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Contents

Section 3.1 - The motion of falling objects

Section 3.2 - Acceleration

Section 3.3 - Positive and negative acceleration

Section 3.4 - Varying acceleration

Section 3.5 - The area under the velocity-time graph

Section 3.6 - Algebraic results for constant acceleration

Section 3.7 - A test of the principle of inertia (optional)

Section 3.8 - Applications of calculus (optional calculus-based section)

Section 3.9 - Summary

Section 3.1 - The motion of falling objects

Section 3.2 - Acceleration

Section 3.3 - Positive and negative acceleration

Section 3.4 - Varying acceleration

Section 3.5 - The area under the velocity-time graph

Section 3.6 - Algebraic results for constant acceleration

Section 3.7 - A test of the principle of inertia (optional)

Section 3.8 - Applications of calculus (optional calculus-based section)

Section 3.9 - Summary

The motion of falling objects is the simplest and most common example of motion with changing velocity. The early pioneers of physics had a correct intuition that the way things drop was a message directly from Nature herself about how the universe worked. Other examples seem less likely to have deep significance. A walking person who speeds up is making a conscious choice. If one stretch of a river flows more rapidly than another, it may be only because the channel is narrower there, which is just an accident of the local geography. But there is something impressively consistent, universal, and inexorable about the way things fall.

Stand up now and simultaneously drop a coin and a bit of paper side by side. The paper takes much longer to hit the ground. That's why Aristotle wrote that heavy objects fell more rapidly. Europeans believed him for two thousand years.

Now repeat the experiment, but make it into a race between the coin and your shoe. My own shoe is about 50 times heavier than the nickel I had handy, but it looks to me like they hit the ground at exactly the same moment. So much for Aristotle! Galileo, who had a flair for the theatrical, did the experiment by dropping a bullet and a heavy cannonball from a tall tower. Aristotle's observations had been incomplete, his interpretation a vast oversimplification.

It is inconceivable that Galileo was the first person to observe a discrepancy with Aristotle's predictions. Galileo was the one who changed the course of history because he was able to assemble the observations into a coherent pattern, and also because he carried out systematic quantitative (numerical) measurements rather than just describing things qualitatively.

Why is it that some objects, like the coin and the shoe, have similar motion, but others, like a feather or a bit of paper, are different? Galileo speculated that in addition to the force that always pulls objects down, there was an upward force exerted by the air. Anyone can speculate, but Galileo went beyond speculation and came up with two clever experiments to probe the issue. First, he experimented with objects falling in water, which probed the same issues but made the motion slow enough that he could take time measurements with a primitive pendulum clock. With this technique, he established the following facts:

- All heavy, streamlined objects (for example a steel rod dropped point-down) reach the bottom of the tank in about the same amount of time, only slightly longer than the time they would take to fall the same distance in air.
- Objects that are lighter or less streamlined take a longer time to reach the bottom.

This supported his hypothesis about two contrary forces. He imagined an idealized situation in which the falling object did not have to push its way through any substance at all. Falling in air would be more like this ideal case than falling in water, but even a thin, sparse medium like air would be sufficient to cause obvious effects on feathers and other light objects that were not streamlined. Today, we have vacuum pumps that allow us to suck nearly all the air out of a chamber, and if we drop a feather and a rock side by side in a vacuum, the feather does not lag behind the rock at all.

Galileo's second stroke of genius was to find a way to make quantitative measurements of how the speed of a falling object increased as it went along. Again it was problematic to make sufficiently accurate time measurements with primitive clocks, and again he found a tricky way to slow things down while preserving the essential physical phenomena: he let a ball roll down a slope instead of dropping it vertically. The steeper the incline, the more rapidly the ball would gain speed. Without a modern video camera, Galileo had invented a way to make a slow-motion version of falling.

Although Galileo's clocks were only good enough to do accurate experiments at the smaller angles, he was confident after making a systematic study at a variety of small angles that his basic conclusions were generally valid. Stated in modern language, what he found was that the velocity-versus-time graph was a line. In the language of algebra, we know that a line has an equation of the form \(y=ax+b\), but our variables are \(v\) and \(t\), so it would be \(v=at+b\). (The constant \(b\) can be interpreted simply as the initial velocity of the object, i.e., its velocity at the time when we started our clock, which we conventionally write as \(v_\text{o}\).)

An object is rolling down an incline. After it has been rolling for a short time, it is found to travel 13 cm during a certain one-second interval. During the second after that, it goes 16 cm. How many cm will it travel in the second after that?

(answer in the back of the PDF version of the book)Galileo's inclined-plane experiment disproved the long-accepted claim by Aristotle that a falling object had a definite “natural falling speed” proportional to its weight. Galileo had found that the speed just kept on increasing, and weight was irrelevant as long as air friction was negligible. Not only did Galileo prove experimentally that Aristotle had been wrong, but he also pointed out a logical contradiction in Aristotle's own reasoning. Simplicio, the stupid character, mouths the accepted Aristotelian wisdom:

Simplicio:

There can be no doubt but that a particular body ... has a fixed velocity which is determined by nature...

Salviati:

If then we take two bodies whose natural speeds are different, it is clear that, [according to Aristotle], on uniting the two, the more rapid one will be partly held back by the slower, and the slower will be somewhat hastened by the swifter. Do you not agree with me in this opinion?

Simplicio:

You are unquestionably right.

Salviati:

But if this is true, and if a large stone moves with a speed of, say, eight [unspecified units] while a smaller moves with a speed of four, then when they are united, the system will move with a speed less than eight; but the two stones when tied together make a stone larger than that which before moved with a speed of eight. Hence the heavier body moves with less speed than the lighter; an effect which is contrary to your supposition. Thus you see how, from your assumption that the heavier body moves more rapidly than the lighter one, I infer that the heavier body moves more slowly.

The physicist Richard Feynman liked to tell a story about how when he was a little kid, he asked his father, “Why do things fall?” As an adult, he praised his father for answering, “Nobody knows why things fall. It's a deep mystery, and the smartest people in the world don't know the basic reason for it.” Contrast that with the average person's off-the-cuff answer, “Oh, it's because of gravity.” Feynman liked his father's answer, because his father realized that simply giving a name to something didn't mean that you understood it. The radical thing about Galileo's and Newton's approach to science was that they concentrated first on describing mathematically what really did happen, rather than spending a lot of time on untestable speculation such as Aristotle's statement that “Things fall because they are trying to reach their natural place in contact with the earth.” That doesn't mean that science can never answer the “why” questions. Over the next month or two as you delve deeper into physics, you will learn that there are more fundamental reasons why all falling objects have \(v-t\) graphs with the same slope, regardless of their mass. Nevertheless, the methods of science always impose limits on how deep our explanation can go.

Galileo's experiment with dropping heavy and light objects from a tower showed that all falling objects have the same motion, and his inclined-plane experiments showed that the motion was described by \(v=at+v_\text{o}\). The initial velocity \(v_\text{o}\) depends on whether you drop the object from rest or throw it down, but even if you throw it down, you cannot change the slope, \(a\), of the \(v-t\) graph.

