Other Science and Math Wiley Self-Teaching Guides:
Science
Biology: A Self-Teaching Guide, Third Edition, by Steven D. Garber
Chemistry: A Self-Teaching Guide, Third Edition, by Clifford C. Houk, Richard Post, and Chad A. Snyder
Math
All the Math You'll Ever Need: A Self-Teaching Guide, by Steve Slavin
Practical Algebra: A Self-Teaching Guide, Second Edition, by Peter H. Selby and Steve Slavin
Quick Algebra Review: A Self-Teaching Guide, by Peter H. Selby and Steve Slavin
Quick Business Math: A Self-Teaching Guide, by Steve Slavin
Quick Calculus: A Self-Teaching Guide, Second Edition, by Daniel Kleppner and Norman Ramsey
Third Edition
Copyright © 2020 by John Wiley & Sons, Inc.
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ISBNs: 978-1-119-62990-0 (paperback), 978-1-119-62991-7 (ePDF), 978-1-119-62989-4 (ePub)
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THIRD EDITION
For Jonah and Sophie:
“The important thing is to not stop questioning. Curiosity has its own reason for existence.”
- Albert Einstein
First, I am grateful to the many John Jay High School students I've taught over the past 20-plus years. They have taught me just as much, if not more, than I taught them. The insights I've gained about how students learn science have informed the revisions for this edition.
Thanks for the tremendous help and understanding from Pete Gaughan, Riley Harding, and the rest of the team at John Wiley and Sons.
Additional thanks to Larry Ferlazzo for connecting me with the Wiley team to work on the third edition.
—Frank Noschese
“If it ain't broke, don't fix it.” This maxim applies to books as well as to many other endeavors, and it explains why such a long time passed between the second and third editions of this book. Elementary physics does not change much in a few decades, and therefore this edition contains very little new subject matter.
Why a new edition, then? First, some new examples have been added, either because of change in fashions or because of new applications of basic physics. An example of the latter is the ubiquity of the smartphone, a device that didn't exist when the second edition was printed. Second, some explanations have been rewritten to improve clarity. Finally, the “look” of the book has been updated to make it more attractive to today's readers.
The fundamental purpose of the book has not changed, however, and if you are considering a study of this book, you probably fit into one of the following categories:
In any case, Basic Physics can help you. It is a complete, self-contained physics book with a programmed format. The chapters are divided into short steps called frames. Each frame presents some new material, and then asks you questions to test your comprehension. By faithfully answering the questions (preferably by actually writing the answers in the spaces provided or on a separate sheet) before you check the answers we supply, you will be able to check your progress. Learning theory tells us that by writing your answers, you understand and retain the material for a longer time. In fact, I suggest that you do not look at the answers until have completed your own. If your answer does not agree with the one we provide, be sure you understand the discrepancy before proceeding to the next frame. To check yourself further, a Self-Test, with answers, is included at the end of each chapter.
Since physics builds from one principle to another, many chapters require an understanding of previous chapters. For this reason, the prerequisites for each chapter are listed on the first page of the chapter. Some frames within chapters, however, may be skipped without disrupting the logic of the development. Such frames, which are often mathematical treatments of the subject at hand, are labeled as “optional.” Some optional frames have a prerequisite of a prior optional frame; such prerequisites are listed at the beginning of the frame.
I suggest that you complete the entire Self-Test at the end of a chapter before checking your answers. In this way, your test will be more similar to classroom testing. Each answer in the Self-Test section includes a reference to the frame (s) to which you should return for help on missed items.
Have fun!
Physics deals with quantities that can be measured. Thus, you won't find concepts such as honesty, love, and courage as primary topics of discussion in a physics book. As you proceed through your study of physics, you will find that every one of the measurable quantities that is discussed can be specified in terms of only four basic dimensions: mass, length, time, and electric charge. In this chapter, we will begin a study of the first three of these.
Speed is defined as the distance traveled divided by the time of travel. Thus, if you travel 12 miles in 3 hours, the average speed is 4 mi/hr (mi/hr or mph).*
Answers: (a) 3 miles; (b) 1.5 ft/s; (c) s = d/t
In practice, we usually use the symbol v to represent speed (because of the similarity between speed and velocity, a concept that will be discussed later). For an object that moves with constant speed, the defining equation for speed is:
Here v is the speed and d is the distance the object travels after time t has elapsed.
