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coefficient and change the ¿2 to #; let us assume that the same thing will happen
this time, and you can check the result yourself. The derivative of 34/2 will
then be 64. Next we diferentiate , a constant term; but by a rule stated
previously, the derivative of Ö is zero; hence this term contributes nothing to
the acceleration. The final result, therefore, is ø = du/dt = 6At.
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, ø,
its velocity 0 at any time £ is given by
U = gỉ.
The distance it covers in the same tỉme is
s= 3 gt2.
'Various mathematical notations are used in writing derivatives. 5ince velocity
1s ds/dt and acceleration is the time derivative of the velocity, we can also write
d (ds d2s
G=_— —— = _—xY (8.10)
đt \ dị d2
which are common ways of writing a second derivative.
--- Trang 174 ---
W© have another law that the velocity is equal to the integral of the acceleration.
This is just the opposite of a = du/di; we have already seen that distance is
the integral of the velocity, so distance can be found by twice integrating the
acceleration.
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 ? which moves in three dimensions in any manner whatsoever. At the
beginning of this chapter, we opened our discussion of the one-dimensional 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 three-
dimensional 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 païir 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 z-distance and a z-distance, 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 ripght angles to the first two, and measuring a third
distance, the z-distance. The distances are now measured from coordinate pÌanes
instead of lines.) Having constructed a table with z- and -distances, how can
we determine the velocity? We first fnd the components of velocity in each
direction. 'Phe horizontal part of the velocity, or z-component, is the derivative
of the z-distance with respect to the time, or
U„ = dø/dt. (8.11)
Similarly, the vertical part of the velocity, or -component, is
uụ = dụ /dt. (8.12)
In the third dimension,
Uy = đz/dl. (8.13)
Now, given the components of velocity, how can we fnd the velocity along
the actual path of motion? In the two-dimensional case, consider two successive
positions of the particle, separated by a short distance As and a short time
--- Trang 175 ---
M As A/(Ax)2 + (Ay)2
Ayv/Atf— XỬ
Ax#ø#v„At
Fig. 8-3. Description of the motion of a body in two dimensions and
the computation of its velocity.
interval f¿ — fq = Ai. In the time A£ the particle moves horizontally a dis-
tance Az % 0„ Af, and vertically a distance A¿ uy At. (The symbol “+” is
read “is approximately.”) The actual distance moved is approximately
Asxz V(Az)2 + (Aø)2, (8.14)
as shown in EFig. 8-3. The approximate velocity during this interval can be
obtained by dividing by A£ and by letting A# go to 0, as at the beginning of the
chapter. We then get the velocity as
U= T= V(dz/đdt)? + (dụ/đE)? = vu + 0. (8.15)
For three dimensions the result is
Đ= \(02 + 02 + 0Ẻ. (8.16)
In the same way as we defned velocities, we can delne accelerations: we have
an #-component of acceleration ø„, which is the derivative of ø„, the z-component
of the velocity (that is, a„ = đ?z/d/2, the second derivative of z with respect
to £), 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 w, and
at the same time goes vertically downward with a constant acceleration —g; what
is the motion? We can say d#/dt = 0u„ = u. Since the velocity 0x is constant,
# = tứ, (8.17)
--- Trang 176 ---
and since the downward acceleration —gø 1s constant, the distance # the objec
falls can be written as
ụ= —39Ÿ. (8.18)
'What is the curve of its path, i.e., what is the relation between and z? We can
eliminate £ from Eq. (8.18), since ý = z/u. When we make this substitution we
fñnd that :
This relation between ø and z 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.
Fig. 8-4. The parabola described by a falling body with an initial
horizontal velocity.
--- Trang 177 ---
NWeosrfore s EL«ttfs ©œŸ` ÏÌggTt(i110fS
9-1 Momentum and force
The discovery of the laws of dynamies, or the laws of motion, was a dramatic
moment in the history ofscience. Before Newton's time, the motions of things like
the planets were a mystery, but after Newton there was complete understanding.
ven the slight deviations from Kepler”s laws, due to the perturbations of the
planets, were computable. "The motions of pendulums, oscillators with springs
and weights in them, and so on, could all be analyzed completely after Newton”s
laws were enunciated. So it is with this chapter: before this chapter we could
not calculate how a mass on a spring would move; much less could we calculate
the perturbations on the planet Uranus due to Jupiter and Saturn. After this