| 2.1� |
Displacement |
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There are two aspects to any motion. In a purely descriptive sense, there is the movement itself. Is it rapid or slow, for instance? Then, there is the issue of what causes the motion or what changes it, which requires that forces be considered. Kinematics deals with the concepts that are needed to describe motion, without any reference to forces. The present chapter discusses these concepts as they apply to motion in one dimension, and the next chapter treats two-dimensional motion. Dynamics deals with the effect that forces have on motion, a topic that is considered in Chapter 4. Together, kinematics and dynamics form the branch of physics known as mechanics. We turn now to the first of the kinematics concepts to be discussed, which is displacement.
To describe the motion of an object, we must be able to specify the location of the object at all times, and Figure 2.1 shows how to do this for one-dimensional motion. In this drawing, the initial position of a car is indicated by the vector labeled
. The length of
is the distance of the car from an arbitrarily chosen origin. At a later time the car has moved to a new position, which is indicated by the vector
. The
displacement of the car
(read as “delta x” or “the change in x”) is a vector drawn from the initial position to the final position. Displacement is a vector quantity in the sense discussed in Section 1.5, for it conveys both a magnitude (the distance between the initial and final positions) and a direction. The displacement can be related to
and
by noting from the drawing that
Thus, the displacement
is the difference between
and
, and the Greek letter delta
is used to signify this difference. It is important to note that the change in any variable is always the final value minus the initial value.
Definition of Displacement
The displacement is a vector that points from an object's initial position to its final position and has a magnitude that equals the shortest distance between the two positions.
SI Unit of Displacement: meter (m)
The SI unit for displacement is the meter (m), but there are other units as well, such as the centimeter and the inch. When converting between centimeters (cm) and inches (in.), remember that
.
Often, we will deal with motion along a straight line. In such a case, a displacement in one direction along the line is assigned a positive value, and a displacement in the opposite direction is assigned a negative value. For instance, assume that a car is moving along an east/west direction and that a positive
sign is used to denote a direction due east. Then,
represents a displacement that points to the east and has a magnitude of 500 meters. Conversely,
is a displacement that has the same magnitude but points in the opposite direction, due west.
The magnitude of the displacement vector is the shortest distance between the initial and final positions of the object. However, this does not mean that displacement and distance are the same physical quantities. In Figure 2.1, for example, the car could reach its final position after going forward and backing up several times. In that case, the total distance traveled by the car would be greater than the magnitude of the displacement vector.
| Check Your Understanding� |
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| 1.�� |
A honeybee leaves the hive and travels a total distance of  before returning to the hive. What is the magnitude of the displacement vector of the bee? Answer:
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| 2.2� |
Speed and Velocity |
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Average Speed
One of the most obvious features of an object in motion is how fast it is moving. If a car travels 200 meters in 10 seconds, we say its average speed is 20 meters per second, the
average speed being the distance traveled divided by the time required to cover the distance:
Equation 2.1 indicates that the unit for average speed is the unit for distance divided by the unit for time, or meters per second (m/s) in SI units. Example 1 illustrates how the idea of average speed is used.
| Example� |
1� |
Distance Run by a Jogger |
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How far does a jogger run in 1.5 hours  if his average speed is  ?
Reasoning
The average speed of the jogger is the average distance per second that he travels. Thus, the distance covered by the jogger is equal to the average distance per second (his average speed) multiplied by the number of seconds (the elapsed time) that he runs.
Solution
To find the distance run, we rewrite Equation 2.1 as
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Speed is a useful idea, because it indicates how fast an object is moving. However, speed does not reveal anything about the direction of the motion. To describe both how fast an object moves and the direction of its motion, we need the vector concept of velocity.
Average Velocity
To define the velocity of an object, we will use two concepts that we have already encountered, displacement and time. The building of new concepts from more basic ones is a common theme in physics. In fact the great strength of physics as a science is that it builds a coherent understanding of nature through the development of interrelated concepts.
Suppose that the initial position of the car in Figure 2.1 is
when the time is
. A little later that car arrives at the final position
at the time
t. The difference between these times is the time required for the car to travel between the two positions. We denote this difference by the shorthand notation
(read as “delta
t”), where
represents the final time
t minus the initial time
:
Note that
is defined in a manner analogous to
, which is the final position minus the initial position
. Dividing the displacement
of the car by the elapsed time
gives the
average velocity of the car. It is customary to denote the average value of a quantity by placing a horizontal bar above the symbol representing the quantity. The average velocity, then, is written as
, as specified in Equation 2.2:
Definition of Average Velocity
SI Unit of Average Velocity: meter per second (m/s)
Equation 2.2 indicates that the unit for average velocity is the unit for length divided by the unit for time, or meters per second (m/s) in SI units. Velocity can also be expressed in other units, such as kilometers per hour (km/h) or miles per hour (mi/h).
Average velocity is a vector that points in the same direction as the displacement in Equation 2.2. Figure 2.2 illustrates that the velocity of an object confined to move along a line can point either in one direction or in the opposite direction. As with displacement, we will use plus and minus signs to indicate the two possible directions. If the displacement points in the positive direction, the average velocity is positive. Conversely, if the displacement points in the negative direction, the average velocity is negative. Example 2 illustrates these features of average velocity.
| Example� |
2� |
The World's Fastest Jet-Engine Car |
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Andy Green in the car ThrustSSC set a world record of  in 1997. The car was powered by two jet engines, and it was the first one officially to exceed the speed of sound. To establish such a record, the driver makes two runs through the course, one in each direction, to nullify wind effects. Figure 2.3 a shows that the car first travels from left to right and covers a distance of  (1 mile) in a time of  . Figure 2.3 b shows that in the reverse direction, the car covers the same distance in  . From these data, determine the average velocity for each run.
Reasoning
Average velocity is defined as the displacement divided by the elapsed time. In using this definition we recognize that the displacement is not the same as the distance traveled. Displacement takes the direction of the motion into account, and distance does not. During both runs, the car covers the same distance of . However, for the first run the displacement is  , while for the second it is  . The plus and minus signs are essential, because the first run is to the right, which is the positive direction, and the second run is in the opposite or negative direction.
Solution
According to Equation 2.2, the average velocities are
In these answers the algebraic signs convey the directions of the velocity vectors. In particular, for run 2 the minus sign indicates that the average velocity, like the displacement, points to the left in Figure 2.3 b. The magnitudes of the velocities are 339.5 and  . The average of these numbers is  , and this is recorded in the record book.
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Instantaneous Velocity
Suppose the magnitude of your average velocity for a long trip was
. This value, being an average, does not convey any information about how fast you were moving or the direction of the motion at any instant during the trip. Both can change from one instant to another. Surely there were times when your car traveled faster than
and times when it traveled more slowly. The
instantaneous velocity 
of the car indicates how fast the car moves and the direction of the motion at each instant of time. The magnitude of the instantaneous velocity is called the
instantaneous speed, and it is the number (with units) indicated by the speedometer.
The instantaneous velocity at any point during a trip can be obtained by measuring the time interval
for the car to travel a
very small displacement
. We can then compute the average velocity over this interval. If the time
is small enough, the instantaneous velocity does not change much during the measurement. Then, the instantaneous velocity
at the point of interest is approximately equal to
the average velocity
computed over the interval, or
(for sufficiently small
). In fact, in the limit that
becomes infinitesimally small, the instantaneous velocity and the average velocity become equal, so that
The notation
means that the ratio
is defined by a limiting process in which smaller and smaller values of
are used, so small that they approach zero. As smaller values of
are used,
also becomes smaller. However, the ratio
does
not become zero but, rather, approaches the value of the instantaneous velocity. For brevity, we will use the word
velocity to mean “instantaneous velocity” and
speed to mean “instantaneous speed.”
| Check Your Understanding� |
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| 2.�� |
Is the average speed of a vehicle a vector or a scalar quantity? Answer:
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| 3.�� |
Two buses depart from Chicago, one going to New York and one to San Francisco. Each bus travels at a speed of  . Do they have equal velocities? Answer:
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| 4.�� |
One of the following statements is incorrect.