Since these experiments show that all falling objects have linear \(v-t\) graphs with the same slope, the slope of such a graph is apparently an important and useful quantity. We use the word acceleration, and the symbol \(a\), for the slope of such a graph. In symbols, \(a=\Delta v/\Delta \)t. The acceleration can be interpreted as the amount of speed gained in every second, and it has units of velocity divided by time, i.e., “meters per second per second,” or m/s/s. Continuing to treat units as if they were algebra symbols, we simplify “m/s/s” to read \(“\text{m}/\text{s}^2\).” Acceleration can be a useful quantity for describing other types of motion besides falling, and the word and the symbol “\(a\)” can be used in a more general context. We reserve the more specialized symbol “\(g\)” for the acceleration of falling objects, which on the surface of our planet equals \(9.8\ \text{m}/\text{s}^2\). Often when doing approximate calculations or merely illustrative numerical examples it is good enough to use \(g=10\ \text{m}/\text{s}^2\), which is off by only 2%.

\(\triangleright\) Approximating \(g\) as \(10\ \text{m}/\text{s}^2\), he will gain 10 m/s of speed each second. After one second, his velocity is 10 m/s, after two seconds it is 20 m/s, and on impact, after falling for three seconds, he is moving at 30 m/s.

\(\triangleright\) Acceleration is defined as the slope of the v-t graph. The graph rises by 3 m/s during a time interval of 3 s, so the acceleration is \((3\ \text{m}/\text{s})/(3\ \text{s})=1\ \text{m}/\text{s}^2\).

Incorrect solution #1: The final velocity is 3 m/s, and acceleration is velocity divided by time, so the acceleration is \((3\ \text{m}/\text{s})/(10\ \text{s})=0.3\ \text{m}/\text{s}^2\).

The solution is incorrect because you can't find the slope of a graph from one point. This person was just using the point at the right end of the v-t graph to try to find the slope of the curve.

Incorrect solution #2: Velocity is distance divided by time so \(v=(4.5\) m)/(3 \(s)=1.5\) m/s. Acceleration is velocity divided by time, so \(a=(1.5\) m/s)/(3 \(s)=0.5\ \text{m}/\text{s}^2\).

The solution is incorrect because velocity is the slope of the tangent line. In a case like this where the velocity is changing, you can't just pick two points on the x-t graph and use them to find the velocity.

\(\triangleright\) What is \(g\) in units of \(\text{cm}/\text{s}^2\)?

\(\triangleright\) The answer is going to be how many cm/s of speed a falling object gains in one second. If it gains 9.8 m/s in one second, then it gains 980 cm/s in one second, so \(g=980\ \text{cm}/\text{s}^2\). Alternatively, we can use the method of fractions that equal one:

\[\begin{equation*}
\frac{9.8\ {\text{m}}}{\text{s}^2}\times\frac{100\ \text{cm}}{1\ {\text{m}}}
=\frac{980\ \text{cm}}{\text{s}^2}
\end{equation*}\]

\(\triangleright\) What is \(g\) in units of \(\text{miles}/\text{hour}^2\)?

\(\triangleright\)

\[\begin{equation*}
\frac{9.8\ \text{m}}{\text{s}^2}
\times \frac{1\ \text{mile}}{1600\ \text{m}}
\times \left(\frac{3600\ \text{s}}{1\ \text{hour}}\right)^2
= 7.9\times 10^4\ \text{mile}/\text{hour}^2
\end{equation*}\]

This large number can be interpreted as the speed, in miles per hour, that you would gain by falling for one hour. Note that we had to square the conversion factor of 3600 s/hour in order to cancel out the units of seconds squared in the denominator.

\(\triangleright\) What is \(g\) in units of miles/hour/s?

\(\triangleright\)

\[\begin{equation*}
\frac{9.8\ \text{m}}{\text{s}^2}
\times \frac{1\ \text{mile}}{1600\ \text{m}}
\times \frac{3600\ \text{s}}{1\ \text{hour}}
= 22\ \text{mile}/\text{hour}/\text{s}
\end{equation*}\]

This is a figure that Americans will have an intuitive feel for. If your car has a forward acceleration equal to the acceleration of a falling object, then you will gain 22 miles per hour of speed every second. However, using mixed time units of hours and seconds like this is usually inconvenient for problem-solving. It would be like using units of foot-inches for area instead of \(\text{ft}^2\) or \(\text{in}^2\).

Everyone knows that gravity is weaker on the moon, but actually it is not even the same everywhere on Earth, as shown by the sampling of numerical data in the following table.

location | latitude | elevation (m) | g textupmtextups2) |

north pole | 90∘N | 0 | 9.8322 |

Reykjavik, Iceland | 64∘N | 0 | 9.8225 |

Guayaquil, Ecuador | 2∘S | 0 | 9.7806 |

Mt. Cotopaxi, Ecuador | 1∘S | 5896 | 9.7624 |

Mt. Everest | 28∘N | 8848 | 9.7643 |

The main variables that relate to the value of \(g\) on Earth are latitude and elevation. Although you have not yet learned how \(g\) would be calculated based on any deeper theory of gravity, it is not too hard to guess why \(g\) depends on elevation. Gravity is an attraction between things that have mass, and the attraction gets weaker with increasing distance. As you ascend from the seaport of Guayaquil to the nearby top of Mt. Cotopaxi, you are distancing yourself from the mass of the planet. The dependence on latitude occurs because we are measuring the acceleration of gravity relative to the earth's surface, but the earth's rotation causes the earth's surface to fall out from under you. (We will discuss both gravity and rotation in more detail later in the course.)

Much more spectacular differences in the strength of gravity can be observed away from the Earth's surface:

location | g textupmtextups2) |

asteroid Vesta (surface) | 0.3 |

Earth’s moon (surface) | 1.6 |

Mars (surface) | 3.7 |

Earth (surface) | 9.8 |

Jupiter (cloud-tops) | 26 |

Sun (visible surface) | 270 |

typical neutron star (surface) | 10 |

black hole (center) | infinite according to some
theories, on the order
of10 |

A typical neutron star is not so different in size from a large asteroid, but is orders of magnitude more massive, so the mass of a body definitely correlates with the \(g\) it creates. On the other hand, a neutron star has about the same mass as our Sun, so why is its \(g\) billions of times greater? If you had the misfortune of being on the surface of a neutron star, you'd be within a few thousand miles of all its mass, whereas on the surface of the Sun, you'd still be millions of miles from most of its mass.

◊

What is wrong with the following definitions of \(g?\)

(1) “\(g\) is gravity.”

(2) “\(g\) is the speed of a falling object.”

(3) “\(g\) is how hard gravity pulls on things.”

◊

When advertisers specify how much acceleration a car is capable of, they do not give an acceleration as defined in physics. Instead, they usually specify how many seconds are required for the car to go from rest to 60 miles/hour. Suppose we use the notation “\(a\)” for the acceleration as defined in physics, and “\(a_\text{car ad}\)” for the quantity used in advertisements for cars. In the US's non-metric system of units, what would be the units of \(a\) and \(a_\text{car ad}\)? How would the use and interpretation of large and small, positive and negative values be different for \(a\) as opposed to \(a_\text{car ad}\)?

