MasteringPhysics Homework 1

11/3/13

Homework 1

Homework 1 Due: 5:00pm on Sunday, September 1, 2013 You will receive no credit for items you complete after the assignment is due. Grading Policy

Converting Units The ability to convert from one system of units to another is important in physics. It is often impractical to measure quantities in the standard meters, kilograms, and seconds, but the laws of physics that you learn will involve constants that are defined in these units. Therefore, you may often have to convert your measured quantities into meters, kilograms, and seconds. The following table lists metric prefixes that come up frequently in physics. Learning these prefixes will help you in the various exercises. mega- (M)

×10

kilo- (k)

×10

6

3

centi- (c)

×10

milli- (m)

×10

micro- (μ)

×10

nano- (n)

×10

−2

−3

−6

−9

When doing unit conversions, you need a relation between the two units. For instance, in converting from millimeters to meters, you need to know that 1 m = 1000 mm. Once you know this, you need to divide one side by the other to obtain a ratio of m to mm: 1=

1 m 1000 mm

.

If you are converting from millimeters to meters, then this is the proper ratio. It has mm in the denominator, so that it will cancel the units of the quantity that you are converting. For instance, if you were converting 63 mm, then you would have 63 mm ⋅

1 m 1000

= 0.063 m.

mm

If you were converting a quantity from meters to millimeters, you would use the reciprocal ratio: 1=

1000 mm 1 m

.

Part A Suppose that you measure a pen to be 10.5 cm long. Convert this to meters. Express your answer in meters.

Hint 1. Relating centimeters and meters To solve this problem, you will need to use the relation 100

cm = 1 m

introduction to this problem. If one centimeter equals

meters, then you need 102 centimeters to equal a whole meter, just as you know

that if one quarter equals

−1

4

−2

10

. You can determine such relations using the metric prefixes given in the

1

US dollars, then you need 4 quarters to equal a whole US dollar.

ANSWER: 10.5 cm = 0.105

m

Correct When converting areas, you must be careful to use the correct ratio. If you were converting from

2

mm

to m2 , it might be tempting to use

1 m 1000 mm 2

again. Be careful! Think of mm2 as (mm) = (mm) ⋅ (mm) . That is to say, think of this as a pair of millimeter units, each of which must be converted separately. To convert 130 mm2 to square meters you would use the following calculation: 2

130 mm

2

⋅(

1 m 1000 mm

)⋅(

1 m 1000 mm

session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2469924

) = 130 mm

2

⋅(

1 m 10

3

mm

)

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Homework 1

Notice that the exponent distributes to both the numbers and the units: 130 m m

2

⋅(

2

1 m 3

10

= 130 mm

)

mm

2

1 m

⋅(

10

6

2

mm

2

).

Now the mm2 will cancel properly: 130 mm

2

1 m

⋅( 10

6

2

mm 2

−4

) = 1.30 × 10

m

2

.

Part B Suppose that, from measurements in a microscope, you determine that a certain bacterium covers an area of 1.50

2

μm

. Convert this to square

meters. Express your answer in square meters.

Hint 1. Find the conversion factor Which of the following gives the proper conversion factor to use? From the table in the introduction, you can see that gives

6

10

μm = 1 m

−6

1 μm = 10

m

, which

.

ANSWER: 6

10

μm

2

1 m2 1 m 6

10

12

10

2

μm 2 μm

1 m 1 m 12

10

2

2 2

μm 2

Correct

ANSWER: 1.50 μm2 = 1.50×10−12

2

m

Correct As with areas, you must be careful when converting between volumes. For volumes, you must cancel off three copies of whatever unit you are converting from.

Part C Suppose that you find the volume of all the oceans to be 1.4 × 109 km3 in a reference book. To find the mass, you can use the density of water, also found in this reference book, but first you must convert the volume to cubic meters. What is this volume in cubic meters? Express your answer in cubic meters.

Hint 1. Find the conversion factor Which of the following gives the proper conversion factor to use? From the table in the introduction, you can see that

1 km = 10

3

m

.

