the other hand, when we have a com
plete mathematical formula, as in the case of a falling body (Eq. 8.1), then it is possible to calculate the velocity, because we can calculate the position at any time
whatsoever. Let us take as an example the problem of determining the velocity of the
falling ball at the particular time 5 seconds. One way to do this is to see from 8—4 Table 8—2 what it did in the 5th second; it went 400 — 256 = 144 ft, so it is going
144 ft/sec; however, that is wrong, because the speed is changing; on the average
it is 144 ft/sec during this interval, but the ball is speeding up and is really going
faster than 144 ft/sec. We want to ﬁnd out exactly how fast. The technique in
volved in this process is the following: We know where the ball was at 5 sec.
At 5.1 see, the distance that it has gone all together is l6(5.1)2 = 416.16 ft (see
Eq. 8.1). At 5 see it had already fallen 400 ft; in the last tenth of a second it fell
416.16 — 400 = 16.16 ft. Since 16.16 ft in 0.1 see is the same as 161.6 ft/sec,
that is the speed more or less, but it is not exactly correct. Is that the speed at
5, or at 5.1, or halfway between at 5.05 sec, or when is that the speed? Never mind
—the problem was to ﬁnd the speed at 5 seconds, and we do not have exactly
that; we have to do a better job. So, we take onethousandth of a second more than
5 sec, or 5.001 sec, and calculate the total fall as s = l6(5.001)2 = l6(25.010001) = 400160016 ft. In the last 0.001 sec the ball fell 0.160016 ft, and if we divide this number by 0.001
sec we obtain the speed as 160.016 ft/sec. That is closer, very close, but it is
still not exact. It should now be evident what we must do to ﬁnd the speed exactly.
To perform the mathematics we state the problem a little more abstractly: to
ﬁnd the velocity at a special time, to, which in the original problem was 5 sec.
Now the distance at to, which we call so, is 16t3, or 400 ft in this case. In order
to ﬁnd the velocity, we ask, “At the time to + (a little bit), or to + e, where is
the body?” The new position is 16(t0 + e)2 = 16t3 + 32toe + 1662. So it is
farther along than it was before, because before it was only 16tg. This distance
we shall call so + (a little bit more), or so + x (if x is the extra bit). Now if we
subtract the distance at to from the distance at to + e, we get x, the extra distance
gone, as x = 32t0  e + 1662. Our ﬁrst approximation to the velocity is v = i: = 3210 + 165. (8.4)
The true velocity is the value of this ratio, x/e, when 6 becomes vanishingly small.
In other words, after forming the ratio, we take the limit as 6 gets smaller and smaller, that is, approaches 0. The equation reduces to,
1) (at time to) = 32to. In our problem, to = 5 sec, so the solution is v = 32 X 5 = 160 ft/sec. A few
lines above, where we took 6 as 0.1 and 0.01 sec successively, the value we got for 1) was a little more than this, but now we see that the actual velocity is precisely
160 ft/sec. 8—3 Speed as a derivative The procedure we have just carried out is performed so often in mathematics
that for convenience special notations have been assigned to our quantities e and x.
In this notation, the 6 used above becomes At and x becomes As. This At means
“an extra bit of t,” and carries an implication that it can be made smaller. The
preﬁx A is not a multiplier, any more than sin 0 means 3  i  n  0—it simply
deﬁnes a time increment, and reminds us of its special character. As has an
analogous meaning for the distance s. Since A is not a factor, it cannot be can
celled in the ratio As/At to give s/ t, any more than the ratio sin 6/sin 20 can be
reduced to 1/2 by cancellation. In this notation, velocity is equal to the limit of
As/At when At gets smaller, or v = lim — (8.5) This is really the same as our previous expression (8.3) with e and x, but it has the
advantage of showing that something is changing, and it keeps track of what is
changing. 85 Incidentally, to a good approximation we have another law, which says that
the change in distance of a moving point is the velocity times the time interval,
or As = 0 At. This statement is true only if the velocity is not changing during
that time interval, and this condition is true only in the limit as A1 goes to 0.