◊

Two people stand on the edge of a cliff. As they lean over the edge, one person throws a rock down, while the other throws one straight up with an exactly opposite initial velocity. Compare the speeds of the rocks on impact at the bottom of the cliff.

Gravity always pulls down, but that does not mean it always speeds things up. If you throw a ball straight up, gravity will first slow it down to \(v=0\) and then begin increasing its speed. When I took physics in high school, I got the impression that positive signs of acceleration indicated speeding up, while negative accelerations represented slowing down, i.e., deceleration. Such a definition would be inconvenient, however, because we would then have to say that the same downward tug of gravity could produce either a positive or a negative acceleration. As we will see in the following example, such a definition also would not be the same as the slope of the \(v-t\) graph

Let's study the example of the rising and falling ball. In the example of the person falling from a bridge, I assumed positive velocity values without calling attention to it, which meant I was assuming a coordinate system whose \(x\) axis pointed down. In this example, where the ball is reversing direction, it is not possible to avoid negative velocities by a tricky choice of axis, so let's make the more natural choice of an axis pointing up. The ball's velocity will initially be a positive number, because it is heading up, in the same direction as the \(x\) axis, but on the way back down, it will be a negative number. As shown in the figure, the \(v-t\) graph does not do anything special at the top of the ball's flight, where \(v\) equals 0. Its slope is always negative. In the left half of the graph, there is a negative slope because the positive velocity is getting closer to zero. On the right side, the negative slope is due to a negative velocity that is getting farther from zero, so we say that the ball is speeding up, but its velocity is decreasing!

To summarize, what makes the most sense is to stick with the original definition of acceleration as the slope of the \(v-t\) graph, \(\Delta v/\Delta t\). By this definition, it just isn't necessarily true that things speeding up have positive acceleration while things slowing down have negative acceleration. The word “deceleration” is not used much by physicists, and the word “acceleration” is used unblushingly to refer to slowing down as well as speeding up: “There was a red light, and we accelerated to a stop.”

\(\triangleright\) In figure i, what happens if you calculate the acceleration between \(t=1.0\) and 1.5 s?

\(\triangleright\) Reading from the graph, it looks like the velocity is about \(-1\) m/s at \(t=1.0\) s, and around \(-6\) m/s at \(t=1.5\) s. The acceleration, figured between these two points, is

\[\begin{equation*}
a = \frac{\Delta v}{\Delta t} = \frac{(-6\ \text{m}/\text{s})-(-1\ \text{m}/\text{s})}{(1.5\ \text{s})-(1.0\ \text{s})} = -10\ \text{m}/\text{s}^2 .
\end{equation*}\]

Even though the ball is speeding up, it has a negative acceleration.

Another way of convincing you that this way of handling the plus and minus signs makes sense is to think of a device that measures acceleration. After all, physics is supposed to use operational definitions, ones that relate to the results you get with actual measuring devices. Consider an air freshener hanging from the rear-view mirror of your car. When you speed up, the air freshener swings backward. Suppose we define this as a positive reading. When you slow down, the air freshener swings forward, so we'll call this a negative reading on our accelerometer. But what if you put the car in reverse and start speeding up backwards? Even though you're speeding up, the accelerometer responds in the same way as it did when you were going forward and slowing down. There are four possible cases:

motion of car | accelerometer swings | slope of v-t graph | direction of force acting on car |

forward, speeding up | backward | + | forward |

forward, slowing down | forward | − | backward |

backward, speeding up | forward | − | backward |

backward, slowing down | backward | + | forward |

Note the consistency of the three right-hand columns --- nature is trying to tell us that this is the right system of classification, not the left-hand column.

Because the positive and negative signs of acceleration depend on the choice of a coordinate system, the acceleration of an object under the influence of gravity can be either positive or negative. Rather than having to write things like “\(g=9.8\ \text{m}/\text{s}^2\) or \(-9.8\ \text{m}/\text{s}^2\)” every time we want to discuss \(g\)'s numerical value, we simply define \(g\) as the absolute value of the acceleration of objects moving under the influence of gravity. We consistently let \(g=9.8 \ \text{m}/\text{s}^2\), but we may have either \(a=g\) or \(a=-g\), depending on our choice of a coordinate system.

\(\triangleright\) A person kicks a ball, which rolls up a sloping street, comes to a halt, and rolls back down again. The ball has constant acceleration. The ball is initially moving at a velocity of 4.0 m/s, and after 10.0 s it has returned to where it started. At the end, it has sped back up to the same speed it had initially, but in the opposite direction. What was its acceleration?

\(\triangleright\) By giving a positive number for the initial velocity, the statement of the question implies a coordinate axis that points up the slope of the hill. The “same” speed in the opposite direction should therefore be represented by a negative number, -4.0 m/s. The acceleration is

\[\begin{align*}
a &= \Delta v/\Delta t \\
&= (v_f-v_\text{o})/10.0\ \text{s} \\
&= [(-4.0 \ \text{m}/\text{s})-(4.0 \ \text{m}/\text{s})]/10.0 s \\
&= -0.80\ \text{m}/\text{s}^2 .
\end{align*}\]

The acceleration was no different during the upward part of the roll than on the downward part of the roll.

Incorrect solution: Acceleration is \(\Delta v/\Delta \)t, and at the end it's not moving any faster or slower than when it started, so \(\Delta \)v=0 and \(a=0\).

The velocity does change, from a positive number to a negative number.

◊

A child repeatedly jumps up and down on a trampoline. Discuss the sign and magnitude of his acceleration, including both the time when he is in the air and the time when his feet are in contact with the trampoline.

◊

The figure shows a refugee from a Picasso painting blowing on a rolling water bottle. In some cases the person's blowing is speeding the bottle up, but in others it is slowing it down. The arrow inside the bottle shows which direction it is going, and a coordinate system is shown at the bottom of each figure. In each case, figure out the plus or minus signs of the velocity and acceleration. It may be helpful to draw a \(v-t\) graph in each case.

◊

Sally is on an amusement park ride which begins with her chair being hoisted straight up a tower at a constant speed of 60 miles/hour. Despite stern warnings from her father that he'll take her home the next time she misbehaves, she decides that as a scientific experiment she really needs to release her corndog over the side as she's on the way up. She does not throw it. She simply sticks it out of the car, lets it go, and watches it against the background of the sky, with no trees or buildings as reference points. What does the corndog's motion look like as observed by Sally? Does its speed ever appear to her to be zero? What acceleration does she observe it to have: is it ever positive? negative? zero? What would her enraged father answer if asked for a similar description of its motion as it appears to him, standing on the ground?

◊

Can an object maintain a constant acceleration, but meanwhile reverse the direction of its velocity?

◊

Can an object have a velocity that is positive and increasing at the same time that its acceleration is decreasing?

