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Physics For Scientists And Engineers 6E - part 8

S E C T I O N 2 . 2 • Instantaneous Velocity and Speed

x(m)

t(s)

(a)

–20

–40

–60

(b)

represents the velocity of the car at the moment we started taking data, at point !.
What we have done is determine the instantaneous velocity at that moment. In other
words,

the instantaneous velocity v

equals the limiting value of the ratio !x(!t

as !t approaches zero:

(2.4)

In calculus notation, this limit is called the derivative of x with respect to t, written dx/dt:

(2.5)

The instantaneous velocity can be positive, negative, or zero. When the slope of the
position–time graph is positive, such as at any time during the first 10 s in Figure 2.3,
v

is positive—the car is moving toward larger values of x. After point ", v

is nega-

tive because the slope is negative—the car is moving toward smaller values of x. At
point ", the slope and the instantaneous velocity are zero—the car is momentarily at
rest.

From here on, we use the word velocity to designate instantaneous velocity. When it

is average velocity we are interested in, we shall always use the adjective average.

The

instantaneous speed of a particle is defined as the magnitude of its instan-

taneous velocity. As with average speed, instantaneous speed has no direction
associated with it and hence carries no algebraic sign. For example, if one particle
has an instantaneous velocity of % 25 m/s along a given line and another particle
has an instantaneous velocity of # 25 m/s along the same line, both have a speed

of 25 m/s.

lim

! lim

t : 0

Active Figure 2.3 (a) Graph representing the motion of the car in Figure 2.1. (b) An

enlargement of the upper-left-hand corner of the graph shows how the blue line

between positions ! and " approaches the green tangent line as point " is moved

closer to point !.

At the Active Figures link at http://www.pse6.com, you can move point "

as suggested in (b) and observe the blue line approaching the green tangent

line.

Instantaneous velocity

Note that the displacement $x also approaches zero as $t approaches zero, so that the ratio

looks like 0/0. As $x and $t become smaller and smaller, the ratio $x/$t approaches a value

equal to the slope of the line tangent to the x-versus-t curve.

As with velocity, we drop the adjective for instantaneous speed: “Speed” means instantaneous

speed.

▲

PITFALL PREVENTION

2.3 Instantaneous Speed

and Instantaneous
Velocity

In  Pitfall  Prevention  2.1,  we  ar-
gued  that  the  magnitude  of  the
average velocity is not the average
speed.  Notice  the  difference
when  discussing  instantaneous
values.  The  magnitude  of  the  in-
stantaneous velocity is the instan-
taneous speed. In an infinitesimal
time  interval,  the  magnitude  of
the  displacement  is  equal  to  the
distance  traveled  by  the  particle.

A particle moves along the x axis. Its position varies with
time according to the expression x ! # 4t % 2t

where x is

in meters and t is in seconds.

The position–time graph for

this motion is shown in Figure 2.4. Note that the particle
moves in the negative x direction for the first second of mo-
tion, is momentarily at rest at the moment t ! 1 s, and
moves in the positive x direction at times t & 1 s.

(A)

Determine the displacement of the particle in the time

intervals t ! 0 to t ! 1 s and t ! 1 s to t ! 3 s.

Solution During  the  first  time  interval,  the  slope  is  nega-
tive  and  hence  the  average  velocity  is  negative.  Thus,  we
know  that  the  displacement  between  ! and  " must  be  a
negative number having units of meters. Similarly, we expect
the displacement between " and $ to be positive.

In the first time interval, we set t

0 and

1 s. Using Equation 2.1, with x ! # 4t % 2t

, we

obtain for the displacement between t ! 0 and t ! 1 s,

To calculate the displacement during the second time inter-
val (t ! 1 s to t ! 3 s), we set t

1 s and t

3 s:

These displacements can also be read directly from the posi-
tion–time graph.

(B)

Calculate the average velocity during these two time in-

tervals.

Solution In the first time interval, $t ! t

1 s. Therefore, using Equation 2.2 and the dis-

placement calculated in (a), we find that

2 m/s

)

2 m

1 s

8 m

! [#4(3) % 2(3)

] # [#4(1) % 2(1)

]

2 m

! [#4(1) % 2(1)

] # [#4(0) % 2(0)

]

! x

C H A P T E R 2 • Motion in One Dimension

Conceptual Example 2.2 The Velocity of Different Objects

Example 2.3 Average and Instantaneous Velocity

Consider the following one-dimensional motions:

(A)

A ball

thrown directly upward rises to a highest point and falls
back into the thrower’s hand.