The car traveled around the circular track at a constant velocity. The car traveled around the circular track at a constant speed. Which statement is incorrect?
Answer:
| a. |
The car traveled around the circular track at a constant velocity. |
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| 5.�� |
A straight track is  in length. A runner begins at the starting line, runs due east for the full length of the track, turns around and runs halfway back. The time for this run is five minutes. What is the runner's average velocity, and what is his average speed? Answer:
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| 6.�� |
The average velocity for a trip has a positive value. Is it possible for the instantaneous velocity at a point during the trip to have a negative value? Answer:
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| 2.3� |
Acceleration |
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In a wide range of motions, the velocity changes from moment to moment. To describe the manner in which it changes, the concept of acceleration is needed. The velocity of a moving object may change in a number of ways. For example, it may increase, as it does when the driver of a car steps on the gas pedal to pass the car ahead. Or it may decrease, as it does when the driver applies the brakes to stop at a red light. In either case, the change in velocity may occur over a short or a long time interval. To describe how the velocity of an object changes during a given time interval, we now introduce the new idea of acceleration. This idea depends on two concepts that we have previously encountered, velocity and time. Specifically, the notion of acceleration emerges when the change in the velocity is combined with the time during which the change occurs.
The meaning of
average acceleration can be illustrated by considering a plane during takeoff. Figure 2.4 focuses attention on how the plane's velocity changes along the runway. During an elapsed time interval
, the velocity changes from an initial value of
to a final velocity of
. The change
in the plane's velocity is its final velocity minus its initial velocity, so that
. The average acceleration
is defined in the following manner, to provide a measure of how much the velocity changes per unit of elapsed time.
Definition of Average Acceleration
SI Unit of Average Acceleration: meter per second squared
The average acceleration
is a vector that points in the same direction as
, the change in the velocity. Following the usual custom, plus and minus signs indicate the two possible directions for the acceleration vector when the motion is along a straight line.
We are often interested in an object's acceleration at a particular instant of time. The
instantaneous acceleration 
can be defined by analogy with the procedure used in Section 2.2 for instantaneous velocity:
Equation 2.5 indicates that the instantaneous acceleration is a limiting case of the average acceleration. When the time interval
for measuring the acceleration becomes extremely small (approaching zero in the limit), the average acceleration and the instantaneous acceleration become equal. Moreover, in many situations the acceleration is constant, so the acceleration has the same value at any instant of time. In the future, we will use the word
acceleration to mean “instantaneous acceleration.” Example 3 deals with the acceleration of a plane during takeoff.
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As this sprinter explodes out of the starting block, his velocity is changing, which means that he is accelerating.�(� Jim Cummins/Taxi/Getty Images, Inc.)
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| Example� |
3� |
Acceleration and Increasing Velocity |
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Suppose the plane in Figure 2.4 starts from rest  when  . The plane accelerates down the runway and at  attains a velocity of  , where the plus sign indicates that the velocity points to the right. Determine the average acceleration of the plane.
Reasoning
The average acceleration of the plane is defined as the change in its velocity divided by the elapsed time. The change in the plane's velocity is its final velocity minus its initial velocity , or  . The elapsed time is the final time t minus the initial time , or  .
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| Problem-Solving Insight.� |
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The change in any variable is the final value minus the initial value: for example, the change in velocity is , and the change in time is  . |
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Solution
The average acceleration is expressed by Equation 2.4 as
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The average acceleration calculated in Example 3 is read as “nine kilometers per hour per second.” Assuming the acceleration of the plane is constant, a value of
means the velocity changes by
during each second of the motion. During the first second, the velocity increases from 0 to
; during the next second, the velocity increases by another
to
, and so on. Figure 2.5 illustrates how the velocity changes during the first two seconds. By the end of the 29th second, the velocity is
.
It is customary to express the units for acceleration solely in terms of SI units. One way to obtain SI units for the acceleration in Example 3 is to convert the velocity units from km/h to m/s:
The average acceleration then becomes
where we have used
. An acceleration of
is read as “2.5 meters per second per second” (or “2.5 meters per second squared”) and means that the velocity changes by
during each second of the motion.
Example 4 deals with a case where the motion becomes slower as time passes.
| Example� |
4� |
Acceleration and Decreasing Velocity |
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A drag racer crosses the finish line, and the driver deploys a parachute and applies the brakes to slow down, as Figure 2.6 illustrates. The driver begins slowing down when  and the car's velocity is  . When  , the velocity has been reduced to  . What is the average acceleration of the dragster?
Reasoning
The average acceleration of an object is always specified as its change in velocity, , divided by the elapsed time, . This is true whether the final velocity is less than the initial velocity or greater than the initial velocity.
Solution
The average acceleration is, according to Equation 2.4,
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| Problem-Solving Insight.� |
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Whenever the acceleration and velocity vectors have opposite directions, the object slows down and is said to be “decelerating.” |
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Figure 2.7 shows how the velocity of the dragster in Example 4 changes during the braking, assuming that the acceleration is constant throughout the motion. The acceleration calculated in Example 4 is negative, indicating that the acceleration points to the left in the drawing. As a result, the acceleration and the velocity point in opposite directions. In contrast, the acceleration and velocity vectors in Figure 2.5 point in the same direction, and the object speeds up.
| Check Your Understanding� |
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| 7.�� |
At one instant of time, a car and a truck are traveling side by side in adjacent lanes of a highway. The car has a greater velocity than the truck has. Does the car necessarily have the greater acceleration? Answer:
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| 8.�� |
Two cars are moving in the same direction (the positive direction) on a straight road. The acceleration of each car also points in the positive direction. Car 1 has a greater acceleration than car 2 has. Which one of the following statements is true?
The velocity of car 1 is always greater than the velocity of car 2. The velocity of car 2 is always greater than the velocity of car 1. In the same time interval, the velocity of car 1 changes by a greater amount than the velocity of car 2 does. In the same time interval, the velocity of car 2 changes by a greater amount than the velocity of car 1 does.
Answer:
| c. |
In the same time interval, the velocity of car 1 changes by a greater amount than the velocity of car 2 does. |
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| 9.�� |
An object moving with a constant acceleration slows down if the acceleration points in the direction opposite to the direction of the velocity. But can an object ever come to a permanent halt if its acceleration truly remains constant? Answer:
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| 10.�� |
A runner runs half the remaining distance to the finish line every ten seconds. She runs in a straight line and does not ever reverse her direction. Does her acceleration have a constant magnitude? Answer:
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As this bald eagle comes in for a landing, it is slowing down. Therefore, it has an acceleration vector that points opposite to its velocity vector.�(� Science Faction/SuperStock)
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| 2.4� |
Equations of Kinematics for Constant Acceleration |
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It is now possible to describe the motion of an object traveling with a constant acceleration along a straight line. To do so, we will use a set of equations known as the equations of kinematics for constant acceleration. These equations entail no new concepts, because they will be obtained by combining the familiar ideas of displacement, velocity, and acceleration. However, they will provide a very convenient way to determine certain aspects of the motion, such as the final position and velocity of a moving object.