So far we have only been discussing examples of motion for which the \(v-t\) graph is linear. If we wish to generalize our definition to v-t graphs that are more complex curves, the best way to proceed is similar to how we defined velocity for curved \(x-t\) graphs:

The acceleration of an object at any instant is the slope of the tangent line passing through its \(v\)-versus-\(t\) graph at the relevant point.

\(\triangleright\) The solution is shown in figure l. I've added tangent lines at the two points in question.

(a) To find the slope of the tangent line, I pick two points on the line (not necessarily on the actual curve): \((3.0\ \text{s},28 \text{m}/\text{s})\) and \((5.0\ \text{s},42\ \text{m}/\text{s})\). The slope of the tangent line is \((42\ \text{m}/\text{s}-28\ \text{m}/\text{s})/(5.0\ \text{s} - 3.0\ \text{s})=7.0\ \text{m}/\text{s}^2\).

(b) Two points on this tangent line are \((7.0\ \text{s},47\ \text{m}/\text{s})\) and \((9.0\ \text{s}, 52\ \text{m}/\text{s})\). The slope of the tangent line is \((52\ \text{m}/\text{s}-47\ \text{m}/\text{s})/(9.0\ \text{s} - 7.0\ \text{s})=2.5\ \text{m}/\text{s}^2\).

Physically, what's happening is that at \(t=3.0\ \text{s}\), the skydiver is not yet going very fast, so air friction is not yet very strong. She therefore has an acceleration almost as great as \(g\). At \(t=7.0\ \text{s}\), she is moving almost twice as fast (about 100 miles per hour), and air friction is extremely strong, resulting in a significant departure from the idealized case of no air friction.

In example 6, the \(x-t\) graph was not even used in the solution of the problem, since the definition of acceleration refers to the slope of the \(v-t\) graph. It is possible, however, to interpret an \(x-t\) graph to find out something about the acceleration. An object with zero acceleration, i.e., constant velocity, has an \(x-t\) graph that is a straight line. A straight line has no curvature. A change in velocity requires a change in the slope of the \(x-t\) graph, which means that it is a curve rather than a line. Thus acceleration relates to the curvature of the \(x-t\) graph. Figure m shows some examples.

In example 6, the \(x-t\) graph was more strongly curved at the beginning, and became nearly straight at the end. If the \(x-t\) graph is nearly straight, then its slope, the velocity, is nearly constant, and the acceleration is therefore small. We can thus interpret the acceleration as representing the curvature of the \(x-t\) graph, as shown in figure m. If the “cup” of the curve points up, the acceleration is positive, and if it points down, the acceleration is negative.

Since the relationship between \(a\) and \(v\) is analogous to the relationship between \(v\) and \(x\), we can also make graphs of acceleration as a function of time, as shown in figure n.

◊ Solved problem: Drawing a \(v-t\) graph. — problem 14

◊ Solved problem: Drawing \(v-t\) and \(a-t\) graphs. — problem 20

Figure o summarizes the relationships among the three types of graphs.

◊

Describe in words how the changes in the \(a-t\) graph in figure n/2 relate to the behavior of the \(v-t\) graph.

◊

In each case, pick a coordinate system and draw \(x-t,v-t\), and \(a-t\) graphs. Picking a coordinate system means picking where you want \(x=0\) to be, and also picking a direction for the positive \(x\) axis.

(1) An ocean liner is cruising in a straight line at constant speed.

(2) You drop a ball. Draw two different sets of graphs (a total of 6), with one set's positive \(x\) axis pointing in the opposite direction compared to the other's.

(3) You're driving down the street looking for a house you've never been to before. You realize you've passed the address, so you slow down, put the car in reverse, back up, and stop in front of the house.

A natural question to ask about falling objects is how fast they fall, but Galileo showed that the question has no answer. The physical law that he discovered connects a cause (the attraction of the planet Earth's mass) to an effect, but the effect is predicted in terms of an acceleration rather than a velocity. In fact, no physical law predicts a definite velocity as a result of a specific phenomenon, because velocity cannot be measured in absolute terms, and only changes in velocity relate directly to physical phenomena.

The unfortunate thing about this situation is that the definitions of velocity and acceleration are stated in terms of the tangent-line technique, which lets you go from \(x\) to \(v\) to \(a\), but not the other way around. Without a technique to go backwards from \(a\) to \(v\) to \(x\), we cannot say anything quantitative, for instance, about the \(x-t\) graph of a falling object. Such a technique does exist, and I used it to make the \(x-t\) graphs in all the examples above.

First let's concentrate on how to get \(x\) information out of a \(v-t\) graph. In example p/1, an object moves at a speed of \(20\ \text{m}/\text{s}\) for a period of 4.0 s. The distance covered is \(\Delta x=v\Delta t=(20\ \text{m}/\text{s})\times(4.0\ \text{s})=80\ \text{m}\). Notice that the quantities being multiplied are the width and the height of the shaded rectangle --- or, strictly speaking, the time represented by its width and the velocity represented by its height. The distance of \(\Delta x=80\ \text{m}\) thus corresponds to the area of the shaded part of the graph.

The next step in sophistication is an example like p/2, where the object moves at a constant speed of \(10\ \text{m}/\text{s}\) for two seconds, then for two seconds at a different constant speed of \(20\ \text{m}/\text{s}\). The shaded region can be split into a small rectangle on the left, with an area representing \(\Delta x=20\ \text{m}\), and a taller one on the right, corresponding to another 40 m of motion. The total distance is thus 60 m, which corresponds to the total area under the graph.

An example like p/3 is now just a trivial generalization; there is simply a large number of skinny rectangular areas to add up. But notice that graph p/3 is quite a good approximation to the smooth curve p/4. Even though we have no formula for the area of a funny shape like p/4, we can approximate its area by dividing it up into smaller areas like rectangles, whose area is easier to calculate. If someone hands you a graph like p/4 and asks you to find the area under it, the simplest approach is just to count up the little rectangles on the underlying graph paper, making rough estimates of fractional rectangles as you go along.

That's what I've done in figure q. Each rectangle on the graph paper is 1.0 s wide and \(2\ \text{m}/\text{s}\) tall, so it represents 2 m. Adding up all the numbers gives \(\Delta x=41\ \text{m}\). If you needed better accuracy, you could use graph paper with smaller rectangles.

It's important to realize that this technique gives you \(\Delta x\), not \(x\). The \(v-t\) graph has no information about where the object was when it started.

The following are important points to keep in mind when applying this technique:

- If the range of \(v\) values on your graph does not extend down to zero, then you will get the wrong answer unless you compensate by adding in the area that is not shown.
- As in the example, one rectangle on the graph paper does not necessarily correspond to one meter of distance.
- Negative velocity values represent motion in the opposite direction, so as suggested by figure r, area under the \(t\) axis should be subtracted, i.e., counted as “negative area.”
- Since the result is a \(\Delta x\) value, it only tells you \(x_{after}-x_{before}\), which may be less than the actual distance traveled. For instance, the object could come back to its original position at the end, which would correspond to \(\Delta x\)=0, even though it had actually moved a nonzero distance.