ANSWER:

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Homework 1 1 m 6

10

9

10

3

km 3 m

3

1 km 3 3

10

m

3

1 km 3 6

10

km

3

1 m3 1 km 9

10

1 km 3

10

3

m3 3

m3

ANSWER: 9

1.4 × 10

3

km

= 1.40×1018

3

m

Correct

Part D In a laboratory, you determine that the density of a certain solid is

−6

5.23 × 10

3

kg/mm

. Convert this density into kilograms per cubic meter.

Notice that the units you are trying to eliminate are now in the denominator. The same principle from the previous parts applies: Pick the conversion factor so that the units cancel. The only change is that now the units you wish to cancel must appear in the numerator of the conversion factor. Express your answer in kilograms per cubic meter.

Hint 1. Find the conversion factor Which of the following gives the proper conversion factor to use? From the table in the introduction, you can see that that you are trying to cancel units out of the denominator of the fraction.

−3

1 mm = 10

m

. Recall

ANSWER: 1 m 9

10

6

10

3

mm 3 m

3

1 mm 3 3

10

m

3

1 mm 3 9

10

mm

3

1 m3 1 mm 6

10

1 mm 3

10

3

m3 3

m3

ANSWER: −6

5.23 × 10

kg mm 3

= 5230

kg m3

Correct You are now ready to do any sort of unit conversion. You may encounter problems that look far more complex than those you've done in this problem, but if you carefully set up conversion factors one at a time to cancel the units you don't want and replace them with the units that you do want, then you will have no trouble.

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Interpreting Graphs Learning Goal: To be able to gain many different types of information from a graph. Being able to read graphs is an important skill in physics. It is also critical in day-to-day life, as information in the news and in business meetings is often presented in graphical form. In this problem, you will consider a single graph and all of the information that can be gained from it. Since the graph axes have no labels, think of it as a graph of something important to you, whether that is GPA, your bank balance, or something else. Specific applications will be noted for each way of analyzing a graph.

The easiest information to obtain from a graph is its value at a point. The height of the graph above the horizontal axis gives the value of the graph. Points above the horizontal axis have positive values, whereas points below the axis have negative values. The vertical axis will usually have specific values marked off so that you can tell exactly what value each height corresponds to. In the graph you've been given, there are no exact values labeled, but you can still tell relative values; you can make statements such as, "At point D, the graph has a greater value than at point C."

Part A At which point(s) does the graph have a positive value? Enter all of the correct letters in alphabetical order. For instance, if you think that the correct choices are B and F, you would enter BF.

Hint 1. Determining positive values from a graph Any value above the horizontal axis is positive; any below is negative. Where the graph intersects the horizontal axis, the value of the graph is zero, which is neither positive nor negative. ANSWER: ABCDEF

Correct The difference between positive and negative is important in many situations, for instance on your bank statement. In physics it makes a big difference in many scenarios. Positive position means to the right or above some reference point; negative position means to the left or below the reference point. Positive velocity means moving to the right, whereas negative velocity means moving to the left. The graph is often more convenient than a table of numbers or an equation, because you can immediately see where the graph takes positive values and where it takes negative values. With an equation or a table of numbers, this would take some algebra or guess work.

Since the value of the graph at a point is indicated by its height above the horizontal axis, the maximum value of the graph is the highest point on the graph. Similarly, the minimum value of the graph is the lowest point, which may be below the horizontal axis.

Part B At which point does the graph have its maximum value? Enter the correct letter. ANSWER: E

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Correct

Rate of change and slope Another type of information that can be gleaned from a graph is the rate of change of the values. Just as you care whether your bank account has a positive or negative value (i.e., if you have money or owe the bank money) you may also be interested in the rate of change of your bank account. If the rate of change is positive, then you are gaining money. If the rate of change is negative, then you are losing money. The rate of change of a graph is given by the slope of the graph. If the graph is a line, then the slope is just the slope that you are accustomed to for lines (i.e., the change in the vertical position divided by the change in the horizontal position). There are a few important things to remember about slope. If the line tilts upward as you follow it to the right (like this: / ) then the graph has a positive slope. We would say that the graph is increasing (becoming more positive). Similarly, if it tilts downward as you follow it to the right (like this: ∖ ) then the slope is negative and we say that the graph is decreasing (becoming more negative). The steeper the tilt, the larger the magnitude of the rate of change."

