1.2 Position and Displacement
One general way of locating a particle (or particle-like object) is with a position vector , which is a vector that extends from a reference point (usually the origin of a coordinate system) to the particle. In the unit-vector notation, can be written
,
(4-1)
where
,
,
and
are the vector components of
,
and the coefficients
,
,
and
are its scalar components.
The coefficients , , and give the particle's location along the coordinate axes and relative to the origin; that is, the particle has the rectangular coordinates ( , , ). For instance, Fig. 4-1 shows a particle with position vector
(4-2)
Fig. 4-1 |
.
Along the
axis the particle is 3 m from the origin, in the
direction. Along the
axis it is 2 m
from the origin, in the
direction. Along the
axis it is 5 m
from the origin, in the
direction.
As a
particle moves, its position vector changes in such a way that the
vector always extends to the particle from the reference point (the
origin). If the position vector changes, say, from
to
during a certain time interval, then the particle's displacement
during that time interval is
(4-2)
Using the unit-vector notation, we can rewrite this displacement as
or as
(4-3)
where
coordinates (
,
,
)
correspond to position vector
and
coordinates (
,
,
)
correspond
to position vector
.
We can also rewrite the displacement by
substituting
for
,
for
,
and
for
:
(4-3)
Example 4-1
In Fig. 4-2, the position vector for a particle is initially
Fig. 4-2 |
.
What is the particle's displacement from to ?
Solution. The Key Idea is that the displacement is obtained by subtracting the initial position vector from the later position vector . That is most easily done by components:
(Answer)
This
displacement vector is parallel to the
plane,
because it lacks any
component,
a fact that is easier to see in the numerical result than in Fig.
4-2.
Sample Problem 4-2
A rabbit runs across a parking lot on which a set of coordinate axes has been drawn. The coordinates of the rabbit's position as functions of time are given by
(4-5)
And
,
(4-6)
with in seconds and and in meters.
(a) At
s, what is the rabbit's position vector
in unit-vector notation and as a magnitude and an angle?
Solution. The Key Idea here is that the and coordinates of the rabbit's position, as given, are the scalar components of the rabbit's position vector . Thus, we can write
|
rather than
because the
components are functions of
,
and thus
is also.)
At s, the scalar components are
m,
m.
Thus, at
s,
(Answer)
To get the magnitude and angle of , we can use a vector-capable calculator, or we can write
m.
(Answer)
and
(Answer)
|
has the same tangent as - 41°, study of the signs of the components
of
rules out 139°.)
(b) Graph
the rabbit's path for
to
s.
Solution: We can repeat part (a) for several values of and then plot the results. Figure 4-36 shows the plots for five values of and the path connecting them. We can also use a graphing calculator to make a. parametric graph; that is, we would have the calculator plot у versus x, where these coordinates are given by Eqs. 4-5 and 4-6 as functions of time .
4-3 Average Velocity and Instantaneous Velocity
If
a particle moves through a displacement
in a time interval
,
then
its average
velocity
is
Or
(4-8)
This tells us the direction of must be the same as that of the displacement . Using Eq. 4-4, we can write Eq. 4-8 in vector components as
(4-9)
For example, if the particle in Sample Problem 4-1 moves from its initial position to its later position in 2.0 s, then its average velocity during that move is
When we
speak of the velocity
of
a particle, we usually mean the particle's instantaneous
velocity
at some instant. This
is the value that
approaches
in
the limit as we shrink the time interval
to
0
about that instant. Using the language of calculus, we may write
as the derivative
(4-10)
Figure 4-4
shows the path of a particle that is restricted to the
plane.
As the particle travels to the right along the curve, its position
vector sweeps to the right. During
time interval
,
the position vector changes from
to
,
and
the particle's displacement
is
.
Fig. 4-4 |
(when the
particle is at position 1), we shrink interval
to 0 about
.
Three things
happen as we do so: (1) Position vector
in Fig. 4-4 moves
toward
so that
shrinks toward
zero. (2) The direction of
(thus of
)
approaches the direction of the tangent line to the particle's path
at position 1. (3) The average velocity
approaches the instantaneous velocity
at
.