Finally, note that one can find \(\Delta v\) from an \(a-t\) graph using an entirely analogous method. Each rectangle on the \(a-t\) graph represents a certain amount of velocity change.

◊

Roughly what would a pendulum's \(v-t\) graph look like? What would happen when you applied the area-under-the-curve technique to find the pendulum's \(\Delta x\) for a time period covering many swings?

Although the area-under-the-curve technique can be applied to any graph, no matter how complicated, it may be laborious to carry out, and if fractions of rectangles must be estimated the result will only be approximate. In the special case of motion with constant acceleration, it is possible to find a convenient shortcut which produces exact results. When the acceleration is constant, the \(v-t\) graph is a straight line, as shown in the figure. The area under the curve can be divided into a triangle plus a rectangle, both of whose areas can be calculated exactly: \(A=bh\) for a rectangle and \(A=bh/2\) for a triangle. The height of the rectangle is the initial velocity, \(v_\text{o}\), and the height of the triangle is the change in velocity from beginning to end, \(\Delta v\). The object's \(\Delta x\) is therefore given by the equation \(\Delta x = v_\text{o} \Delta t + \Delta v\Delta t/2\). This can be simplified a little by using the definition of acceleration, \(a=\Delta v/\Delta t\), to eliminate \(\Delta v\), giving

\[\begin{multline*}
\Delta x = v_\text{o} \Delta t + \frac{1}{2}a\Delta t^2 .
\shoveright{\text{[motion with}}\\
\text{constant acceleration]}
\end{multline*}\]

Since this is a second-order polynomial in \(\Delta t\), the graph of \(\Delta x\) versus \(\Delta t\) is a parabola, and the same is true of a graph of \(x\) versus \(t\) --- the two graphs differ only by shifting along the two axes. Although I have derived the equation using a figure that shows a positive \(v_\text{o}\), positive \(a\), and so on, it still turns out to be true regardless of what plus and minus signs are involved.

Another useful equation can be derived if one wants to relate the change in velocity to the distance traveled. This is useful, for instance, for finding the distance needed by a car to come to a stop. For simplicity, we start by deriving the equation for the special case of \(v_\text{o}=0\), in which the final velocity \(v_f\) is a synonym for \(\Delta v\). Since velocity and distance are the variables of interest, not time, we take the equation \(\Delta x=\frac{1}{2}a\Delta t^2\) and use \(\Delta t=\Delta v/a\) to eliminate \(\Delta t\). This gives \(\Delta x=(\Delta v)^2/2a\), which can be rewritten as

\[\begin{equation*}
v_f^2 = 2a\Delta x . \shoveright{\text{[motion with constant acceleration, $v_\text{o}=0$]}}
\end{equation*}\]

For the more general case where \(v_\text{o}\ne 0\), we skip the tedious algebra leading to the more general equation,

\[\begin{equation*}
v_f^2 = v_\text{o}^2 + 2a\Delta x . \shoveright{\text{[motion with constant acceleration]}}
\end{equation*}\]

To help get this all organized in your head, first let's categorize the variables as follows:

Variables that change during motion with constant acceleration:

\(x\) ,\(v\), \(t\)

Variable that doesn't change:

\(a\)

If you know one of the changing variables and want to find another, there is always an equation that relates those two:

The symmetry among the three variables is imperfect only because the equation relating \(x\) and \(t\) includes the initial velocity.

There are two main difficulties encountered by students in applying these equations:

- The equations apply only to motion with constant acceleration. You can't apply them if the acceleration is changing.
- Students are often unsure of which equation to use, or may cause themselves unnecessary work by taking the longer path around the triangle in the chart above. Organize your thoughts by listing the variables you are given, the ones you want to find, and the ones you aren't given and don't care about.

\(\triangleright\) You are trying to pull an old lady out of the way of an oncoming truck. You are able to give her an acceleration of \(20\ \text{m}/\text{s}^2\). Starting from rest, how much time is required in order to move her 2 m?

\(\triangleright\) First we organize our thoughts:

Variables given: \(\Delta x\), \(a\), \(v_\text{o}\)

Variables desired: \(\Delta t\)

Irrelevant variables: \(v_f\)

Consulting the triangular chart above, the equation we need is clearly \(\Delta x=v_\text{o}\Delta t+\frac{1}{2}a\Delta t^2\), since it has the four variables of interest and omits the irrelevant one. Eliminating the \(v_\text{o}\) term and solving for \(\Delta t\) gives \(\Delta t=\sqrt{2\Delta x/a}=0.4\ \text{s}\).

◊ Solved problem: A stupid celebration — problem 15

◊ Solved problem: Dropping a rock on Mars — problem 16

◊ Solved problem: The Dodge Viper — problem 18

◊ Solved problem: Half-way sped up — problem 22

◊

In chapter 1, I gave examples of correct and incorrect reasoning about proportionality, using questions about the scaling of area and volume. Try to translate the incorrect modes of reasoning shown there into mistakes about the following question: If the acceleration of gravity on Mars is 1/3 that on Earth, how many times longer does it take for a rock to drop the same distance on Mars?

◊

Check that the units make sense in the three equations derived in this section.

Historically, the first quantitative and well
documented experimental test
of the principle of inertia (p. 80) was performed by Galileo
around 1590 and published decades later when he managed to find a publisher in
the Netherlands that was beyond the reach of
the Roman Inquisition.^{1} It was ingenious but somewhat indirect, and required a layer of
interpretation and extrapolation on top of the actual observations. As described
on p. 93, he established that objects rolling
on inclined planes moved according to mathematical laws that we would today describe
as in section 3.6. He knew that his rolling balls were subject
to friction, as well as random errors due to the limited precision of the water clock that
he used, but he took the approximate agreement of his equations with experiment to indicate
that they gave the results that would be exact in the absence of friction. He also showed,
purely empirically, that when a ball went up or down a ramp inclined at an angle \(\theta\),
its acceleration was proportional to \(\sin\theta\). Again, this required extrapolation to
idealized conditions of zero friction. He then reasoned that if a ball was rolled on a
*horizontal* ramp, with \(\theta=0\), its acceleration would be zero. This is exactly
what is required by the principle of inertia: in the absence of friction, motion continues
indefinitely.

In section 2.7, I discussed how the slope-of-the-tangent-line idea related to the calculus concept of a derivative, and the branch of calculus known as differential calculus. The other main branch of calculus, integral calculus, has to do with the area-under-the-curve concept discussed in section 3.5. Again there is a concept, a notation, and a bag of tricks for doing things symbolically rather than graphically. In calculus, the area under the \(v-t\) graph between \(t=t_1\) and \(t=t_2\) is notated like this:

\[\begin{equation*}
\text{area under curve} = \Delta x = \int_{t_1}^{t_2}v dt .
\end{equation*}\]

The expression on the right is called an integral, and the s-shaped symbol, the integral sign, is read as “integral of ...”

Integral calculus and differential calculus are closely related. For instance, if you take the derivative of the function \(x(t)\), you get the function \(v(t)\), and if you integrate the function \(v(t)\), you get \(x(t)\) back again. In other words, integration and differentiation are inverse operations. This is known as the fundamental theorem of calculus.