Physicists like to write it ds = 2) dt, because by dt they mean At in circumstances
in which it is very small; with this understanding, the expression is valid to a close
approximation. If At is too long, the velocity might change during the interval,
and the approximation would become less accurate. For a time dt, approaching
zero, ds = 1) dt precisely. In this notation we can write (8.5) as v= lim§_d_s. At—»0 At _ dt The quantity ds/dt which we found above is called the “derivative of s with
respect to t” (this language helps to keep track of What was changed), and the com
plicated process of ﬁnding it is called ﬁnding a derivative, or differentiating.
The ds’s and dt’s which appear separately are called diﬂerentz'als. To familiarize
you with the words, we say we found the derivative of the function l6t2, or the
derivative (with respect to t) of 1622 is 32t. When we get used to the words, the
ideas are more easily understood. For practice, let us ﬁnd the derivative of a more
complicated function. We shall consider the formula 3 = At3 + Bt + C, which
might describe the motion of a point. The letters A, B, and C represent constant
numbers, as in the familiar general form of a quadratic equation. Starting from
the formula for the motion, we wish to ﬁnd the velocity at any time. To ﬁnd
the velocity in the more elegant manner, we change I to t + At and note that
s is then changed to s + some As; then we ﬁnd the As in terms of At. That is to say, s+As=A(t+At)3+B(t+At)+ C
= A13 + Bt + C + 3At2 At + BAt + 3Az(At)2 + A(At)3,
but since
3 = A13 + Bt + C,
we ﬁnd that As = 3At2 At + BAt + 3At(Az)2 + A(At)3. But we do not want As—we want As divided by Al. We divide the preceding equa
tion by At, getting 2—: = 3At2 + B + 3At(At) + A(At)2. Table 83. A Short Table of Derivatives s, u, v, w are arbitrary functions of t; a, b, c, and n are arbitrary constants Function Derivative
__ n dS_ n—l
s—t dt—n
_ iii_ a
5—6" (It—cdt
ds du dv dw
S—u—ivlwl Et—E+E+—d_t_+
ds
s—c E_0
s _ uavbwc ds s ﬂdu+édv c dw,
_ dt _ u dt 1) dt w dt‘ 8—6 As At goes toward 0 the limit of As/At is ds/dt and is equal to d5 2 It? — 3/1! + B. This is the fundamental process of calculus, differentiating functions. The process
is even more simple than it appears. Observe that when these expansions con
tain any term with a square or a cube or any higher power of At, such terms may be
dropped at once, since they will go to 0 when the limit is taken. After a little prac
tice the process gets easier because one knows what to leave out. There are many
rules or formulas for differentiating various; types of functions. These can be memorized, or can be found in tables. A short list is found in Table 8—3. 8—4 Distance as an integral Now we have to discuss the inverse problem. Suppose that instead of a table of
distances, we have a table of speeds at different times, starting from zero. For the
falling ball, such speeds and times are shown in Table 8—4. A similar table could
be constructed for the velocity of the car, by recording the speedometer reading
every minute or halfminute. If we know how fast the car is going at any time, can
we determine how far it goes? This problem is just the inverse of the one solved Velocity Of a Falling Ball
above; we are given the velocity and asked to ﬁnd the distance. How can we ﬁnd
the distance if we know the speed? If the speed of the car is not constant, and the
lady goes sixty miles an hour for a moment, then slows down, speeds up, and so 0 0
on, how can we determine how far she has gone? That is easy. We use the same 1 32
idea, and express the distance in terms of inﬁnitesimals. Let us say, “In the ﬁrst 2 64 3
4 Table 84 1 (sec) v (ft/sec) second her speed was such and such, and from the formula As = vAt we can 96
calculate how far the car went the ﬁrst second at that speed.” Now in the next 128
second her speed is nearly the same, but slightly different; we can calculate how far she went in the next second by taking the new speed times the time. We pro ceed similarly for each second, to the end of the run. We now have a number of little distances, and the total distance will be the sum of all these little pieces. That is, the distance will be the sum of the velocities times the times, or s = 21) At, Where the Greek letter Z (sigma) is used to denote addition. To be more precise, it is the sum of the velocity at a certain time, let us say the ith time, multiplied by At. s = Z v(t,) At. (8.6) The rule for the times is that n+1 = t, + At. However, the distance we obtain
by this method will not be correct, because the velocity changes during the time
interval At. If we take the times short enough, the sum is precise, so we take them
smaller and smaller until we obtain the desired accuracy. The true s is s = lim 2 0(1)) At. (8.7)
At—>0 z
The mathematicians have invented a symbol for this limit, analogous to the symbol