The above equation holds true for objects that move with constant speed (like a car in cruise control). However, if the speed of the object changes while moving, the above equation gives us just the object's average speed. For example, if it takes you 30 minutes to drive to a friend's house 12 miles away, your average speed is 24 mi/hr. That doesn't mean your speed was 24 mi/hr for the entire trip. Due to traffic lights, road speed limits, and traffic conditions, there were times during your trip when you traveled slower than 24 mi/hr and times you were traveling faster. An average speed doesn't tell us how an object is moving at any particular instant. Only when an object is moving with constant speed does its average speed equal its actual speed.
Answers: (a) 50 mi/hr; (b) 400 mi/hr; (c) 24 m
As a car gains speed, we say that it accelerates. The change in speed divided by the time it takes to make the change is called the acceleration. In equation form:
Here v0 represents the speed at the start and vf is the speed after time t has passed. Notice that although in everyday language the word “acceleration” is used only to refer to a gain in speed, if v0 is larger than v, the acceleration is negative and is actually a deceleration. In physics, we use the word “acceleration” to include this negative case.
Answers: (a) vf – v0; (b) 4 miles/hour per second (4 mi/hr/s)
The unit of acceleration in the problem you just solved is “miles per hour per second,” or “mi/hr/s.” If the speed is measured in meters per second and its change is measured over a few seconds, a natural unit for acceleration is “m/s/s,” which is sometimes called meters per second squared and written “m/s2.”
Suppose a car starts from rest and accelerates at 2 m/s2 for 3 seconds. This means that the car gains __________ of speed for every __________ of time
Answer: 2 m/s … 1 second
Suppose that, starting from rest, your new truck accelerates at a rate of 5 mi/hr/s and it continues this acceleration for 8 seconds. What will be its speed at the end of the 8 seconds? (If you don't know how to find the answer, read the rest of this frame; otherwise, just calculate the answer.)
To solve this problem, first rewrite the acceleration equation from frame 3 as:
This is a useful form of the acceleration equation because it says that the speed of the object after some time (vf) depends on its starting speed (v0), how quickly its speed is changing (a), and how much time has elapsed (t).
Let's apply this equation to your truck. In this case, v0 is zero since the truck started from rest. Substitute in the acceleration and time, and solve for vf .
Answer: 40 mi/hr (0 + 5 mi/hr/s · 8 s = 40 mi/hr)
It is helpful to visualize the motion of an object using motion diagrams. Suppose we took a time-lapse photo of a car that first increased in speed, then moved at constant speed, and finally slowed down. The photo would look like figure (1) here:
We can represent the car's motion using a motion diagram, as shown in figure (2). The dots represent the position of the car at each instance in time. At the beginning, when the car is increasing in speed, the spacing between the dots increases because the car travels a greater distance in each successive instance. In the middle, when the car's speed is constant, the dots are evenly spaced because the car travels the same distance in each successive instance. Finally, when the car slows down, the spacing between the dots decreases because the car travels a greater distance in each successive instance.
The motion diagram also has arrows attached to the dots to represent both the speed of the car and the direction the car is moving. The longer the arrow, the greater the speed. We can also show the acceleration (or deceleration) of the car. When the car is increasing in speed, its acceleration is in the same direction as the car is moving. When the car is slowing down, its acceleration is in the direction opposite its motion. When the car is moving at a constant speed, its acceleration is zero.
Answers:
The acceleration of an object falling freely near the surface of the earth is 9.8 m/s2. An object dropped from a very high platform achieves a speed of 9.8 m/s by the end of the first second and 19.6 m/s by the end of the next second.
Answers: (a) 29.4 m/s; (b) 49 m/s
The next figure shows a motion diagram of a ball being dropped from a building, and it indicates the speed of the ball at various times while it falls.
Since the ball is moving downward and speeding up, the acceleration is also downward. For each second that passes, the ball's speed increases by 9.8 m/s.
Now suppose that a ball is thrown upward from the base of the building at a starting speed of 29.4 m/s, as in figure (2). Notice that the speed of the ball decreases by 9.8 m/s for every second that passes. Since the ball is moving upward, but slowing down, the acceleration of the ball is downward (opposite the direction of its motion). We see that, in both cases, the acceleration of the ball is 9.8 m/s2 downward.
Answers: (a) (i) 9.8 m/s downward, and (ii) 19.6 m/s downward; (b) 6 seconds
The equation relating acceleration, the distance traveled, and the time elapsed is:
where v0 is the starting speed, a is the acceleration, and t is the time elapsed. Thus, in the figure in frame 8, when the ball has fallen 1 second (and its speed is 9.8 m/s), it has fallen a distance of 4.9 m (d = 0 m/s · 1 s + ½ · 9.8 m/s2 · 1 s2 = 4.9 m, since v0 was zero).