Part C Look at the graph from the introduction. The three points C, D, and F are all on straight segments. Rank them from greatest rate of change to least rate of change. Remember to take into account the sign of the rate of change.

Hint 1. Slope at point C When the graph is horizontal, the slope of the graph is zero. You can see this by noticing that the change in vertical position as you move from left to right across a horizontal graph is zero. No matter what the change in horizontal position along a horizontal graph, when you divide to find the slope it will be zero. ANSWER:

Correct

For more complex curves, you will have to draw the tangent line at a point to determine the rate of change of the graph at that point. The tangent is a line that just touches the curve. To do this, instead of passing through two nearby points, the line has to align itself so that its rate of change is the same as the rate of change of the graph at that point. Therefore, once you've drawn the tangent line, its slope is the same as the slope of the curve at that point. In general, you will be able to rely on your intuitive sense of "Is the graph growing higher or lower at this point?" but it's good to keep in mind this more precise definition in terms of the tangent line. It will help you out in situations that are hard to figure out by simple examination. In the following two parts, consider again the graph shown in the introduction to the problem.

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Part D At which point is the graph increasing at the greatest rate? For now, ignore point E. We will discuss it after this part. Enter the correct letter.

Hint 1. Drawing the tangent To find the tangent line at a particular point, you should draw a dot at that point on the graph and then draw pairs of points, one on either side of the point you care about, that are the same distance from the point you care about.

If you then connect those pairs of dots, the lines connecting them will get closer and closer to the proper slope as you move to pairs that are closer and closer to the point you care about. Once you get pretty close to that point, you should be pretty confident of the slope for the tangent line.

With practice, you will gain an intuitive ability to see roughly how the slope of the tangent at a point should look. ANSWER: D

Correct You were told to ignore point E for this part. This is because the rate of change is not well defined at sharp corners. You won't ever be asked for the rate of change of a graph at a sharp corner, though points near the corner should have well-defined rates of change. Points B and C are also special, because the slope at those points is zero. This should be easy to see at C, since the graph is actually a horizontal line in the area near C. If you carefully work out the tangent at point B using the method described in the hint for this part, you will see that the tangent is horizontal there as well. Since a horizontal tangent has a slope of zero, which is neither negative nor positive, the graph is neither increasing nor decreasing at points B and C.

Part E At which point(s) is the graph decreasing? Enter all of the correct letters in alphabetical order. For instance, if you think that the correct choices are B and F, you would enter BF. ANSWER: session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2469924

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Homework 1

FGH

Correct

Area under a graph The other piece of information important to physics that can be found from a graph is the area under the graph between two points on the graph. The light blue region in the figure shows the area under the graph between two points. The area under a graph is important if you have graphed the rate of change in some quantity. In physics, you might have a graph of the velocity of some object vs. time. Since velocity is the rate of change of position, the area under the velocity graph between two times gives the total change in position between those two times. An important point is that if the graph dips beneath the horizontal axis, then the area below the axis is subtracted from the area above the axis. In this figure , the same graph is shown, but now the area is between two more widely separated points. However, if you compare this figure to the previous one, you'll see that equal areas were added above and below the horizontal axis. Therefore, the "area under the graph" in the two figures is the same, even though you see more shading in the second figure. In the following two parts, consider again the graph shown in the introduction to the problem.

Part F You wish to find the area under the graph between the origin and some point on the graph. Which point will yield the greatest area? Enter the correct letter. ANSWER: G

Correct

Part G You are looking at the area under the graph between two points. The area is zero. Which two points are you looking at? Enter the two letters in alphabetical order. For instance, if you think that the correct choice is B and F, you would enter BF.