(B)

A race car starts from rest

and speeds up to 100 m/s.

(C)

A spacecraft drifts through

space at constant velocity. Are there any points in the mo-
tion of these objects at which the instantaneous velocity has
the same value as the average velocity over the entire mo-
tion? If so, identify the point(s).

Solution (A)  The  average  velocity  for  the  thrown  ball  is
zero  because  the  ball  returns  to  the  starting  point;  thus  its
displacement  is  zero.  (Remember  that  average  velocity  is
defined as $x/$t.) There is one point at which the instanta-
neous velocity is zero—at the top of the motion.

(B)  The  car’s  average  velocity  cannot  be  evaluated  unam-
biguously with the information given, but it must be some
value  between  0  and  100 m/s.  Because  the  car  will  have
every  instantaneous  velocity  between  0  and  100 m/s  at
some time during the interval, there must be some instant
at which the instantaneous velocity is equal to the average
velocity.

(C) Because the spacecraft’s instantaneous velocity is con-
stant, its instantaneous velocity at any time and its average
velocity over any time interval are the same.

–2

–4

t(s)

x(m)

Slope = 4 m/s

Slope = –2 m/s

Figure 2.4 (Example 2.3) Position–time graph for a particle

having an x coordinate that varies in time according to the

expression x ! # 4t % 2t

In the second time interval, $t ! 2 s; therefore,

These values are the same as the slopes of the lines joining
these points in Figure 2.4.

(C)

Find the instantaneous velocity of the particle at t ! 2.5 s.

Solution We can guess that this instantaneous velocity must
be of the same order of magnitude as our previous results,
that is, a few meters per second. By measuring the slope of
the green line at t ! 2.5 s in Figure 2.4, we find that

6 m/s

4 m/s

)

8 m

2 s

Simply to make it easier to read, we write the expression as
x ! # 4t % 2t

rather than as x ! (# 4.00 m/s)t % (2.00 m/s

2.00

When  an  equation  summarizes  measurements,  consider  its  coeffi-
cients  to  have  as  many  significant  digits  as  other  data  quoted  in  a
problem. Consider its coefficients to have the units required for di-
mensional consistency. When we start our clocks at t ! 0, we usually
do  not  mean  to  limit  the  precision  to  a  single  digit.  Consider  any
zero  value  in  this  book  to  have  as  many  significant  figures  as  you
need.

∆t

∆v

∆t

(b)

(a)

v = v

–

S E C T I O N 2 . 3 • Acceleration

Figure 2.5 (a) A car, modeled as a particle, moving along the x

axis from ! to " has velocity v

at t ! t

and velocity v

at t ! t

(b) Velocity–time graph (rust) for the particle moving in a

straight line. The slope of the blue straight line connecting !

and " is the average acceleration in the time interval
$

t ! t

The average acceleration a–

of the particle is defined as the change in velocity

divided by the time interval $t during which that change occurs:

2.3 Acceleration

In the last example, we worked with a situation in which the velocity of a particle
changes while the particle is moving. This is an extremely common occurrence. (How
constant is your velocity as you ride a city bus or drive on city streets?) It is possible to
quantify changes in velocity as a function of time similarly to the way in which we quan-
tify changes in position as a function of time. When the velocity of a particle changes
with time, the particle is said to be accelerating. For example, the magnitude of the
velocity of a car increases when you step on the gas and decreases when you apply the
brakes. Let us see how to quantify acceleration.

Suppose an object that can be modeled as a particle moving along the x axis has an

initial velocity v

at time t

and a final velocity v

at time t

, as in Figure 2.5a.

Average acceleration

(2.6)

As with velocity, when the motion being analyzed is one-dimensional, we can use

positive and negative signs to indicate the direction of the acceleration. Because the di-
mensions of velocity are L/T and the dimension of time is T, acceleration has dimen-
sions of length divided by time squared, or L/T

. The SI unit of acceleration is meters

per second squared (m/s

). It might be easier to interpret these units if you think of

them as meters per second per second. For example, suppose an object has an acceler-
ation of % 2 m/s

. You should form a mental image of the object having a velocity that

is along a straight line and is increasing by 2 m/s during every interval of 1 s. If the ob-
ject starts from rest, you should be able to picture it moving at a velocity of % 2 m/s af-
ter 1 s, at % 4 m/s after 2 s, and so on.