On an unrelated topic, there is a special notation for taking the derivative of a function twice. The acceleration, for instance, is the second (i.e., double) derivative of the position, because differentiating \(x\) once gives \(v\), and then differentiating \(v\) gives \(a\). This is written as

\[\begin{equation*}
a = \frac{d^2 x}{dt^2} .
\end{equation*}\]

The seemingly inconsistent placement of the twos on the top and bottom confuses all beginning calculus students. The motivation for this funny notation is that acceleration has units of \(\text{m}/\text{s}^2\), and the notation correctly suggests that: the top looks like it has units of meters, the bottom \(\text{seconds}^2\). The notation is not meant, however, to suggest that \(t\) is really squared.

*gravity* — A general term for the phenomenon of attraction
between things having mass. The attraction between our
planet and a human-sized object causes the object to fall.

*acceleration* — The rate of change of velocity; the slope of
the tangent line on a \(v-t\) graph.

\(v_o\) — initial velocity \(v_f\) — final velocity \(a\) — acceleration \(g\) — the acceleration of objects in free fall; the strength of the local gravitational field

{}

Galileo showed that when air resistance is negligible all falling bodies have the same motion regardless of mass. Moreover, their \(v-t\) graphs are straight lines. We therefore define a quantity called acceleration as the slope, \(\Delta v/\Delta \)t, of an object's \(v-t\) graph. In cases other than free fall, the \(v-t\) graph may be curved, in which case the definition is generalized as the slope of a tangent line on the \(v-t\) graph. The acceleration of objects in free fall varies slightly across the surface of the earth, and greatly on other planets.

Positive and negative signs of acceleration are defined according to whether the \(v-t\) graph slopes up or down. This definition has the advantage that a force with a given sign, representing its direction, always produces an acceleration with the same sign.

The area under the \(v-t\) graph gives \(\Delta x\), and analogously the area under the \(a-t\) graph gives \(\Delta v\).

For motion with constant acceleration, the following three equations hold:

\[\begin{align*}
\Delta x &= v_\text{o}\Delta t + \frac{1}{2}a\Delta t^2 \\
v_f^2 &= v_\text{o}^2 + 2 a \Delta x \\
a &= \frac{\Delta v}{\Delta t}
\end{align*}\]

They are not valid if the acceleration is changing.

The following form can be used for the homework problems that require sketching a set of graphs.

\begin{homeworkforcelabel}{honeybee}{1}{}{1}The graph represents the velocity of a bee along a straight line. At \(t=0\), the bee is at the hive. (a) When is the bee farthest from the hive? (b) How far is the bee at its farthest point from the hive? (c) At \(t=13\) s, how far is the bee from the hive? [Hint: Try problem 19 first.] (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{pond}{1}{}{2}A rock is dropped into a pond. Draw plots of its position versus time, velocity versus time, and acceleration versus time. Include its whole motion, starting from the moment it is dropped, and continuing while it falls through the air, passes through the water, and ends up at rest on the bottom of the pond. Do your work on a photocopy or a printout of page 121. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{warships}{1}{}{3}In an 18th-century naval battle, a cannon ball is shot horizontally, passes through the side of an enemy ship's hull, flies across the galley, and lodges in a bulkhead. Draw plots of its horizontal position, velocity, and acceleration as functions of time, starting while it is inside the cannon and has not yet been fired, and ending when it comes to rest. There is not any significant amount of friction from the air. Although the ball may rise and fall, you are only concerned with its horizontal motion, as seen from above. Do your work on a photocopy or a printout of page 121. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{bunjee}{1}{}{4}Draw graphs of position, velocity, and acceleration as functions of time for a person bunjee jumping. (In bunjee jumping, a person has a stretchy elastic cord tied to his/her ankles, and jumps off of a high platform. At the bottom of the fall, the cord brings the person up short. Presumably the person bounces up a little.) Do your work on a photocopy or a printout of page 121. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{ramp}{1}{}{5}A ball rolls down the ramp shown in the figure, consisting of a curved knee, a straight slope, and a curved bottom. For each part of the ramp, tell whether the ball's velocity is increasing, decreasing, or constant, and also whether the ball's acceleration is increasing, decreasing, or constant. Explain your answers. Assume there is no air friction or rolling resistance. Hint: Try problem 20 first. [Based on a problem by Hewitt.] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{endofarc}{1}{}{6}A toy car is released on one side of a piece of track that is bent into an upright \(U\) shape. The car goes back and forth. When the car reaches the limit of its motion on one side, its velocity is zero. Is its acceleration also zero? Explain using a \(v-t\) graph. [Based on a problem by Serway and Faughn.] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{caracceleration}{1}{}{7}What is the acceleration of a car that moves at a steady velocity of 100 km/h for 100 seconds? Explain your answer. [Based on a problem by Hewitt.] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{auntwanda}{1}{}{8}A physics homework question asks, “If you start from rest and accelerate at 1.54 \(\ \text{m}/\text{s}^2\) for 3.29 s, how far do you travel by the end of that time?” A student answers as follows:

\[\begin{equation*}
1.54 \times 3.29 = 5.07\ \text{m}
\end{equation*}\]

His Aunt Wanda is good with numbers, but has never taken physics. She doesn't know the formula for the distance traveled under constant acceleration over a given amount of time, but she tells her nephew his answer cannot be right. How does she know? \end{homeworkforcelabel}

\begin{homeworkforcelabel}{well}{1}{}{9}You are looking into a deep well. It is dark, and you cannot see the bottom. You want to find out how deep it is, so you drop a rock in, and you hear a splash 3.0 seconds later. How deep is the well? (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{alphac}{2}{}{10}You take a trip in your spaceship to another star.
Setting off, you increase your speed at a constant
acceleration. Once you get half-way there, you start
decelerating, at the same rate, so that by the time you get
there, you have slowed down to zero speed. You see the
tourist attractions, and then head home by the same method.

(a) Find a formula for the time, \(T\), required for the round
trip, in terms of \(d\), the distance from our sun to the
star, and \(a\), the magnitude of the acceleration. Note that
the acceleration is not constant over the whole trip, but
the trip can be broken up into constant-acceleration parts.

(b) The nearest star to the Earth (other than our own sun)
is Proxima Centauri, at a distance of \(d=4\times10^{16}\ \text{m}\).
Suppose you use an acceleration of \(a=10\ \text{m}/\text{s}^2\), just enough
to compensate for the lack of true gravity and make you feel
comfortable. How long does the round trip take, in years?