In discussing the equations of kinematics, it will be convenient to assume that the object is located at the origin
when
. With this assumption, the displacement
becomes
. Furthermore, in the equations that follow, as is customary, we dispense with the use of boldface symbols overdrawn with small arrows for the displacement, velocity, and acceleration vectors. We will, however, continue to convey the directions of these vectors with plus or minus signs.
Consider an object that has an initial velocity of
at time
and moves for a time
t with a constant acceleration
a. For a complete description of the motion, it is also necessary to know the final velocity and displacement at time
t. The final velocity
v can be obtained directly from Equation 2.4:
The displacement
x at time
t can be obtained from Equation 2.2, if a value for the average velocity
can be obtained. Considering the assumption that
at
, we have
Because the acceleration is constant, the velocity increases at a constant rate. Thus, the average velocity
is midway between the initial and final velocities:
Equation 2.6, like Equation 2.4, applies only if the acceleration is constant and cannot be used when the acceleration is changing. The displacement at time
t can now be determined as
Notice in Equations 2.4
and 2.7
that there are five kinematic variables:
Each of the two equations contains four of these variables, so if three of them are known, the fourth variable can always be found. Example 5 illustrates how Equations 2.4 and 2.7 are used to describe the motion of an object.
| Analyzing Multiple-Concept Problems� |
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| Example� |
5� |
The Displacement of a Speedboat |
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The speedboat in Figure 2.8 has a constant acceleration of  . If the initial velocity of the boat is  , find the boat's displacement after 8.0 seconds.
Reasoning
As the speedboat accelerates, its velocity is changing. The displacement of the speedboat during a given time interval is equal to the product of its average velocity during that interval and the time. The average velocity, on the other hand, is just one-half the sum of the boat's initial and final velocities. To obtain the final velocity, we will use the definition of constant acceleration, as given in Equation 2.4.
Knowns and Unknowns The numerical values for the three known variables are listed in the table:
Description |
Symbol |
Value |
Comment |
Acceleration |
a |
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Positive, because the boat is accelerating to the right, which is the positive direction. See Figure 2.8b. |
Initial velocity |
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Positive, because the boat is moving to the right, which is the positive direction. See Figure 2.8b. |
Time interval |
t |
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Unknown Variable |
� |
� |
� |
Displacement of boat |
x |
? |
� |
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Modeling the Problem
 Displacement Since the acceleration is constant, the displacement x of the boat is given by Equation 2.7, in which and v are the initial and final velocities, respectively. In this equation, two of the variables, and t, are known (see the Knowns and Unknowns table), but the final velocity v is not. However, the final velocity can be determined by employing the definition of acceleration, as Step 2 shows. |
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 Acceleration According to Equation 2.4, which is just the definition of constant acceleration, the final velocity v of the boat is
All the variables on the right side of the equals sign are known, and we can substitute this relation for v into Equation 2.7, as shown at the right. |
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Solution
Algebraically combining the results of Steps 1 and 2, we find that
The displacement of the boat after  is
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In Example 5, we combined two equations
into a single equation by algebraically eliminating the final velocity
v of the speedboat (which was not known). The result was the following expression for the displacement
x of the speedboat:
The first term
on the right side of this equation represents the displacement that would result if the acceleration were zero and the velocity remained constant at its initial value of
. The second term
gives the additional displacement that arises because the velocity changes (
a is not zero) to values that are different from its initial value. We now turn to another example of accelerated motion.
| Analyzing Multiple-Concept Problems� |
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| Example� |
6� |
The Physics of Catapulting a Jet |
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A jet is taking off from the deck of an aircraft carrier, as Figure 2.9 shows. Starting from rest, the jet is catapulted with a constant acceleration of  along a straight line and reaches a velocity of  . Find the displacement of the jet.
Reasoning
When the jet is accelerating, its velocity is changing. The displacement of the jet during a given time interval is equal to the product of its average velocity during that interval and the time, the average velocity being equal to one-half the sum of the jet's initial and final velocities. The initial and final velocities are known, but the time is not. However, the time can be determined from a knowledge of the jet's acceleration.
Knowns and Unknowns The data for this problem are listed below:
Description |
Symbol |
Value |
Comment |
Explicit Data |
� |
� |
� |
Acceleration |
a |
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Positive, because the acceleration points in the positive direction. See Figure 2.9b. |
Final velocity |
v |
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Positive, because the final velocity points in the positive direction. See Figure 2.9b. |
Implicit Data |
� |
� |
� |
Initial velocity |
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The jet starts from rest. |
Unknown Variable |
� |
� |
� |
Displacement of jet |
x |
? |
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| Problem-Solving Insight.� |
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Implicit data are important. For instance, in Example 6 the phrase “starting from rest” means that the initial velocity is zero  . |
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Modeling the Problem
Displacement The displacement x of the jet is given by Equation 2.7, since the acceleration is constant during launch. Referring to the Knowns and Unknowns table, we see that the initial and final velocities, and v, in this relation are known. The unknown time t can be determined by using the definition of acceleration (see Step 2). |
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Acceleration The acceleration of an object is defined by Equation 2.4 as the change in its velocity,  , divided by the elapsed time t:
Solving for t yields
All the variables on the right side of the equals sign are known, and we can substitute this result for t into Equation 2.7, as shown at the right. |
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Solution
Algebraically combining the results of Steps 1 and 2, we find that
The displacement of the jet is
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In Example 6 we were able to determine the displacement
x of the jet from a knowledge of its acceleration
a and its initial and final velocities,
and
v. The result was
. Solving for
gives
Equation 2.9 is often used when the time involved in the motion is unknown.
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Table�2.1��� |
Equations of Kinematics for Constant Acceleration |
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Variables |
Equation Number |
Equation |
x |
a |
v |
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t |
2.4 |
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— |
✓ |
✓ |
✓ |
✓ |
2.7 |
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✓ |
— |
✓ |
✓ |
✓ |
2.8 |
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✓ |
✓ |
— |
✓ |
✓ |
2.9 |
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✓ |
✓ |
✓ |
✓ |
— |
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Table 2.1 presents a summary of the equations that we have been considering. These equations are called the equations of kinematics. Each equation contains four of the five kinematic variables, as indicated by the check marks (✓) in the table. The next section shows how to apply the equations of kinematics.
| Check Your Understanding� |
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| 11.�� |
The muzzle velocity of a gun is the velocity of the bullet when it leaves the barrel. The muzzle velocity of one rifle with a short barrel is greater than the muzzle velocity of another rifle that has a longer barrel. In which rifle is the acceleration of the bullet larger? Answer:
the rifle with the short barrel |
| 12.�� |
A motorcycle starts from rest and has a constant acceleration. In a time interval t, it undergoes a displacement x and attains a final velocity v. Then t is increased so that the displacement is  . In this same increased time interval, what final velocity does the motorcycle attain? Answer:
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| 2.5� |
Applications of the Equations of Kinematics |
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The equations of kinematics can be applied to any moving object, as long as the acceleration of the object is constant. However, remember that each equation contains four variables. Therefore, numerical values for three of the four must be available if an equation is to be used to calculate the value of the remaining variable. To avoid errors when using these equations, it helps to follow a few sensible guidelines and to be alert for a few situations that can arise during your calculations.
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| Problem-Solving Insight.� |
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Decide at the start which directions are to be called positive  and negative  relative to a conveniently chosen coordinate origin.