Hint 1. How to approach the problem Since you want the area under the curve between two points to be zero, the graph must define equal-sized shaded regions above and below the horizontal axis. Look for two points that are near each other, one above and one below the horizontal axis. ANSWER: FH

Correct

Units and Dimensions Conceptual Question d session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2469924

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In the SI unit system, time (t) is measured in seconds, distance (d) is measured in meters, velocity (v) is measured in meters per second, and acceleration (a) is measured in meters per second squared.

Part A Based on these units, select any of the following expressions that could be valid expressions?

Hint 1. Consistent units The left-hand side and the right-hand side of any equation must have the same units. For example, only a quantity measured in meters can be equal to another quantity measured in meters. Just as you can’t compare apples and oranges, you can’t compare meters and seconds. To see if an equation is consistent, substitute in the units of all of the terms in the equation. If the two sides of the equation have the same units, the equation may express some valid physical relationship. ANSWER: a

v=

t d

a=

2

t

− −

t=

3

√d

2

va

v = at d = at − − −

a=

√

vd 3

t

Correct

Consistency of Units In physics, every physical quantity is measured with respect to a unit. Time is measured in seconds, length is measured in meters, and mass is measured in kilograms. Knowing the units of physical quantities will help you solve problems in physics.

Part A Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes the moon to orbit the earth. The force of gravitational attraction is represented by the equation F =

where F is the magnitude of the gravitational attraction on either body, G

m1

Gm1m2 r2

,

and m2 are the masses of the bodies,

r

is the distance between them, and

is the gravitational constant. In SI units, the units of force are kg ⋅ m/s2 , the units of mass are kg, and the units of distance are m. For this

equation to have consistent units, the units of G must be which of the following?

Hint 1. How to approach the problem To solve this problem, we start with the equation F =

Gm1m2 r2

.

For each symbol whose units we know, we replace the symbol with those units. For example, we replace m1 with kg. We now solve this equation for G .

ANSWER:

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Homework 1 3

kg

m⋅s 2 kg⋅s

2

m3 m

3

kg⋅s 2 m kg⋅s 2

Correct

Part B One consequence of Einstein's theory of special relativity is that mass is a form of energy. This mass-energy relationship is perhaps the most famous of all physics equations: 2 E = mc , where m is mass, c is the speed of the light, and E is the energy. In SI units, the units of speed are m/s. For the preceding equation to have consistent units (the same units on both sides of the equation), the units of E must be which of the following?

Hint 1. How to approach the problem To solve this problem, we start with the equation . For each symbol whose units we know, we replace the symbol with those units. For example, we replace m with kg. We now solve this E = mc

2

equation for E .

ANSWER: kg⋅m s kg⋅m

2

s2 kg⋅s

2

m2 kg⋅m

2

s

Correct To solve the types of problems typified by these examples, we start with the given equation. For each symbol whose units we know, we replace the symbol with those units. For example, we replace m with kg. We now solve this equation for the units of the unknown variable.

Significant Figures

Part A To seven significant figures, the mass of a proton is

−27

1.672623 × 10

kg

. Which of the following choices demonstrates correct rounding?

Check all that apply. ANSWER: 1.672 × 10

−27

−27

1.67 × 10

kg

kg −27

1.67263 × 10

kg

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Homework 1

Correct −27

The number 1.672 × 10 is incorrect because when we round to four significant figures we get 1.673, not 1.672. Similarly, is incorrect because when we round to six significant figures we get 1.67262, not 1.67263.

−27

1.67263 × 10

Part B To eight significant figures, Avogadro's constant is

23

6.0221367 × 10

−1

mo l

. Which of the following choices demonstrates correct rounding?

Check all that apply. ANSWER:

6.022 × 10 23

6.0 × 10

23

−1

mo l −1

mo l

23

6.02214 × 10

−1

mo l

Correct All these options are correct; they represent different levels of precision, even though the numerical value is the same.

Significant Figures Conceptual Question In the parts that follow select whether the number presented in statement A is greater than, less than, or equal to the number presented in statement B. Be sure to follow all of the rules concerning significant figures.