After 3 seconds of fall, how far has it fallen? _________________________
Answer: 44.1 m (d = 0 m/s · 3 s + ½ · 9.8 m/s2 · (3 s)2 = 44.1 m)
Although its speed after 1 second is 9.8 m/s, the ball fell only 4.9 m. This is because the ball wasn't moving at a constant speed of 9.8 m/s during that first second—its speed was slower than 9.8 m/s. The average speed of the ball while it falls is distance/time = 4.9 m/1 second = 4.9 m/s. For an object that is speeding up or slowing down at a constant rate (i.e., constant acceleration), the average speed of the object during a time t can also be determined with the following equation:
which is the simple mathematical average of its starting speed and its speed at time t. When the average speed is calculated this way, we can use that average speed to figure out how far an object travels using d = vavg · t (rearranging the equation from frame 2).
Take the example in the previous frame. We calculated the distance fallen after 3 seconds to be 44.1 m. Let's calculate this a different way, using average velocity. We know that v0 = 0 (because the object is dropped) and that after 3 seconds, vf = 29.4 m/s. Using the equation above, the average speed during those 3 seconds is (0 m/s + 29.4 m/s)/2 = 14.7 m/s. Therefore, since the ball falls with an average speed of 14.7 m/s in those 3 seconds, the distance traveled in that time is 14.7 m/s · 3 s = 44.1 m. This is the same result we calculated before.
A car traveling at 20 m/s accelerates at a rate of 5 m/s2 for 10 seconds.
Answers:
Often speed does not tell us all we want to know about a motion. If you were told that while you slept your sheepdog left you at a speed of 5 mi/hr, and that he traveled in a straight line, you would not have enough information to go looking for him. You would also need to know in which direction he went. Although the term speed tells us how fast something is moving, the term velocity specifies both speed and direction. Identify the two descriptions below as speed or velocity.
Answers: (a) speed; (b) velocity; (c) velocity includes a direction
The defining equation for average velocity is:
Notice that velocity is defined in the same way as speed, except that in this case we added arrows over some of the symbols. The quantities with arrows are vectors. A vector is a quantity that has direction as well as magnitude (size). The symbol , called the object's displacement, is defined as the straight-line distance from the object's starting position to its final position, including the direction from start to finish. Thus, an object's displacement might be stated as 4.2 m east.
Answers: (a) average velocity and displacement ( and ); (b) time (t)
We cannot assign a special direction to time, so it is not a vector. All quantities that are not vectors are called scalars. Identify each of the following quantities as a vector or a scalar.
Answers: (a) vector; (b) scalar; (c) scalar; (d) vector
As you might have guessed, acceleration is also a vector, since we must take into account its direction. (We have already seen in motion diagrams that an object will speed up or slow down depending on the direction of the acceleration.)
To see the importance of vectors in a simple situation, consider the following example: Suppose a jogger leaves their house and jogs north 300 ft to the streetlamp on the corner. They turn west, jog another 400 ft, and get to the oak tree 2 minutes after leaving home. We will calculate the jogger's average velocity for the 2 minutes. (See next figure.)
Answers: (a) 500 ft; (b) 250 ft/min in a direction between north and west; (c) 700 ft; (d) 350 ft/min
This example shows that velocity and speed are definitely different things. It may seem to you that speed is the more useful of the two, but the following examples should show the advantage of the concept of velocity.
Imagine a moving hot-air balloon that drops a sandbag. Since the balloon is moving along with the air, the effect of air resistance can be neglected as the sandbag falls. An interesting thing occurs: as the sandbag falls, it continues moving forward at the speed that the balloon had when the sandbag was dropped. The sandbag does fall, of course, but its downward speed is independent of its forward speed. The sandbag continues forward as if it had not been dropped, and it falls downward with an ever-increasing speed just as if it were not moving forward. The figure assumes that the balloon is traveling at 12 m/s, and it shows the sandbag at intervals of 1 second beginning when it is dropped. In each second the sandbag moves forward 12 m. Now note its downward motion. At the end of the first second after being released, its downward speed is 9.8 m/s.
Answers: (a) 12 m/s; (b) 19.6 m/s; (c) 12 m/s; (d) 29.4 m/s
In this example, we have seen that it is sometimes advantageous to consider motion as having two components, each perpendicular to the other. Although the actual velocity of the sandbag is neither straight down nor horizontal, we can consider its motion to be a combination of a downward motion and a horizontal motion. When we do this, each of the separate motions is a simple one. We will return to projectile motion later in this chapter and again in the next.