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This decision is arbitrary, but important because displacement, velocity, and acceleration are vectors, and their directions must always be taken into account. In the examples that follow, the positive and negative directions will be shown in the drawings that accompany the problems. It does not matter which direction is chosen to be positive. However, once the choice is made, it should not be changed during the course of the calculation.
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| Problem-Solving Insight.� |
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As you reason through a problem before attempting to solve it, be sure to interpret the terms “decelerating” or “deceleration” correctly, should they occur in the problem statement. |
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These terms are the source of frequent confusion, and Conceptual Example 7 offers help in understanding them.
| Conceptual Example� |
7� |
Deceleration Versus Negative Acceleration |
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A car is traveling along a straight road and is decelerating. Which one of the following statements correctly describes the car's acceleration?
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(a)�� |
It must be positive. |
(b)�� |
It must be negative. |
(c)�� |
It could be positive or negative. |
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Reasoning
The term “decelerating” means that the acceleration vector points opposite to the velocity vector and indicates that the car is slowing down. One possibility is that the velocity vector of the car points to the right, in the positive direction, as Figure 2.10a shows. The term “decelerating” implies, then, that the acceleration vector of the car points to the left, which is the negative direction. Another possibility is that the car is traveling to the left, as in Figure 2.10b. Now, since the velocity vector points to the left, the acceleration vector would point opposite, or to the right, which is the positive direction.
Answers (a) and (b) are incorrect. The term “decelerating” means only that the acceleration vector points opposite to the velocity vector. It is not specified whether the velocity vector of the car points in the positive or negative direction. Therefore, it is not possible to know whether the acceleration is positive or negative.
Answer (c) is correct. As shown in Figure 2.10, the acceleration vector of the car could point in the positive or the negative direction, so that the acceleration could be either positive or negative, depending on the direction in which the car is moving.
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| Problem-Solving Insight.� |
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Sometimes there are two possible answers to a kinematics problem, each answer corresponding to a different situation. |
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Example 8 discusses one such case.
| Example� |
8� |
The Physics of Spacecraft Retrorockets |
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The spacecraft shown in Figure 2.11 a is traveling with a velocity of  . Suddenly the retrorockets are fired, and the spacecraft begins to slow down with an acceleration whose magnitude is  . What is the velocity of the spacecraft when the displacement of the craft is  , relative to the point where the retrorockets began firing?
Reasoning
Since the spacecraft is slowing down, the acceleration must be opposite to the velocity. The velocity points to the right in the drawing, so the acceleration points to the left, in the negative direction; thus,  . The three known variables are listed as follows:
Spacecraft Data |
x |
a |
v |
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t |
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The final velocity v of the spacecraft can be calculated using Equation 2.9, since it contains the four pertinent variables.
Solution
From Equation 2.9  , we find that
Both of these answers correspond to the same displacement  , but each arises in a different part of the motion. The answer  corresponds to the situation in Figure 2.11 a, where the spacecraft has slowed to a speed of  , but is still traveling to the right. The answer  arises because the retrorockets eventually bring the spacecraft to a momentary halt and cause it to reverse its direction. Then it moves to the left, and its speed increases due to the continually firing rockets. After a time, the velocity of the craft becomes , giving rise to the situation in Figure 2.11 b. In both parts of the drawing the spacecraft has the same displacement, but a greater travel time is required in part b compared to part a.
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| Problem-Solving Insight.� |
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The motion of two objects may be interrelated, so that they share a common variable. The fact that the motions are interrelated is an important piece of information. In such cases, data for only two variables need be specified for each object. |
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| Problem-Solving Insight.� |
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Often the motion of an object is divided into segments, each with a different acceleration. When solving such problems, it is important to realize that the final velocity for one segment is the initial velocity for the next segment, |
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as Example 9 illustrates.
| Analyzing Multiple-Concept Problems� |
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| Example� |
9� |
A Motorcycle Ride |
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A motorcycle ride consists of two segments, as shown in Figure 2.12 a. During segment 1, the motorcycle starts from rest, has an acceleration of  , and has a displacement of  . Immediately after segment 1, the motorcycle enters segment 2 and begins slowing down with an acceleration of  until its velocity is  . What is the displacement of the motorcycle during segment 2?
Reasoning
We can use an equation of kinematics from Table 2.1 to find the displacement  for segment 2. To do this, it will be necessary to have values for three of the variables that appear in the equation. Values for the acceleration  and final velocity  are given. A value for a third variable, the initial velocity  , can be obtained by noting that it is also the final velocity of segment 1. The final velocity of segment 1 can be found by using an appropriate equation of kinematics, since three variables (  ,  , and  ) are known for this part of the motion, as the following table reveals.
Knowns and Unknowns The data for this problem are listed in the table:
Description |
Symbol |
Value |
Comment |
Explicit Data |
� |
� |
� |
Displacement for segment 1 |
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� |
Acceleration for segment 1 |
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Positive, because the motorcycle moves in the  direction and speeds up. |
Acceleration for segment 2 |
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Negative, because the motorcycle moves in the direction and slows down. |
Final velocity for segment 2 |
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� |
Implicit Data |
� |
� |
� |
Initial velocity for segment 1 |
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The motorcycle starts from rest. |
Unknown Variable |
� |
� |
� |
Displacement for segment 2 |
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� |
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Modeling the Problem
Displacement During Segment 2 Figure 2.12 b shows the situation during segment 2. Two of the variables—the final velocity  and the acceleration  —are known, and for convenience we choose Equation 2.9 to find the displacement of the motorcycle:
Solving this relation for yields Equation 1 at the right. Although the initial velocity of segment 2 is not known, we will be able to determine its value from a knowledge of the motion during segment 1, as outlined in Steps 2 and 3. |
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Initial Velocity of Segment 2 Since the motorcycle enters segment 2 immediately after leaving segment 1, the initial velocity of segment 2 is equal to the final velocity  of segment 1, or  . Squaring both sides of this equation gives
This result can be substituted into Equation 1 as shown at the right. In the next step will be determined. |
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 Final Velocity of Segment 1 Figure 2.12 c shows the motorcycle during segment 1. Since we know the initial velocity , the acceleration , and the displacement , we can employ Equation 2.9 to find the final velocity at the end of segment 1:
This relation for  can be substituted into Equation 2, as shown at the right. |
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Solution
Algebraically combining the results of each step, we find that
The displacement of the motorcycle during segment 2 is
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Figure 2.12 a shows which direction has been chosen as the positive direction, and the choice leads to a value of  for the displacement of the motorcycle during segment 2. This answer means that displacement is  in the positive direction, which is upward and to the right in the drawing. The choice for the positive direction is completely arbitrary, however. The meaning of the answer to the problem will be the same no matter what choice is made. Suppose that the choice for the positive direction were opposite to that shown in Figure 2.12 a. Then, all of the data listed in the Knowns and Unknowns table would appear with algebraic signs opposite to those specified, and the calculation of the displacement would appear as follows:
The value for has the same magnitude of as before, but it is now negative. The negative sign means that the displacement of the motorcycle during segment 2 is in the negative direction. But the negative direction is now upward and to the right in Figure 2.12 a, so the meaning of the result is exactly the same as it was before. |
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Now that we have seen how the equations of kinematics are applied to various situations, it's a good idea to summarize the reasoning strategy that has been used. This strategy, which is outlined below, will also be used when we consider freely falling bodies in Section 2.6 and two-dimensional motion in Chapter 3.