for the differential. The A turns into a d to remind us that the time is as small as
it can be; the velocity is then called 1) at the time t, and the addition is written
as a sum with a great “3,” I (from the Latin summa), which has become distorted
and is now unfortunately just called an integral sign. Thus we write 5 = / 0(1) dt. (8.8) This process of adding all these terms together is called integration, and it is the
opposite process to differentiation. The derivative of this integral is 2), so one
operator (d) undoes the other (I). One can get formulas for integrals by taking
the formulas for derivatives and running them backwards, because they are re
lated to each other inversely. Thus one can work out his own table of integrals
by differentiating all sorts of functions. For every formula with a differential,
we get an integral formula if we turn it around. 8—7 Every function can be differentiated analytically, i.e., the process can be carried
out algebraically, and leads to a deﬁnite function. But it is not possible in a simple
manner to write an analytical value for any integral at will. You can calculate it,
for instance, by doing the above sum, and then doing it again with a ﬁner interval
At and again with a ﬁner interval until you have it nearly right. In general, given
some particular function, it is not possible to ﬁnd, analytically, what the integral
is. One may always try to ﬁnd a function which, when differentiated, gives some
desired function; but one may not ﬁnd it, and it may not exist, in the sense of being
expressible in terms of functions that have already been given names. 8—5 Acceleration The next step in developing the equations of motion is to introduce another
idea which goes beyond the concept of velocity to that of change of velocity,
and we now ask, “How does the velocity change?” In previous chapters we have
discussed cases in which forces produce changes in velocity. You may have heard
with great excitement about some car that can get from rest to 60 miles an hour
in ten seconds ﬁat. From such a performance we can see how fast the speed
changes, but only on the average. What we shall now discuss is the next level of
complexity, which is how fast the velocity is changing. In other words, by how
many feet per second does the velocity change in a second, that is, how many feet
per second, per second? We previously derived the formula for the velocity of
a falling body as v = 32t, which is charted in Table 8—4, and now we want to
ﬁnd out how much the velocity changes per second; this quantity is called the
acceleration. Acceleration is deﬁned as the time rate of change of velocity. From the
preceding discussion we know enough already to write the acceleration as the
derivative dv/dt, in the same way that the velocity is the derivative of the distance.
If we now differentiate the formula 1) = 321 we obtain, for a falling body, dv a — —d—t — 32. (8.9)
[To differentiate the term 321 we can utilize the result obtained in a previous
problem, where we found that the derivative of BI is simply B (a constant). So
by letting B = 32, we have at once that the derivative of 321 is 32.] This means
that the velocity of a falling body is changing by 32 feet per second, per second
always. We also see from Table 8—4 that the velocity increases by 32 ft/sec in
each second. This is a very simple case, for accelerations are usually not constant.
The reason the acceleration is constant here is that the force on the falling body
is constant, and Newton’s law says that the acceleration is proportional to the force. As a further example, let us ﬁnd the acceleration in the problem we have
already solved for the velocity. Starting with s = A13 + Bt + C
we obtained, for v = d5/ dt, 1) == 3A12 + B. Since acceleration is the derivative of the velocity with respect to the time, we need
to differentiate the last expression above. Recall the rule that the derivative of the
two terms on the right equals the sum of the derivatives of the individual terms.
To differentiate the ﬁrst of these terms, instead of going through the fundamental
process again we note that we have already differentiated a quadratic term when
we differentiated 16t2, and the effect was to double the numerical coefﬁcient and
change the 12 to I; let us assume that the same thing will happen this time, and you
can check the result yourself. The derivative of 3A:2 will then be 6A1. Next we
differentiate B, a constant term; but by a rule stated previously, the derivative of
B is zero; hence this term contributes nothing to the acceleration. The ﬁnal
result, therefore, is a dv/dt = 6At. 8—8 H For reference, we state two very useful formulas, which can be obtained by
integration. If a body starts from rest and moves with a constant acceleration,
g, its velocity v at any time t is given by v = gt.