In
the limit as
,
we have
and,
most important here,
takes
on
the direction of the tangent line. Thus,
has that direction as well:
► The direction of the instantaneous velocity of a particle is always tangent to the particle's path at the particle's position.
The result is the same in three dimensions: is always tangent to the particle's path.
To write Eq. 4-10 in unit-vector form, we substitute for from Eq. 4-1:
.
This equation can be simplified somewhat by writing it as
(4-11)
where the scalar components of are
(4-12)
For
example,
is
the scalar component of
along the
axis.
Thus, we can find the scalar components of
by
differentiating the scalar components of
.
|
Sample Problem 4-3
For the rabbit in Sample Problem 4-2, find the velocity at time s, in unit-vector notation and as a magnitude and an angle.
Solution There are two Key Ideas here: (1) We can find the rabbit's velocity by first finding the velocity components. (2) We can find those components by taking derivatives of the components of the rabbit's position vector. Applying the first of Eqs. 4-12 to Eq. 4-5, we find the x component of to be
(4-13)
At
s, this gives
m/s.
Similarly, applying the second of Eqs. 4-12 to Eq. 4-6, we find that the component is
At
s, this gives
m/s.
Equation 4-11 then yields
(Answer)
which is shown in Fig. 4-6, tangent to the rabbit's path and in the direction the rabbit is running at s.
To get the magnitude and angle of , either we use a vector-capable calculator or we follow Eq. 3-6 to write
m/s(Answer)
and
(Answer)
(Although 50° has the same tangent as -130°, inspection of the signs of the velocity components indicates that the desired angle is in the third quadrant, given by 50° - 180° = -130°.)
4-4 Average Acceleration and Instantaneous Acceleration
When a
particle's velocity changes from
to
in
a time interval
,
its
average
acceleration
during
is
Or
If we
shrink
to
zero about some instant, then in the limit
approaches
the instantaneous
acceleration
(or
acceleration)
at
that instant; that is,
(4-16)
If the velocity changes in either magnitude or direction (or both), the particle must have an acceleration.
We can write Eq. 4-16 in unit-vector form by substituting for v*from Eq. 4-11 to obtain
|
(4-17)
where the scalar components of are
(4-18)
We can rewrite this as
(4-17)
where the scalar components of are
,
(4-18)
Sample Problem 4-4
For the rabbit in Sample Problems 4-2 and 4-3, find the acceleration at time s, in unit-vector notation and as a magnitude and an angle.
Solution: There are two Key Ideas here: (1) We can find the rabbit's acceleration by first finding the acceleration components. (2) We can find those components by taking derivatives of the rabbit's velocity components. Applying the first of Eqs. 4-18 to Eq. 4-13, we find the component of to be
m/s2
Similarly, applying the second of Eqs. 4-18 to Eq. 4-14 yields the component as
m/s2
We see that the acceleration does not vary with time (it is a constant) because the time variable does not appear in the expression for either acceleration component. Equation 4-17 then yields
, (Answer)
which is shown superimposed on the rabbit's path in Fig. 4-8.
To get the magnitude and angle of , either we use a vector-capable calculator or we follow Eq. 3-6. For the magnitude we have
m/s2
(Answer)
For the angle we have
For each description, determine whether the x and у components of the puck's acceleration are constant, and whether the acceleration a* is constant.
|
-35° + 180° = 145c
This is consistent with the components of . Note that has the same magnitude and direction throughout the rabbit's run because, as we noted previously, the acceleration is constant.
Exercises
1. The
position of an object moving only an
-axis
is given by
,
where
is in meters and
in seconds. (a) What is the position of the object at
s? (b) What is the displacement’s between
and
s?
(c) What is the average velocity for the time interval from
and
?
2. (a) If
particle’s position is given by
,
(where
is in meters and
in seconds), what its velocity at
?
(b) Is it moving in positive or negative direction of
just then? (c) What is its speed just then? (d) Is the speed larger
or smaller at later time? (e) Is there ever an instant when the
velocity is zero? (f) Is there a time after
when the particle is moving in the negative direction of
?