| Reasoning Strategy� |
Applying the Equations of Kinematics |
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1.�� |
Make a drawing to represent the situation being studied. A drawing helps us to see what's happening. |
2.�� |
Decide which directions are to be called positive and negative  relative to a conveniently chosen coordinate origin. Do not change your decision during the course of a calculation. |
3.�� |
In an organized way, write down the values (with appropriate plus and minus signs) that are given for any of the five kinematic variables (x, a, v, , and t). Be on the alert for implicit data, such as the phrase “starts from rest,” which means that the value of the initial velocity is  . The data summary tables used in the examples in the text are a good way to keep track of this information. In addition, identify the variables that you are being asked to determine. |
4.�� |
Before attempting to solve a problem, verify that the given information contains values for at least three of the five kinematic variables. Once the three known variables are identified along with the desired unknown variable, the appropriate relation from Table 2.1 can be selected. Remember that the motion of two objects may be interrelated, so they may share a common variable. The fact that the motions are interrelated is an important piece of information. In such cases, data for only two variables need be specified for each object. |
5.�� |
When the motion of an object is divided into segments, as in Example 9, remember that the final velocity of one segment is the initial velocity for the next segment. |
6.�� |
Keep in mind that there may be two possible answers to a kinematics problem as, for instance, in Example 8. Try to visualize the different physical situations to which the answers correspond. |
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| 2.6� |
Freely Falling Bodies |
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Everyone has observed the effect of gravity as it causes objects to fall downward. In the absence of air resistance, it is found that all bodies at the same location above the earth fall vertically with the same acceleration. Furthermore, if the distance of the fall is small compared to the radius of the earth, the acceleration remains essentially constant throughout the descent. This idealized motion, in which air resistance is neglected and the acceleration is nearly constant, is known as free-fall. Since the acceleration is constant in free-fall, the equations of kinematics can be used.
The acceleration of a freely falling body is called the
acceleration due to gravity, and its magnitude (without any algebraic sign) is denoted by the symbol
g. The acceleration due to gravity is directed downward, toward the center of the earth. Near the earth's surface,
g is approximately
Unless circumstances warrant otherwise, we will use either of these values for
g in subsequent calculations. In reality, however,
g decreases with increasing altitude and varies slightly with latitude.
Figure 2.13a shows the well-known phenomenon of a rock falling faster than a sheet of paper. The effect of air resistance is responsible for the slower fall of the paper, for when air is removed from the tube, as in Figure 2.13b, the rock and the paper have exactly the same acceleration due to gravity. In the absence of air, the rock and the paper both exhibit free-fall motion. Free-fall is closely approximated for objects falling near the surface of the moon, where there is no air to retard the motion. A nice demonstration of free-fall was performed on the moon by astronaut David Scott, who dropped a hammer and a feather simultaneously from the same height. Both experienced the same acceleration due to lunar gravity and consequently hit the ground at the same time. The acceleration due to gravity near the surface of the moon is approximately one-sixth as large as that on the earth.
When the equations of kinematics are applied to free-fall motion, it is natural to use the symbol y for the displacement, since the motion occurs in the vertical or y direction. Thus, when using the equations in Table 2.1 for free-fall motion, we will simply replace x with y. There is no significance to this change. The equations have the same algebraic form for either the horizontal or vertical direction, provided that the acceleration remains constant during the motion. We now turn our attention to several examples that illustrate how the equations of kinematics are applied to freely falling bodies.
| Example� |
10� |
A Falling Stone |
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A stone is dropped from rest from the top of a tall building, as Figure 2.14 indicates. After  of free-fall, what is the displacement y of the stone?
Reasoning
The upward direction is chosen as the positive direction. The three known variables are shown in the table below. The initial velocity of the stone is zero, because the stone is dropped from rest. The acceleration due to gravity is negative, since it points downward in the negative direction.
Equation 2.8 contains the appropriate variables and offers a direct solution to the problem. Since the stone moves downward, and upward is the positive direction, we expect the displacement y to have a negative value.
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| Problem-Solving Insight.� |
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It is only when values are available for at least three of the five kinematic variables ( y, a, v, , and t) that the equations in Table 2.1 can be used to determine the fourth and fifth variables. |
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Solution
Using Equation 2.8, we find that
The answer for y is negative, as expected.
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| Example� |
11� |
The Velocity of a Falling Stone |
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After of free-fall, what is the velocity v of the stone in Figure 2.14?
Reasoning
Because of the acceleration due to gravity, the magnitude of the stone's downward velocity increases by  during each second of free-fall. The data for the stone are the same as in Example 10, and Equation 2.4 offers a direct solution for the final velocity. Since the stone is moving downward in the negative direction, the value determined for v should be negative.
Solution
Using Equation 2.4, we obtain
The velocity is negative, as expected.
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The acceleration due to gravity is always a downward-pointing vector. It describes how the speed increases for an object that is falling freely downward. This same acceleration also describes how the speed decreases for an object moving upward under the influence of gravity alone, in which case the object eventually comes to a momentary halt and then falls back to earth. Examples 12 and 13 show how the equations of kinematics are applied to an object that is moving upward under the influence of gravity.
| Example� |
12� |
How High Does It Go? |
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A football game customarily begins with a coin toss to determine who kicks off. The referee tosses the coin up with an initial speed of  . In the absence of air resistance, how high does the coin go above its point of release?
Reasoning
The coin is given an upward initial velocity, as in Figure 2.15. But the acceleration due to gravity points downward. Since the velocity and acceleration point in opposite directions, the coin slows down as it moves upward. Eventually, the velocity of the coin becomes  at the highest point. Assuming that the upward direction is positive, the data can be summarized as shown below:
With these data, we can use Equation 2.9  to find the maximum height y.
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| Problem-Solving Insight.� |
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Implicit data are important. In Example 12, for instance, the phrase “how high does the coin go” refers to the maximum height, which occurs when the final velocity  in the vertical direction is  . |
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Solution
Rearranging Equation 2.9, we find that the maximum height of the coin above its release point is
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The rearrangement of algebraic equations is a problem-solving step that occurs often. The guiding rule in such a procedure is that whatever you do to one side of an equation, you must also do to the other side. Here in Example 12, for instance, we need to rearrange  (Equation 2.9) in order to determine y. The part of the equation that contains y is  , and we begin by isolating that part on one side of the equals sign. To do this we subtract  from each side of the equation:
Thus, we see that  . Then, we eliminate the term  from the left side of the equals sign by dividing both sides of the equation by .
The term occurs both in the numerator and the denominator on the left side of this result and can be eliminated algebraically, leaving the desired expression for y. |
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| Example� |
13� |
How Long Is It in the Air? |
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In Figure 2.15, what is the total time the coin is in the air before returning to its release point?
Reasoning
During the time the coin travels upward, gravity causes its speed to decrease to zero. On the way down, however, gravity causes the coin to regain the lost speed. Thus, the time for the coin to go up is equal to the time for it to come down. In other words, the total travel time is twice the time for the upward motion. The data for the coin during the upward trip are the same as in Example 12. With these data, we can use Equation 2.4  to find the upward travel time.
Solution
Rearranging Equation 2.4, we find that
The total up-and-down time is twice this value, or  . It is possible to determine the total time by another method. When the coin is tossed upward and returns to its release point, the displacement for the entire trip is  . With this value for the displacement, Equation 2.8  can be used to find the time for the entire trip directly.
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Examples 12 and 13 illustrate that the expression “freely falling” does not necessarily mean an object is falling down. A freely falling object is any object moving either upward or downward under the influence of gravity alone. In either case, the object always experiences the same downward acceleration due to gravity, a fact that is the focus of the next example.
| Conceptual Example� |
14� |
Acceleration Versus Velocity |
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(a) Do the direction and magnitude of the acceleration vector behave in the same fashion as the direction and magnitude of the velocity vector or (b) does the acceleration vector have a constant direction and a constant magnitude throughout the motion?