The distance it covers in the same time is
s = %gt2. Various mathematical notations are used in writing derivatives. Since velocity
is ds/dt and acceleration is the time derivative of the velocity, we can also write d ds dzs which are common ways of writing a second derivative. We have another law that the velocity is equal to the integral of the accelera
tion. This is just the opposite of a = dv/dt; we have already seen that distance is
the integral of the velocity, so distance can be found by twice integrating the ac
celeration. In the foregoing discussion the motion was in only one dimension, and space
permits only a brief discussion of motion in three dimensions. Consider a particle
P which moves in three dimensions in any manner whatsoever. At the beginning
of this chapter, we opened our discussion of the onedimensional case of a moving
car by observing the distance of the car from its starting point at various times.
We then discussed velocity in terms of changes of these distances with time, and
acceleration in terms of changes in velocity. We can treat threedimensional motion
analogously. It will be simpler to illustrate the motion on a two—dimensional
diagram, and then extend the ideas to three dimensions. We establish a pair of
axes at right angles to each other, and determine the position of the particle at any
moment by measuring how far it is from each of the two axes. Thus each position
is given in terms of an xdistance and a ydistance, and the motion can be described
by constructing a table in which both these distances are given as functions of time.
(Extension of this process to three dimensions requires only another axis, at right
angles to the ﬁrst two, and measuring a third distance, the zdistance. The dis
tances are now measured from coordinate planes instead of lines.) Having con
structed a table With x and y—distances, how can we determine the velocity?
We ﬁrst ﬁnd the components of velocity in each direction. The horizontal part of the velocity, or x—component, is the derivative of the xdistance with respect to
the time, or v, = a’x/dt. (8.11)
Similarly, the vertical part of the velocity, or ycomponent, is 121, = dy/dt. (8.12)
In the third dimension, v, = dz/dt. (8.13) Now, given the components of velocity, how can we ﬁnd the velocity along the
actual path of motion? In the twodimensional case, consider two successive
positions of the particle, separated by a short distance As and a short time in
terval t2 — t1 = At. In the time At the particle moves horizontally a distance
Ax ~ 22, At, and vertically a distance Ay ~ 2),, At. (The symbol “~” is read
“is approximately.") The actual distance moved is approximately As ~ «73252 + (AW, (8.14) as shown in Fig. 8—3. The approximate velocity during this interval can be obtained
by dividing by At and by letting At go to 0, as at the beginning of the chapter. 8—9 V Ask: , /(Ax)2 +(Ay)2 Fig. 8—3. Description of the motion
of a body in two dimensions and the
computation of its velocity. \ Fig. 8—4. The parabola described by
a falling body with an initial horizontal
velocity. We then get the velocity as _£1§
”—dt = x/(dx/alt)2 + (dy/dtz) = x/vg+vg (8.15) For three dimensions the result is v = x/vg + I); + v3. (8.16) In the same way as we deﬁned velocities, we can deﬁne accelerations: we have
an xcomponent of acceleration at, which is the derivative of 0:, the xcomponent
of the velocity (that is, a1 = dzx/dtz, the second derivative of x with respect to
t), and so on. Let us consider one nice example of compound motion in a plane. We shall
take a motion in which a ball moves horizontally with a constant velocity u, and
at the same time goes vertically downward with a constant acceleration —g;
what is the motion? We can say dx/dt = or = u. Since the velocity v; is constant, x = ut, (8.17) and since the downward acceleration —g is constant, the distance y the object
falls can be written as y = —%gt2. (8.18) What is the curve of its path, i.e., what is the relation between y and x? We can eliminate I from Eq. (8.18), since I = x/u. When we make this substitution we
ﬁnd that _ _ i 2
y — 2112 x . (8.19)
This relation between y and x may be considered as the equation of the path of
the moving ball. When this equation is plotted ‘we obtain a curve that is called a parabola; any freely falling body that is shot out in any direction will travel in
a parabola, as shown in Fig. 8—4. 8—10 ...
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 Spring '09
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