(c) Using the same numbers for \(d\) and \(a\), find your
maximum speed. Compare this to the speed of light, which is
\(3.0\times10^8\) \ \textup{m}/\textup{s}. (Later in this course, you will learn
that there are some new things going on in physics when one
gets close to the speed of light, and that it is impossible
to exceed the speed of light. For now, though, just use the
simpler ideas you've learned so far.) (answer check available at lightandmatter.com)
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{halftree}{1}{}{11}You climb half-way up a tree, and drop a rock. Then you climb to the top, and drop another rock. How many times greater is the velocity of the second rock on impact? Explain. (The answer is not two times greater.) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{dropandthrow}{1}{}{12}Alice drops a rock off a cliff. Bubba shoots a gun straight down from the edge of the same cliff. Compare the accelerations of the rock and the bullet while they are in the air on the way down. [Based on a problem by Serway and Faughn.] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{parachutecalc}{1}{1}{13}A person is parachute jumping. During the time between when she leaps out of the plane and when she opens her chute, her altitude is given by an equation of the form

\[\begin{equation*}
y = b - c\left(t+ke^{-t/k}\right) ,
\end{equation*}\]

where \(e\) is the base of natural logarithms, and \(b\), \(c\), and
\(k\) are constants. Because of air resistance, her velocity
does not increase at a steady rate as it would for an
object falling in vacuum.

(a) What units would \(b\), \(c\), and \(k\) have to have for the
equation to make sense?

(b) Find the person's velocity, \(v\), as a function of time.
[You will need to use the chain rule, and the fact that
\(d(e^x)/dx=e^x\).] (answer check available at lightandmatter.com)

(c) Use your answer from part (b) to get an interpretation
of the constant \(c\). [Hint: \(e^{-x}\) approaches zero for
large values of \(x\).]

(d) Find the person's acceleration, \(a\), as a function of time.(answer check available at lightandmatter.com)

(e) Use your answer from part (d) to show that if she waits
long enough to open her chute, her acceleration will become very small.
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{x-graph-to-v-graph}{1}{}{14}(solution in the pdf version of the book) The top part of the figure shows the position-versus-time graph for an object moving in one dimension. On the bottom part of the figure, sketch the corresponding v-versus-t graph. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{stupid}{1}{}{15}(solution in the pdf version of the book) On New Year's Eve, a stupid person fires a pistol straight up. The bullet leaves the gun at a speed of 100 \ \textup{m}/\textup{s}. How long does it take before the bullet hits the ground? \end{homeworkforcelabel}

\begin{homeworkforcelabel}{mars-drop-time}{1}{}{16}(solution in the pdf version of the book) If the acceleration of gravity on Mars is 1/3 that on Earth, how many times longer does it take for a rock to drop the same distance on Mars? Ignore air resistance. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{honeybeeaccel}{1}{1}{17}(solution in the pdf version of the book) A honeybee's position as a function of time is given by \(x=10t-t^3\), where \(t\) is in seconds and \(x\) in meters. What is its acceleration at \(t=3.0\) s? \end{homeworkforcelabel}

\begin{homeworkforcelabel}{dodge-viper}{1}{}{18}(solution in the pdf version of the book) In July 1999, Popular Mechanics carried out tests to
find which car sold by a major auto maker could cover a
quarter mile (402 meters) in the shortest time, starting
from rest. Because the distance is so short, this type of
test is designed mainly to favor the car with the greatest
acceleration, not the greatest maximum speed (which is
irrelevant to the average person). The winner was the Dodge
Viper, with a time of 12.08 s. The car's top (and
presumably final) speed was 118.51 miles per hour (52.98
\ \textup{m}/\textup{s}). (a) If a car, starting from rest and moving with
*constant* acceleration, covers a quarter mile in this
time interval, what is its acceleration? (b) What would be
the final speed of a car that covered a quarter mile with
the constant acceleration you found in part a? (c) Based on
the discrepancy between your answer in part b and the
actual final speed of the Viper, what do you conclude about
how its acceleration changed over time?
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{bounce-graph}{1}{}{19}(solution in the pdf version of the book) The graph represents the motion of a ball that rolls up a hill and then back down. When does the ball return to the location it had at \(t=0?\) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{two-ramps}{1}{}{20}(solution in the pdf version of the book) (a) The ball is released at the top of the ramp shown in the figure. Friction is negligible. Use physical reasoning to draw \(v-t\) and \(a-t\) graphs. Assume that the ball doesn't bounce at the point where the ramp changes slope. (b) Do the same for the case where the ball is rolled up the slope from the right side, but doesn't quite have enough speed to make it over the top. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{graph-bouncing}{1}{}{21}(solution in the pdf version of the book) You throw a rubber ball up, and it falls and bounces several times. Draw graphs of position, velocity, and acceleration as functions of time. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{ramp-half-speed}{1}{}{22}(solution in the pdf version of the book) Starting from rest, a ball rolls down a ramp, traveling a distance \(L\) and picking up a final speed \(v\). How much of the distance did the ball have to cover before achieving a speed of \(v/2?\) [Based on a problem by Arnold Arons.] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{chipmunk}{1}{}{23}The graph shows the acceleration of a chipmunk in a TV cartoon. It consists of two circular arcs and two line segments. At \(t=0\).00 \(s\), the chipmunk's velocity is \(-3.10\ \text{m}/\text{s}\). What is its velocity at \(t=10.00\) s? (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{boguscardistance}{1}{}{24}Find the error in the following calculation. A student wants to find the distance traveled by a car that accelerates from rest for 5.0 s with an acceleration of \(2.0\ \text{m}/\text{s}^2\). First he solves \(a=\Delta v/\Delta t\) for \(\Delta v=10 \ \text{m}/\text{s}\). Then he multiplies to find \((10\ \text{m}/\text{s})(5.0\ \text{s})=50\ \text{m}\). Do not just recalculate the result by a different method; if that was all you did, you'd have no way of knowing which calculation was correct, yours or his. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{betteracceldef}{1}{}{25}Acceleration could be defined either as \(\Delta v/\Delta t\) or as the slope of the tangent line on the \(v-t\) graph. Is either one superior as a definition, or are they equivalent? If you say one is better, give an example of a situation where it makes a difference which one you use. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{justifyvx}{1}{}{26}If an object starts accelerating from rest, we have \(v^2=2a\Delta x\) for its speed after it has traveled a distance \(\Delta x\). Explain in words why it makes sense that the equation has velocity squared, but distance only to the first power. Don't recapitulate the derivation in the book, or give a justification based on units. The point is to explain what this feature of the equation tells us about the way speed increases as more distance is covered. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{droppingballs}{1}{}{27}The figure shows a practical, simple experiment for
determining \(g\) to high precision. Two steel balls are
suspended from electromagnets, and are released simultaneously
when the electric current is shut off. They fall through
unequal heights \(\Delta x_1\) and \(\Delta x_2\). A computer
records the sounds through a microphone as first one ball
and then the other strikes the floor. From this recording,
we can accurately determine the quantity \(T\) defined as
\(T=\Delta t_2-\Delta t_1\), i.e., the time lag between the
first and second impacts. Note that since the balls do not
make any sound when they are released, we have no way of
measuring the individual times \(\Delta t_2\) and \(\Delta t_1\).

(a) Find an equation for \(g\) in terms of the measured
quantities \(T\), \(\Delta x_1\) and \(\Delta x_2\).(answer check available at lightandmatter.com)

(b) Check the
units of your equation.

(c) Check that your equation gives
the correct result in the case where \(\Delta x_1\) is very
close to zero. However, is this case realistic?