Reasoning
Since air resistance is being ignored, the coin is in free-fall motion. This means that the acceleration vector of the coin is the acceleration due to gravity. Acceleration is the rate at which velocity changes and is not the same concept as velocity itself.
Answer (a) is incorrect. During the upward and downward parts of the motion, and also at the top of the path, the acceleration due to gravity has a constant downward direction and a constant magnitude of  . In other words, the acceleration vector of the coin does not behave as the velocity vector does. In particular, the acceleration vector is not zero at the top of the motional path just because the velocity vector is zero there. Acceleration is the rate at which the velocity is changing, and the velocity is changing at the top even though at one instant it is zero.
Answer (b) is correct. The acceleration due to gravity has a constant downward direction and a constant magnitude of at all times during the motion.
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The motion of an object that is thrown upward and eventually returns to earth has a symmetry that is useful to keep in mind from the point of view of problem solving. The calculations just completed indicate that a time symmetry exists in free-fall motion, in the sense that the time required for the object to reach maximum height equals the time for it to return to its starting point.
A type of symmetry involving the speed also exists. Figure 2.16 shows the coin considered in Examples 12 and 13. At any displacement
y above the point of release, the coin's speed during the upward trip equals the speed at the same point during the downward trip. For instance, when
, Equation 2.9 gives two possible values for the final velocity
v, assuming that the initial velocity is
:
The value
is the velocity of the coin on the upward trip, and
is the velocity on the downward trip. The speed in both cases is identical and equals
. Likewise, the speed just as the coin returns to its point of release is
, which equals the initial speed. This symmetry involving the speed arises because the coin loses
in speed each second on the way up and gains back the same amount each second on the way down. In Conceptual Example 15, we use just this kind of symmetry to guide our reasoning as we analyze the motion of a pellet shot from a gun.
| Conceptual Example� |
15� |
Taking Advantage of Symmetry |
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Figure 2.17 a shows a pellet that has been fired straight upward from a gun at the edge of a cliff. The initial speed of the pellet is . It goes up and then falls back down, eventually hitting the ground beneath the cliff. In Figure 2.17 b the pellet has been fired straight downward at the same initial speed. In the absence of air resistance, would the pellet in Figure 2.17 b strike the ground with
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(a)�� |
a smaller speed than, |
(b)�� |
the same speed as, or |
(c)�� |
a greater speed than the pellet in Figure 2.17a? |
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Reasoning
In the absence of air resistance, the motion is that of free-fall, and the symmetry inherent in free-fall motion offers an immediate answer.
Answers (a) and (c) are incorrect. These answers are incorrect, because they are inconsistent with the symmetry that is discussed next in connection with the correct answer.
Answer (b) is correct. Figure 2.17 c shows the pellet after it has been fired upward and has fallen back down to its starting point. Symmetry indicates that the speed in Figure 2.17 c is the same as in Figure 2.17 a—namely, , as is also the case when the pellet has been actually fired downward. Consequently, whether the pellet is fired as in Figure 2.17 a or b, it begins to move downward from the cliff edge at a speed of . In either case, there is the same acceleration due to gravity and the same displacement from the cliff edge to the ground below. Given these conditions, when the pellet reaches the ground, it has the same speed in both Figures 2.17 a and 2.17 b.
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| Check Your Understanding� |
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| 13.�� |
An experimental vehicle slows down and comes to a halt with an acceleration whose magnitude is . After reversing direction in a negligible amount of time, the vehicle speeds up with an acceleration of . Except for being horizontal, is this motion
the same as or different from the motion of a ball that is thrown straight upward, comes to a halt, and falls back to earth? Ignore air resistance.
Answer:
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| 14.�� |
A ball is thrown straight upward with a velocity and in a time t reaches the top of its flight path, which is a displacement  above the launch point. With a launch velocity of  , what would be the time required to reach the top of its flight path and what would be the displacement of the top point above the launch point?
Answer:
| b. |
 and 
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| 15.�� |
Two objects are thrown vertically upward, first one, and then, a bit later, the other. Is it
possible or impossible that both objects reach the same maximum height at the same instant of time?
Answer:
| b. |
impossible that both objects reach the same maximum height at the same instant of time? |
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| 16.�� |
A ball is dropped from rest from the top of a building and strikes the ground with a speed  . From ground level, a second ball is thrown straight upward at the same instant that the first ball is dropped. The initial speed of the second ball is  , the same speed with which the first ball eventually strikes the ground. Ignoring air resistance, decide whether the balls cross paths
at half the height of the building, above the halfway point, or below the halfway point.
Answer:
| b. |
above the halfway point, or |
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| 2.7� |
Graphical Analysis of Velocity and Acceleration |
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Graphical techniques are helpful in understanding the concepts of velocity and acceleration. Suppose a bicyclist is riding with a constant velocity of
. The position
of the bicycle can be plotted along the vertical axis of a graph, while the time
is plotted along the horizontal axis. Since the position of the bike increases by
every second, the graph of
versus
is a straight line. Furthermore, if the bike is assumed to be at
when
, the straight line passes through the origin, as Figure 2.18 shows. Each point on this line gives the position of the bike at a particular time. For instance, at
the position is
, while at
the position is
.
In constructing the graph in Figure 2.18, we used the fact that the velocity was
. Suppose, however, that we were given this graph, but did not have prior knowledge of the velocity. The velocity could be determined by considering what happens to the bike between the times of 1 and
, for instance. The change in time is
. During this time interval, the position of the bike changes from
to
, and the change in position is
. The ratio
is called the
slope of the straight line.
Notice that the slope is equal to the velocity of the bike. This result is no accident, because
is the definition of average velocity (see Equation 2.2).
Thus, for an object moving with a constant velocity, the slope of the straight line in a position–time graph gives the velocity. Since the position–time graph is a straight line, any time interval
can be chosen to calculate the velocity. Choosing a different
will yield a different
, but the velocity
will not change. In the real world, objects rarely move with a constant velocity at all times, as the next example illustrates.
| Example� |
16� |
A Bicycle Trip |
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A bicyclist maintains a constant velocity on the outgoing leg of a trip, zero velocity while stopped, and another constant velocity on the way back. Figure 2.19 shows the corresponding position–time graph. Using the time and position intervals indicated in the drawing, obtain the velocities for each segment of the trip.
Reasoning
The average velocity is equal to the displacement divided by the elapsed time ,  . The displacement is the final position minus the initial position, which is a positive number for segment 1 and a negative number for segment 3. Note for segment 2 that  , since the bicycle is at rest. The drawing shows values for and for each of the three segments.
Solution
The average velocities for the three segments are
In the second segment of the journey the velocity is zero, reflecting the fact that the bike is stationary. Since the position of the bike does not change, segment 2 is a horizontal line that has a zero slope. In the third part of the motion the velocity is negative, because the position of the bike decreases from  to  during the 400-s interval shown in the graph. As a result, segment 3 has a negative slope, and the velocity is negative.
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If the object is accelerating, its velocity is changing. When the velocity is changing, the position–time graph is not a straight line, but is a curve, perhaps like that in Figure 2.20. This curve was drawn using Equation 2.8
, assuming an acceleration of
and an initial velocity of
. The velocity at any instant of time can be determined by measuring the slope of the curve at that instant. The slope at any point along the curve is defined to be the slope of the tangent line drawn to the curve at that point. For instance, in Figure 2.20 a tangent line is drawn at
. To determine the slope of the tangent line, a triangle is constructed using an arbitrarily chosen time interval of
. The change in
associated with this time interval can be read from the tangent line as
. Therefore,
The slope of the tangent line is the instantaneous velocity, which in this case is
. This graphical result can be verified by using Equation 2.4 with
.