In some situations, the value of the average acceleration may be different over

different time intervals. It is therefore useful to define the instantaneous acceleration
as the limit of the average acceleration as $t approaches zero. This concept is analo-
gous to the definition of instantaneous velocity discussed in the previous section. If
we imagine that point ! is brought closer and closer to point " in Figure 2.5a and
we take the limit of $v

/$t as $t approaches zero, we obtain the instantaneous

acceleration:

(2.7)

lim

Instantaneous acceleration

That is,

the instantaneous acceleration equals the derivative of the velocity

with respect to time, which by definition is the slope of the velocity–time graph.
The slope of the green line in Figure 2.5b is equal to the instantaneous acceleration
at point ". Thus, we see that just as the velocity of a moving particle is the slope at a
point on the particle’s x -t graph, the acceleration of a particle is the slope at a point
on the particle’s v

-t graph. One can interpret the derivative of the velocity with re-

spect to time as the time rate of change of velocity. If a

is positive, the acceleration

is in the positive x direction; if a

is negative, the acceleration is in the negative x

direction.

For the case of motion in a straight line, the direction of the velocity of an object

and the direction of its acceleration are related as follows.

When the object’s velocity

and acceleration are in the same direction, the object is speeding up. On the
other hand, when the object’s velocity and acceleration are in opposite direc-
tions, the object is slowing down.

To help with this discussion of the signs of velocity and acceleration, we can relate

the acceleration of an object to the force exerted on the object. In Chapter 5 we for-
mally establish that

force is proportional to acceleration:

This proportionality indicates that acceleration is caused by force. Furthermore, force
and acceleration are both vectors and the vectors act in the same direction. Thus, let
us think about the signs of velocity and acceleration by imagining a force applied to an
object and causing it to accelerate. Let us assume that the velocity and acceleration are
in the same direction. This situation corresponds to an object moving in some direc-
tion that experiences a force acting in the same direction. In this case, the object
speeds up! Now suppose the velocity and acceleration are in opposite directions. In
this situation, the object moves in some direction and experiences a force acting in the
opposite direction. Thus, the object slows down! It is very useful to equate the direc-
tion of the acceleration to the direction of a force, because it is easier from our every-
day experience to think about what effect a force will have on an object than to think
only in terms of the direction of the acceleration.

F ) a

C H A P T E R 2 • Motion in One Dimension

Quick Quiz 2.2

If a car is traveling eastward and slowing down, what is the

direction of the force on the car that causes it to slow down? (a) eastward (b) westward
(c) neither of these.

(b)

(a)

Figure 2.6 The instantaneous

acceleration can be obtained from

the velocity–time graph (a). At

each instant, the acceleration in

the a

versus t graph (b) equals the

slope of the line tangent to the v

versus t curve (a).

▲

PITFALL PREVENTION

2.4 Negative

Acceleration

Keep in mind that negative acceler-
ation does not necessarily mean that
an object is slowing down. If the ac-
celeration is negative, and the ve-
locity is negative, the object is
speeding up!

▲

PITFALL PREVENTION

2.5 Deceleration

The word deceleration has the com-
mon popular connotation of slow-
ing down. We will not use this word
in this text, because it further con-
fuses the definition we have given
for negative acceleration.

From now on we shall use the term acceleration to mean instantaneous acceleration.

When we mean average acceleration, we shall always use the adjective average.

Because v

dx/dt, the acceleration can also be written

(2.8)

That is, in one-dimensional motion, the acceleration equals the second derivative of x
with respect to time.

Figure 2.6 illustrates how an acceleration–time graph is related to a velocity–time

graph. The acceleration at any time is the slope of the velocity–time graph at that time.
Positive values of acceleration correspond to those points in Figure 2.6a where the ve-
locity is increasing in the positive x direction. The acceleration reaches a maximum at
time t

, when the slope of the velocity–time graph is a maximum. The acceleration

then goes to zero at time t

, when the velocity is a maximum (that is, when the slope of

the v

-t graph is zero). The acceleration is negative when the velocity is decreasing in

the positive x direction, and it reaches its most negative value at time t

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