A common trick in skateboarding is called a hippy jump. The skateboarder is rolling along on a flat surface and then jumps straight upwards. As shown in the figure, the skateboarder continues moving forward at the speed the skateboard had before the jump, remaining over the skateboard as it rolls forward, before landing back on the board.* Just like the previous example with the sandbag and hot‐air balloon, the skateboarder's vertical motion (rising and falling) is independent of their horizontal motion (moving forward at constant speed).
If the skateboard is rolling forward at 5 m/s and the skateboarder jumps up with an initial vertical speed of 2.45 m/s:
Answers:
Suppose a person is walking on a treadmill at 3 mi/hr. The person's legs are walking forward at 3 mi/hr, and yet the person stays in place. The person's forward-walking velocity of 3 mi/hr is canceled by the treadmill belt's backward velocity of 3 mi/hr. In other words, the two velocities add to zero because they are equal in magnitude (size) but opposite in direction.
Answer: The person will move backward at 1 mi/hr and eventually fall off the treadmill.
Since velocity is a vector quantity, we must take into account their directions in order to add two or more velocities. Addition of vectors is most easily accomplished by using scale drawings—the “graphical” method of vector addition. Suppose you are rowing a boat across a river, as shown in part 1 of the figure below. Part 2 uses arrows to represent the velocities of the boat and the river. Since the river's velocity is 0.8 m/s and your rowing velocity is 1.6 m/s, the arrow representing the river's velocity is twice as long as the arrow for the rowing velocity (4 and 2 cm, respectively). In part 3, the arrows are shown hooked end-to-end, but they each keep the same length and are pointed in the same direction as the velocities.
Now you draw a line from the tail of the first arrow (the long one) to the head of the second one. This vector will represent the resulting velocity of the boat, due to both your rowing and the river current. So place an arrowhead at its upper end. Go ahead and draw it now, then measure it. What is its length? __________
Answer :
Since we used 1 cm to represent 0.40 m/s, your resultant vector represents a velocity of 2.2 m/s (5.5 times 0.4), pointing in a direction between the velocity of the river current and your rowing velocity. Note that it is, as should be expected, a little closer to the direction of your rowing velocity. (This is because your rowing velocity was faster than the river current and thus affects the resultant velocity of the boat more than does the river current.)
Answers: (a)
(b) 3 m/s (If your resultant vector measures between 7.0 and 8.0 cm, that is OK. You should get between 2.85 and 3.20 m/s.)
Consider an airplane flying in still air at a velocity of 150 mi/hr northward. In part 1 of the figure below, a 3-cm arrow represents this velocity. Now a wind starts blowing toward the southeast at a speed of 50 mi/hr—quite a wind! The second part of the figure shows this 50-mi/hr wind as a 1-cm arrow. Just as was done previously, one arrow is moved so that its tail falls on the head of the other, shown in part 3.
Answers: (a) It is drawn from the tail of the first to the head of the second; (b) 2.4 cm; (c) 120 mi/hr (because each centimeter represents 50 mi/hr) (Note that the plane is blown off course as well as being slowed down.)
Let us return once again to projectile motion and the example of the sandbag being dropped from the hot-air balloon. Suppose the balloon is high enough that the sandbag hits the ground 2 seconds after it is released. At that time the forward speed of the sandbag is still 12 m/s. Its downward speed is now 19.6 m/s. On a separate sheet of paper, use the graphical method to combine both vectors, and determine the velocity of the sandbag at impact.
Answer:
This drawing uses a scale of 1 cm to represent 3 m/s. The arrow that shows the total is 7.6 cm long, so the total velocity is 23 m/s.
The questions below will test your understanding of this chapter. Use a separate sheet of paper for your diagrams or calculations. Compare your answers with the answers provided following the test.
(b) How far does it travel during the 3 seconds of its accelerated motion? ___________________
If your answers do not agree with those given below, review the sections indicated in parentheses before you go on to the next chapter.
Distance traveled divided by the time used to cover that distance. (frame 1)
6 2/3 hours
Solution:
vavg = d/t
t = d/vavg = 300 miles/45 m/hr = 6 2/3 (frames 1, 2)
4 m/s/s, or 4 m/s2 (frames 3–5)
9.8 m/s2 (frame 6)
(a) 8 m/s; (b) 24 meters (frame 10)
49 m/s; 122.5 m (frames 6–9)
(a) 10 seconds; (b) 25 m/s; (c) 250 m
Velocity includes the direction of an object as well as the object’s speed (which does not include direction). (frames 11–14)
Its forward speed is still 55 miles/hour; its downward speed is increasing at the acceleration of gravity. (frame 15)
1 m/s south
5 miles/hour; the direction is east of north, or if you have a protractor you can measure it to be 37° east of north. (frame 19)