Insight into the meaning of acceleration can also be gained with the aid of a graphical representation. Consider an object moving with a constant acceleration of
. If the object has an initial velocity of
, its velocity at any time is represented by Equation 2.4 as
This relation is plotted as the velocity–time graph in Figure 2.21. The graph of
versus
is a straight line that intercepts the vertical axis at
. The slope of this straight line can be calculated from the data shown in the drawing:
The ratio
is, by definition, equal to the average acceleration (Equation 2.4), so
the slope of the straight line in a velocity–time graph is the average acceleration.
and then runs back to her starting point. The time for this round-trip is 
. Which one of the following statements is true?
, due west. One and a half hours later its velocity is 
, due west. What is the train's average acceleration?
above the ground the engines shut down, but the rocket continues straight upward, losing speed as it goes. It reaches the top of its flight path and then falls back to earth. Ignoring air resistance, decide which one of the following statements is true.
above the ground. At a distance 
there is a branch that juts out from the side of the cliff, and on this branch a bird's nest is located. Two children throw stones at the nest with the same initial speed, one stone straight downward from the top of the cliff and the other stone straight upward from the ground. In the absence of air resistance, which stone hits the nest in the least amount of time?
�The space shuttle travels at a speed of about 
. The blink of an astronaut's eye lasts about 
. How many football fields 
does the shuttle cover in the blink of an eye?
�For each of the three pairs of positions listed in the following table, determine the magnitude and direction (positive or negative) of the displacement.
(a little less than a mile)?
�You step onto a hot beach with your bare feet. A nerve impulse, generated in your foot, travels through your nervous system at an average speed of 
. How much time does it take for the impulse, which travels a distance of 
, to reach your brain?
. Review the concept of average velocity in Section 2.2 and then determine the average velocity (magnitude and direction) for each of the three pairs. Note that the algebraic sign of your answers will convey the direction.
. They start at the west side of the lake and head due south to begin with.
, considerably slower than the giant tortoise, which walks at 
. After 12 minutes of walking, how much further would the tortoise have gone relative to the sloth?
. A 50-year-old runner can cover the same distance with an average speed of 
. How much later (in seconds) should the younger runner start in order to finish the course at the same time as the older runner?
. The car is a distance 
away. The bear is 
behind the tourist and running at 
. The tourist reaches the car safely. What is the maximum possible value for 
�In reaching her destination, a backpacker walks with an average velocity of 
, due west. This average velocity results because she hikes for 
with an average velocity of 
, due west, turns around, and hikes with an average velocity of 
, due east. How far east did she walk?
. During the second part, she rides for 36 minutes at an average speed of 
. Finally, during the third part, she rides for 8.0 minutes at an average speed of 
.
, and the average southward velocity has a magnitude of 
. What is the average velocity (magnitude and direction) for the entire trip?
along a level straight track. Very near and parallel to the track is a wall that slopes upward at a 
angle with the horizontal. As you face the window (
high, 
wide) in your compartment, the train is moving to the left, as the drawing indicates. The top edge of the wall first appears at window corner A and eventually disappears at window corner B. How much time passes between appearance and disappearance of the upper edge of the wall?
, and during a 
interval, it changes to a final speed of
to 
. Find
. Both the velocity and acceleration of the motorcycle point in the same direction. How much time is required for the motorcycle to change its speed from
, which she sustains for 
. Then, her acceleration drops to zero for the rest of the race. What is her velocity
.
and slows down to a speed of 
in 
.
in 
. Find the magnitude of the car's acceleration.
�A car is traveling along a straight road at a velocity of 
when its engine cuts out. For the next twelve seconds the car slows down, and its average acceleration is 
. For the next six seconds the car slows down further, and its average acceleration is 
. The velocity of the car at the end of the eighteen-second period is 
. The ratio of the average acceleration values is 
. Find the velocity of the car at the end of the initial twelve-second interval.
due east, while cycle B has an average acceleration of 
due east. By how much did the speeds differ at the beginning of the four-second interval, and which motorcycle was moving faster?
(Cycle A was initially traveling faster.)
. Assuming that the player accelerates uniformly, determine the distance he runs.
also in 
with an acceleration of 
.
.
. Once the jet touches down, it has 
of runway in which to reduce its speed to 
. Compute the average acceleration (magnitude and direction) of the plane during landing.
, directed southward
for 
. Taking the direction of motion as positive, we have
. His velocity at the start of the final mile 
was about 
. The acceleration, although small, was very important to his victory. To assess its effect, determine the difference between the time he would have taken to run the final mile at a constant velocity of 
and a constant acceleration due to the fan. The direction to the right is positive. The cart reaches a maximum position of 
, where it begins to travel in the negative direction. Find the acceleration of the cart.
, so initially it is moving to the right, which is the positive direction. It eventually reaches a point where the displacement is 
, and it begins to move to the left. This must mean that the cart comes to a momentary halt at this point (final velocity is 
), before beginning to move to the left. In other words, the cart is decelerating, and its acceleration must point opposite to the velocity, or to the left. Thus, the acceleration is negative. Since the initial velocity, the final velocity, and the displacement are known, Equation 2.9 
, while rocket B has an initial velocity of 
. After a time 
. What is the acceleration of rocket B?
, and the driver sees a traffic light turn red. After 
(the reaction time), the driver applies the brakes, and the car decelerates at 
. What is the stopping distance of the car, as measured from the point where the driver first sees the red light?
; after 
overtakes and passes the entering car. The entering car maintains its acceleration. How much time is required for the entering car to catch the other car?
apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 
, while Sir Alfred's has a magnitude of 
. Relative to Sir George's starting point, where do the knights collide?
apart. They run directly toward each other, both players accelerating. The first player's acceleration has a magnitude of 
. The second player's acceleration has a magnitude of 
.
on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two cars meet for the first time at the next exit, which is 
away?
, in the same direction as the second car's velocity
, with the 
sign located midway between the other two. Obeying these speed limits, the smallest possible time 
that a driver can spend on this part of the road is to travel between the first and second signs at 
and between the second and third signs at 
. More realistically, a driver could slow down from 55 to 
. Find the ratio 
.
. The car's deceleration has a magnitude of 
during this time. What is the car's displacement?
. The runway on which the plane lands intersects another runway. The width of the intersection is 
. The plane decelerates through the intersection at a rate of 
and clears it with a final speed of 
. How much time is needed for the plane to clear the intersection?
. It passes through a 20.0-m-wide crossing in a time of 
. After the locomotive leaves the crossing, how much time is required until its speed reaches 
?
and 2.7 
, which can be rearranged to give
when the train reaches the end of the crossing. Once 
.
and starts from rest with a constant acceleration at time 
, the car just reaches the front of the train. Ultimately, however, the train pulls ahead of the car, and at time 
, the car is again at the rear of the train. Find the magnitudes of
by stuntman Dan Koko. In 1948 he jumped from rest from the top of the Vegas World Hotel and Casino. He struck the airbag at a speed of 
. To assess the effects of air resistance, determine how fast he would have been traveling on impact had air resistance been absent.
is zero since the stunt man falls from rest. If the origin is chosen at the top of the hotel and the upward direction is positive, then the displacement is 
. Solving for 
. Assuming air resistance has no effect on the rock, calculate its speed
, 
, and 
.
to each pellet. Gun A is fired straight upward, with the pellet going up and then falling back down, eventually hitting the ground beneath the cliff. Gun B is fired straight downward. In the absence of air resistance, how long after pellet B hits the ground does pellet A hit the ground?
and measures a time of 
before the rock returns to his hand. What is the acceleration (magnitude and direction) due to gravity on this planet?