3.
The position of a particle moving along the
axis
is given in centimeters by
,
where
is
in seconds. Calculate (a)
the average velocity during the time interval
s
to
s;
(b) the instantaneous velocity at
s; (c) the instantaneous velocity at
s; (d) the instantaneous velocity at
s;
and (e) the instantaneous velocity when the particle is midway
between its positions at f
s
and
s. (f) Graph
versus
and
indicate your answers graphically.
4. A proton
moves along the
axis
according to the equation
,
where
is
in meters and
is
in seconds. Calculate (a)
average
velocity
of the proton during the first 3.0 s of its motion.
(b) the
instantaneous
velocity of the proton at
s, (c) the
instantaneous acceleration
of the proton at
s.
5. An
electron moving along the
axis
has a position given by
m. where
is
in seconds. How far is the electron from the
origin when it
momentarily stops?
6.
The position of a particle
moving along the
axis
depends on the
time according to the equation
,
where
is
in meters and
in
seconds, (a) What units must с
and
b
have?
Let there numerical values be 3.0 and 2.0, respectively. (b) At what
time does the particle reach its maximum positive
position? From
and
,
(c)
what distance does the particle move and (d) what is its
displacement? At
s what are (e)
its velocity and (f) its acceleration?
Exercises
The
and
coordinates of a particle at any time
are given by
and
,
where
and
are in meters and
in seconds. The acceleration of the particle at
s is
(A) zero (B) 8 m/ s2 (C) 20 m/s2 (D) 40 m·s2
2-34 A jet-propelled motorcycle starts from rest, moves in a straight line with constant acceleration, and covers a distance of 64 m in 4 s.
What is the final velocity?
How much time was required to cover half the total distance?
What is the distance covered in one-half the total time?
What is the velocity when half the total distance has been covered?
What is the velocity after one-half the total time?
When will the instantaneous velocity equal the average velocitv for the 0-to-4 s time interval?
Q. 2.09. If the displacement of a body is proportional to square of time, state whether the body is moving with uniform velocity or uniform acceleration.
Q. 2.10. If
the distance traveled by a body
in time
is given by
,
then what will be the acceleration of the body?
Q. 2.11. An
object is covering distance in direct proportion to
,
where
is
the time elapsed.
(a) What conclusions might you draw about acceleration? Is it constant? Increasing? Decreasing?
(b) What might you conclude about the force acting on the object?
2.
The position of a particle is given by
,
where
t
is
in seconds and the coefficients have the proper units
for
to be in meters.
(a) Find
and
of the particle.
(b) Find the magnitude and direction of at s.
[Ans.
(a)
;
;
(b)
12.37
m/s; 76°]
Section 2—5 Velocity and Coordinate by Integration
2-20 The motion of a particle along a straight line is described by the function
x = (6 m) + (5 m-s-2)/2 - (1 m-s-4)/4.
Assume that t is positive.
Find the position, velocity, and acceleration at time t = 2 s.
During what time interval is the velocity positive?
During what time interval is x positive?
What is the maximum positive velocity attained by the particle?
2—21 The acceleration of a motorcycle is given by a - (1.2 m-s-3)/ - (0.12 m-s-4)/2. It is at rest at the origin
at time t — 0.
Find its position and velocity as functions of time.
Calculate the maximum velocity it attains.
2-22 The acceleration of a bus is given by a = (2 m-s-3)t.
) If the bus's velocity at time t = 1 s is 5 m-s-1, what is its velocity at time ( = 2s?
b) If the bus's position at time t = 1 s is 6 m, what is its position at time t = 2 s?
56 The
acceleration of an object suspended from a spring and oscillating
vertically is
,
where
is
a constant and
is
the coordinate measured from the equilibrium position. Suppose that
an object moving in this way is given an initial velocity
at
the coordinate
.
Find
the expression for the velocity
of
the object as a function of its coordinate
.
2-57 The motion of an object falling from rest in a resisting medium is described by the equation
where A and В are constants. In terms of A and B, find
the initial acceleration;
the velocity at which the acceleration becomes zero (the terminal velocity).