Part A Statement A: 2.567 km, to two significant figures. Statement B: 2.567 km, to three significant figures. Determine the correct relationship between the statements.

Hint 1. Rounding and significant figures Rounding to a different number of significant figures changes a number. For example, consider the number 3.4536. This number has five significant figures. The following table illustrates the result of rounding this number to different numbers of significant figures: Four significant figures

: 3.454

Three significant figures : 3.45 Two significant figures One significant figure

: 3.5 :3

Notice that, when rounding 3.4536 to one significant figure, since 0.4536 is less than 0.5, the result is 3, even though if you first rounded to two significant figures (3.5), the result would be 4. ANSWER: greater than Statement A is

less than

Statement B.

equal to

Correct

Part B session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2469924

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Statement A: (2.567 km + 3.146 km), to two significant figures. Statement B: (2.567 km, to two significant figures) + (3.146 km, to two significant figures). Determine the correct relationship between the statements. ANSWER: greater than Statement A is

less than

Statement B.

equal to

Correct Evaluate statement A as follows: (2.567 km + 3.146 km ) = 5.713 km to two significant figures is 5.7 km . Statement B evaluates as 2.6 km + 3.1 km = 5.7 km . Therefore, the two statements are equal.

Part C Statement A: Area of a rectangle with measured length = 2.536 m and width = 1.4 m. Statement B: Area of a rectangle with measured length = 2.536 m and width = 1.41 m. Since you are not told specific numbers of significant figures to round to, you must use the rules for multiplying numbers while respecting significant figures. If you need a reminder, consult the hint. Determine the correct relationship between the statements.

Hint 1. Significant figures and multiplication When you multiply two numbers, the result should be rounded to the number of significant figures in the less accurate of the two numbers. For instance, if you multiply 2.413 (four significant figures) times 3.81 (three significant figures), the result should have three significant figures: 2.413 × 3.81 = 9.19. Similarly, 2 × 7.664323 = 20, when significant figures are respected (i.e., 15.328646 rounded to one significant figure).

ANSWER: greater than Statement A is

less than

Statement B.

equal to

Correct Evaluate statement A as follows: (2.536 m ) (1.4 m ) = 3.5504 m2 to two significant figures is 3.6 m2 . Statement B evaluates as (2.536 m ) (1.41 m ) = 3.57576 m2 to three significant figures is 3.58 m2 . Therefore, statement A is greater than statement B.

Converting between Different Units Unit conversion problems can seem tedious and unnecessary at times. However, different systems of units are used in different parts of the world, so when dealing with international transactions, documents, software, etc., unit conversions are often necessary. Here is a simple example. The inhabitants of a small island begin exporting beautiful cloth made from a rare plant that grows only on their island. Seeing how popular the small quantity that they export has been, they steadily raise their prices. A clothing maker from New York, thinking that he can save money by "cutting out the middleman," decides to travel to the small island and buy the cloth himself. Ignorant of the local custom of offering strangers outrageous prices and then negotiating down, the clothing maker accepts (much to everyone's surpise) the initial price of 400 tepizes/m2 . The price of this cloth in 2

New York is 120 dollars/yard .

Part A If the clothing maker bought 500 m2 of this fabric, how much money did he lose? Use 1

tepiz = 0.625 dollar

and 0.9144

m = 1 yard

.

Express your answer in dollars using two significant figures. session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2469924

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Hint 1. How to approach the problem To find how much money the clothing maker loses, you must find how much money he spent and how much he would have spent in New York. Furthermore, since the problem asks how much he lost in dollars, you need to determine both in dollars. This will require unit conversions.

Hint 2. Find how much he paid 2

If the clothing maker bought 500 m2 at a cost of 400 tepizes/m , then simple multiplication will give how much he spent in tepizes. Once you've found that, convert to dollars. How much did the clothing maker spend in dollars? Express your answer in dollars to three significant figures.