. When the balloon is 
above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground?
. The initial position of the compass is 3.00 m and its final position is 0 m when it strikes the ground. The displacement of the compass is the final position minus the initial position, or 
. As the compass falls to the ground, its acceleration is the acceleration due to gravity, 
. Equation 2.8 
can be used to find how much time elapses before the compass hits the ground.
and 
. The second solution, being a negative time, is discarded.
above its launch point. At what height above its launch point has the speed of the ball decreased to one-half of its initial value?
from a 3.0-m board.
above its launch point, the ball's speed is one-half its launch speed. What maximum height above its launch point does the ball attain?
below. If the balloon is released from rest, how long is it in the air?
can be found from Equation 2.8 
. Assuming upward to be the positive direction, we find
above the ground. The pellet strikes the ground with a speed of 
, where the minus sign indicates that her motion is directly downward. What is her displacement during the last 
of the dive?
. It hits the pavement, then bounces back up, rising just 
before falling back down again. A boy then catches the ball on the way down when it is 
above the pavement. Ignoring air resistance, calculate the total amount of time that the ball is in the air, from drop to catch.
high sees a raft floating at a constant speed on the river below. Trying to hit the raft, she drops a stone from rest when the raft has 
more to travel before passing under the bridge. The stone hits the water 
. The stones are thrown with the same speed of 
. Find the location (above the base of the cliff) of the point where the stones cross paths.
and that the second arrow is fired 
above the ground, a man, 
tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way?
. Its velocity is given by Equation 2.9 
. Since the block is moving down, its velocity has a negative value,
which satisfies Equation 2.8:
and 
. Thus, 
.
for 1.70 seconds, at which point its fuel abruptly runs out. Air resistance has no effect on its flight. What maximum altitude (above the ground) will the rocket reach?
above the ground, you drop a stone from rest. When the stone has fallen 
, you throw a second stone straight down. What initial velocity must you give the second stone if they are both to reach the ground at the same instant? Take the downward direction to be the negative direction.
for the tile to pass her window, which has a height of 
. How far above the top of this window is the roof?
at time 
to travel 
, 
, and 
?
for segment 
for segment 
for segment 
) of the bus for the entire 3.5-h period shown in the graph?
.
(3 min and 
). In 1999 the Moroccan runner Hicham el-Guerrouj set a record of 
for the mile. If these two runners had run in the same race, each running the entire race at the average speed that earned him a place in the record books, el-Guerrouj would have won. By how many meters?
after getting on. Clifford is in a real hurry, however, and skips the speed ramp. Starting from rest with an acceleration of 
, he covers the same distance as the ramp does, but in one-fourth the time. What is the speed at which the belt of the ramp is moving?
covered by the belt in a time 
is giving by Equation 2.2 (with 
assumed to be zero) as
with which the belt of the ramp is moving can be found by eliminating 
between Equations 1 and 2.
, we have
. A player cannot touch the ball until after it reaches its maximum height and begins to fall down. What is the minimum time that a player must wait before touching the ball?
. How long (in minutes) does it take an electron to traverse 
of wire in the filament of a light bulb?
per day.
at a constant velocity of 
away, on a straight route. The driver accelerates to the speed limit and, on reaching it, begins to decelerate at once. The magnitude of the deceleration is three times the magnitude of the acceleration. Find the lengths of the acceleration and deceleration phases.
away. The woman runs at 
between his owner and the river, until the woman reaches the river. What is the total distance run by the dog?
and is 
ahead of the second-place cyclist. The second-place cyclist has a velocity of 
and an acceleration of 
. How much time elapses before he catches the leader?
for 
. She then gets out of the cart and starts walking at an average speed of 
. For how long (in seconds) must she walk if her average speed for the entire trip, riding and walking, is 
?
in 
. Assume that they are moving in the 
due north, when a car zooms by at a constant velocity of 
due north. After a reaction time of 
the policeman begins to pursue the speeder with an acceleration of 
. Including the reaction time, how long does it take for the police car to catch up with the speeder?
. When the balloon is 
above the ground, a gun fires a pellet straight up from ground level with an initial speed of 
and a maximum speed of 
. Car B has an acceleration of 
and a maximum speed of 
. Which car wins the race, and by how many seconds?
.
long. If a car crosses the finish line before reaching its maximum speed, then there is only one interval of constant acceleration to consider. We will first determine whether this is true by calculating the car's displacement 
while accelerating from rest to top speed from Equation 2.9 
:
, then the car crosses the finish line before reaching top speed, and the total time for its race is found from Equation 2.8 
, with 
and 
to reach the maximum speed is most easily found from Equation 2.4 
, with 
that elapses during the rest of the race is found by solving Equation 2.8 
represent the displacement for this part of the race. With the aid of Equation 1, this becomes 
. Then, since the car is at its maximum speed, the acceleration is 
, and the displacement is
to travel at its maximum speed. Equation 2 gives the time for car A to reach the finish line as
. With this acceleration, he continues in the same direction for another 
. What is the value of his acceleration (assumed to be constant) during the initial 
period?
to the final position 
” direction for 
means the velocity decreases by 
during each second of elapsed time.
, the spacecraft changes its velocity from 
and 
.
. The initial velocity is 
direction.
in 
in 
, as in part 
and her final velocity is 
. Her average acceleration 
is the change in the velocity divided by the elapsed time:
because of air resistance.
and 
, respectively. The average acceleration for this phase of the motion is
. What are the
, and the elapsed time 
, since 
. The displacements at the two times are then
is four times the displacement at 
.
during each second of travel. Therefore, since the dragster starts from rest, the velocity is 
at the end of the 2nd second, 
at the end of the 3rd second, and so on. Thus, when the time doubles, the velocity also doubles.
. Since the initial velocity is zero, 
. However, as we have seen, the final velocity is proportional to the elapsed time, since when the time doubles, the final velocity also doubles. Therefore, the displacement, being the product of the average velocity and the time, is proportional to the time squared, or 
. Consequently, as the time doubles, the displacement does not double, but increases by a factor of four.
, as indicated in Equation 2.5. Acceleration is the rate at which the velocity is changing.
, the acceleration 
at 
, but there is not enough information to determine her average velocity.
.
, due west
, due west
, due east.
, the displacement is proportional to the acceleration.
. Therefore,
, due north
and
. In each case, find the acceleration (magnitude and algebraic sign) and state whether or not the car is decelerating.
. What is the average acceleration of the car?
) during this interval. Include the correct algebraic sign with your answers to convey the directions of the velocity and the acceleration.
, and
?
and
:
further than the jogger.
farther.
. Find the acceleration (in 
when going down a slope for 
?
? Express your answer in terms of 
. Taking downward as the positive direction, we find
, she must fall an additional distance 
. According to Equation 2.9 
would 
.
) indicate positive, negative, and zero average velocities.
and 
, 
, and 
.
for segment 
for segment 
for segment 
. According to the graph, therefore, this position corresponds to a time of 
.
.
. The pebble is not decelerating. Since its velocity and acceleration both point downward, the magnitude of the pebble's velocity is increasing, not decreasing.
beneath the cliff-top.
?
by 
.