Show that the velocity at any given t is given by
2-58 After
the engine of a moving motorboat is cut off, the boat has an
acceleration in the opposite direction to its velocity and directly
proportional to the square of its velocity. That is,
,
where
is
constant.
a) Show that the magnitude of the velocity at a time after the engine is cut off is given by
.
b) Show that the distance x traveled in a time is
c) Show that the velocity after traveling a distance is
As a numerical example, suppose the engine is cut off when the velocity is 6 m/s, and that the velocity decreases to 3 m/s in a time of 15 s.
d) Find the numerical value of the constant , and the unit in which it is expressed.
e) Find the acceleration at the instant the engine is cut off.
f) Calculate x, v, and a at 10-s intervals for the first 60 s after the engine is cut off. Sketch graphs of x, v. and a as
functions oi t, tor the first 60 s of the motion.
4-7 Uniform Circular Motion
A particle is in uniform circular motion if it travels around a circle or a circular arc at constant (uniform) speed. Although the speed does not vary, the particle is accelerating. That fact may be surprising because we often think of acceleration (a change in velocity) as an increase or decrease in speed. However, actually velocity is a vector, not a scalar. Thus, even if a velocity changes only in direction, there is still an acceleration, and that is what happens in uniform circular motion.
|
(centripetal
acceleration) (4-32)
where
is
the radius of the circle and
is the speed of the particle.
In
addition, during this acceleration at constant speed, the particle
travels the circumference of the circle (a distance of
)
in
time
(period).
(4-33)
is called
the period
of revolution, or
simply the period,
of
the motion. It is, in general, the time for a particle to go around a
closed path exactly once.
Proof of Eq. 4-32
To find the
magnitude and direction of the acceleration for uniform circular
motion, we consider Fig. 4-19. In Fig. 4-l9a,
particle
p
moves
at constant speed
around a circle of radius
.
At
the instant shown, p
has
coordinates
and
.
|
that
makes with a vertical at p
equals the angle
that radius
makes with the
axis.
The scalar components of are shown in Fig. 4-19b. With them, we can write the velocity as
|
(4-34)
Now, using
the right triangle in Fig. 4-19a, we can replace
with
and
with
to write
(4-35)
To find the acceleration of particle p, we must take the time derivative of this equation. Noting that speed and radius do not change with time, we obtain
(4-36)
Now
note that the rate
at
which
changes
is equal to the velocity component
.
Similarly,
,
and,
again from Fig. 4-19b, we see that
and
.
Making
these substitutions in Eq. 4-36, we find
(4-37)
This vector and its components are shown in Fig. 4-19c. Following Eq. 3-6, we find that the magnitude of is
as
we wanted to prove. To orient
,
we
can find the angle
shown
in Fig. 4-19c:
Thus,
,
which means that
is
directed along the radius
of
Fig. 4-19a
toward
the
circle's center, as we wanted to prove.
Sample Problem 4-9
"Top gun" pilots have long worried about taking a turn too tightly. As a pilot's body undergoes centripetal acceleration, with the head toward the center of curvature, the blood pressure in the brain decreases, leading to loss of brain function.
There
are several warning signs to signal a pilot to ease up: when
the centripetal acceleration is
or
,
the pilot feels heavy. At
about
,
the
pilot's vision switches to black and white and narrows
to "tunnel vision." If that acceleration is sustained or
increased,
vision ceases and, soon after, the pilot is unconscious - a condition
known as
-LOC
for "
-induced
loss of consciousness." What
is the centripetal acceleration, in
units, of a pilot flying an
F-22 at speed
= 2500 km/h (694 m/s) through a circular arc with
radius of curvature
=
5.80 km?
SOLUTION: The Key Idea here is that although the pilot's speed is constant, the circular path requires a (centripetal) acceleration, with magnitude given by Eq. 4-32:
(Answer)
If an unwary pilot caught in a dogfight puts the aircraft into such a tight turn, the pilot goes into -LOC almost immediately, with no warning signs to signal the danger.