Hint 1. Find how much he paid in

tepizes 2

If the clothing maker bought 500 m2 at a cost of 400 tepizes/m , then how much did he pay in total, in tepizes? Express your answer in

.

tepizes

ANSWER: 2.00×105

tepizes

ANSWER: 1.25×105

dollars

Hint 3. Find the price in New York You know that the price of the fabric in New York is 120 dollars/yard2 . Thus, you need only to find the number of square yards that the clothing maker purchased and then multiply to find the price in New York. What would it have cost him to buy the fabric in New York? Express your answer in dollars to three significant figures.

Hint 1. Determine how much cloth he bought in You are given that

0.9144 m = 1 yard

Express your answer in

yard

2

2

yard

. Squaring both sides, you would get that

2

0.8361 m

= 1 yard

2

. How much is

2

500 m

?

to three significant figures.

ANSWER: 598

yard

2

ANSWER: 7.18×104

dollars

ANSWER: 5.3×104

dollars

Correct

Still think that unit conversion isn't important? Here is a widely publicized, true story about how failing to convert units resulted in a huge loss. In 1998, the Mars Climate Orbiter probe crashed into the surface of Mars, instead of entering orbit. The resulting inquiry revealed that NASA navigators had been making minor course corrections in SI units, whereas the software written by the probe's makers implicitly used British units. In the United States, most scientists use SI units, whereas most engineers use the British, or Imperial, system of units. (Interestingly, British units are not used in Britain.) For these two groups to be able to communicate to one session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2469924

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another, unit conversions are necessary. The unit of force in the SI system is the newton (N), which is defined in terms of basic SI units as 1 N = 1 kg ⋅ m/s2 . The unit of force in the British system is the pound (lb), which is defined in terms of the slug (British unit of mass), foot (f t ), and second (s) as 1 lb = 1 slug ⋅ f t/s2 .

Part B Find the value of 15.0 N in pounds. Use the conversions

1 slug = 14.59 kg

and 1

f t = 0.3048 m

.

Express your answer in pounds to three significant figures.

Hint 1. How to approach the problem When doing a unit conversion, you should begin by comparing the units you are starting with and the units you need to finish with. In this problem, we have the following: Starting units Final units kg⋅m

slug⋅ft

s2

s2

Notice that both have seconds squared in the denominator. You will only have to change the units in the numerator. Match up the units that measure the same quantity (e.g., kilograms and slugs both measure mass). Once you've done this, create a fraction (e.g., 1 hour/60 minutes) based on conversion factors such that the old unit is canceled out of the expression and the new unit appears in the position (i.e., numerator or denominator) of the old unit. In this problem, there are two pairs within the starting and final units that must be converted in this way (i.e., kilograms/slugs and meters/feet).

Hint 2. Calculate the first conversion The first step is to eliminate kilograms from the expression for newtons in favor of slugs. What is the value of 15 kg ⋅ m/s2 in slug ⋅ m/s2 ? Express your answer in slug-meters per second squared to four significant figures. ANSWER: 1.028

ANSWER: 15.0 N = 3.37

lb

Correct Thus, if the NASA navigators believed that they were entering a force value of 15 N (3.37 lb), they were actually entering a value nearly four and a half times higher, 15 lb. Though these errors were only in tiny course corrections, they added up during the trip of many millions of kilometers. In the end, the blame for the loss of the 125-million-dollar probe was placed on the lack of communication between people at NASA that allowed the units mismatch to go unnoticed. Nonetheless, this story makes apparent how important it is to carefully label the units used to measure a number.

Problem 1.1 Spiderman The movie Spiderman brought in $114,000,000 in its opening weekend.

Part A Express this amount in gigadollars. ANSWER: 0.114 gigadollars

Correct

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Part B Express this amount in teradollars. ANSWER: 1.14×10−4 teradollars

Correct

Problem 1.3 The speed of light in a vacuum is approximately 0.3 Gm/s.

Part A Express the speed of light in meters per second. ANSWER: 3×108

m/s

Correct

Conceptual Exercise 1.5

Part A Which of the following equations are dimensionally consistent? Check all that apply. ANSWER:

x =

1 2

at

2

1/2

t = (2x/a) x = vt

Correct

Conceptual Exercise 1.7 Which of the following quantities have the dimensions of a speed?