(d) What
happens when \(\Delta x_1=\Delta x_2\)? Discuss this both
mathematically and physically.
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{railgun}{1}{}{28}The speed required for a low-earth orbit is \(7.9\times10^3\ \text{m}/\text{s}\) (see ch. 10). When a rocket is launched into orbit, it goes up a little at first to get above almost all of the atmosphere, but then tips over horizontally to build up to orbital speed. Suppose the horizontal acceleration is limited to \(3g\) to keep from damaging the cargo (or hurting the crew, for a crewed flight). (a) What is the minimum distance the rocket must travel downrange before it reaches orbital speed? How much does it matter whether you take into account the initial eastward velocity due to the rotation of the earth? (b) Rather than a rocket ship, it might be advantageous to use a railgun design, in which the craft would be accelerated to orbital speeds along a railroad track. This has the advantage that it isn't necessary to lift a large mass of fuel, since the energy source is external. Based on your answer to part a, comment on the feasibility of this design for crewed launches from the earth's surface. \end{homeworkforcelabel}

{Problem 29. This spectacular series of
photos from a 2011 paper by Burrows and Sutton (“Biomechanics of jumping in the flea,” J. Exp. Biology 214:836) shows the flea jumping at about a 45-degree
angle, but for the sake of this estimate just consider the case of a flea jumping vertically.}}

\begin{homeworkforcelabel}{estimate-flea-accel}{1}{}{29}Some fleas can jump as high as 30 cm. The flea only has a short time to build up speed --- the time during which its center of mass is accelerating upward but its feet are still in contact with the ground. Make an order-of-magnitude estimate of the acceleration the flea needs to have while straightening its legs, and state your answer in units of \(g\), i.e., how many “\(g\)'s it pulls.” (For comparison, fighter pilots black out or die if they exceed about 5 or 10 \(g\)'s.) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{alice}{1}{}{30}Consider the following passage from Alice in Wonderland, in which Alice has been falling for a long time down a rabbit hole:

Down, down, down. Would the fall *never* come to an
end? “I wonder how many miles I've fallen by this time?”
she said aloud. “I must be getting somewhere near the
center of the earth. Let me see: that would be four thousand
miles down, I think” (for, you see, Alice had learned
several things of this sort in her lessons in the schoolroom,
and though this was not a *very* good opportunity for
showing off her knowledge, as there was no one to listen to
her, still it was good practice to say it over)...

Alice doesn't know much physics, but let's try to calculate the amount of time it would take to fall four thousand miles, starting from rest with an acceleration of 10 \(\ \text{m}/\text{s}^2\). This is really only a lower limit; if there really was a hole that deep, the fall would actually take a longer time than the one you calculate, both because there is air friction and because gravity gets weaker as you get deeper (at the center of the earth, \(g\) is zero, because the earth is pulling you equally in every direction at once). (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{astronaut-jumping}{1}{}{31}The photo shows Apollo 16 astronaut John Young jumping on the moon and saluting at the top of his jump. The video footage of the jump shows him staying aloft for 1.45 seconds. Gravity on the moon is 1/6 as strong as on the earth. Compute the height of the jump.(answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{stride}{1}{}{32}Most people don't know that *Spinosaurus aegyptiacus*, not *Tyrannosaurus rex*,
was the biggest theropod dinosaur. We can't put a dinosaur on a track and time it in the 100 meter
dash, so we can only infer from physical models how fast it could have run. When an
animal walks at a normal pace, typically its legs swing more or less like pendulums of
the same length \(\ell\). As a further simplification of this model, let's imagine that the leg
simply moves at a fixed acceleration as it falls to the ground.
That is, we model the time for a quarter of a stride cycle as being
the same as the time required for free fall from a height \(\ell\).
*S. aegyptiacus* had legs about four times longer than those of a human.
(a) Compare the time required for a human's stride cycle to that for *S. aegyptiacus*.(answer check available at lightandmatter.com)

(b) Compare their running speeds.(answer check available at lightandmatter.com)
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{punch}{1}{}{33}Engineering professor Qingming Li used sensors and video cameras to study punches delivered in the lab by British former welterweight boxing champion Ricky “the Hitman” Hatton. For comparison, Li also let a TV sports reporter put on the gloves and throw punches. The time it took for Hatton's best punch to arrive, i.e., the time his opponent would have had to react, was about \(0.47\) of that for the reporter. Let's assume that the fist starts from rest and moves with constant acceleration all the way up until impact, at some fixed distance (arm's length). Compare Hatton's acceleration to the reporter's.(answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{aircraft-carrier}{1}{}{34}Aircraft carriers originated in World War I, and the first landing on a carrier was performed by E.H. Dunning in a Sopwith Pup biplane, landing on HMS Furious. (Dunning was killed the second time he attempted the feat.) In such a landing, the pilot slows down to just above the plane's stall speed, which is the minimum speed at which the plane can fly without stalling. The plane then lands and is caught by cables and decelerated as it travels the length of the flight deck. Comparing a modern US F-14 fighter jet landing on an Enterprise-class carrier to Dunning's original exploit, the stall speed is greater by a factor of 4.8, and to accomodate this, the length of the flight deck is greater by a factor of 1.9. Which deceleration is greater, and by what factor?(answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{baseball-pitch-ratios}{1}{}{35}In college-level women's softball in the U.S., typically a pitcher is expected to be at least 1.75 m tall, but Virginia Tech pitcher Jasmin Harrell is 1.62 m. Although a pitcher actually throws by stepping forward and swinging her arm in a circle, let's make a simplified physical model to estimate how much of a disadvantage Harrell has had to overcome due to her height. We'll pretend that the pitcher gives the ball a constant acceleration in a straight line, and that the length of this line is proportional to the pitcher's height. Compare the acceleration Harrell would have to supply with the acceleration that would suffice for a pitcher of the nominal minimum height, if both were to throw a pitch at the same speed.(answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{high-speed-chase}{1}{}{36}When the police engage in a high-speed chase on city streets, it can be extremely dangerous both to the police and to other motorists and pedestrians. Suppose that the police car must travel at a speed that is limited by the need to be able to stop before hitting a baby carriage, and that the distance at which the driver first sees the baby carriage is fixed. Tests show that in a panic stop from high speed, a police car based on a Chevy Impala has a deceleration 9% greater than that of a Dodge Intrepid. Compare the maximum safe speeds for the two cars.(answer check available at lightandmatter.com) \end{homeworkforcelabel}

(c) 1998-2013 Benjamin Crowell, licensed under the Creative Commons Attribution-ShareAlike license. Photo credits are given at the end of the Adobe Acrobat version.

[1] Galileo, *Discourses and Mathematical Demonstrations
Relating to Two New Sciences*, 1638. The experiments are described in the Third Day,
and their support for the principle of inertia is discussed
in the Scholium following
Theorems I-XIV. Another experiment involving a ship is described
in Galileo's 1624 reply to a letter from Fr. Ingoli, but although Galileo vigorously
asserts that he really did carry it out, no detailed description or quantitative results
are given.