Part A Which of the following quantities have the dimensions of a speed? ANSWER:

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Correct

Problem 1.9 Acceleration is related to distance and time by the following expression,

a = 2xt

p

.

Part A Find the power p that makes this equation dimensionally consistent. ANSWER: p

= -2

Correct

Problem 1.13 The first several digits of π are known to be π = 3.14159265358979....

Part A What is

π

to three significant figures?

ANSWER: 3.14

Correct

Part B What is

π

to five significant figures?

ANSWER: session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2469924

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Homework 1

3.1416

Correct

Part C What is

π

to seven significant figures?

ANSWER: 3.141593

Correct

Problem 1.15 A parking lot is 144.0m long and 45.06m wide.

Part A What is the perimeter of the lot? Express your answer using four significant figures. ANSWER: P

= 378.1

m

Correct

Problem 1.17

Part A How many significant figures are there in 0.000054? ANSWER: 2 4 6 7

Correct

Part B How many significant figures are there in 3.001 × 105 ? ANSWER:

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1 2 3 4

Correct

Problem 1.21 The Ark of the Covenant is described as a chest of acacia wood 2.5 cubits in length and 1.5 cubits in width and height.

Part A Given that a cubit is equivalent to 17.7 in., find the volume of the ark in cubic feet. Express your answer using two significant figures. ANSWER: V

= 18

ft

3

Correct

Problem 1.23 Water going over Angel Falls, in Venezuela, the world's highest waterfall, drops through a distance of 3212 f t .

Part A What is this distance in km? Express your answer using four significant figures. ANSWER: 0.9788

km

Correct

Problem 1.27 IP Many highways have a speed limit of 65 mi/h.

Part A Is this speed greater than, less than, or equal to 65 km/h? ANSWER: 65 mi/h is greater than 65 km/h. 65 mi/h is less than 65 km/h. 65 mi/h is equal to 65 km/h.

Correct

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Part B Explain. ANSWER: 3680 Character(s) remaining If you convert 65 miles per hour to kilometers per hours, you obtain 105 km/h which is significantlt more than 65 km/h.

Graded, see 'My Answers' for details

Part C Find the speed limit in km/h that corresponds to 65 mi/h. Express your answer using two significant figures. ANSWER: v

= 100

km/h

Correct

Conceptual Exercise 1.41

Part A Which of the following quantities have the dimensions of an acceleration? ANSWER: 2

v /x v/t x/t xt

2

2

Correct

Problem 1.43 Glacial Speed On June 9, 1983, the lower part of the Variegated Glacier in Alaska was observed to be moving at a rate of 210 feet per day.

Part A What is this speed in meters per second? ANSWER: 7.41×10−4

m/s

Correct

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Problem 1.47 Human Nerve Fibers Type A nerve fibers in humans can conduct nerve impulses at speeds up to 140 m/s.

Part A How fast are the nerve impulses in miles per hour? Express your answer using two significant figures. ANSWER: 310

mi/h

Correct

Part B How far (in meters) can the impulses travel in 5.0 ms? Express your answer using two significant figures. ANSWER: 0.70

m

Correct

Problem 1.49 NASA's Cassini mission to Saturn released a probe on December 25, 2004, that landed on the Saturnian moon Titan on January 14, 2005. The probe, which was named Huygens, was released with a gentle relative speed of 31 cm/s. As Huygens moved away from the main spacecraft, it rotated at a rate of seven revolutions per minute.

Part A How many revolutions had Huygens completed when it was 130 yards from the mother ship? Express your answer using two significant figures. ANSWER: N

= 44

Correct

Part B How far did Huygens move away from the mother ship during each revolution? Give your answer in feet. Express your answer using two significant figures. ANSWER: d

= 8.7

ft

Correct

Problem 1.51 session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2469924

L

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The period T of a simple pendulum is the amount of time required for it to undergo one complete oscillation. If the length of the pendulum is acceleration of gravity is g, then T is given by T = 2πL

p

g

L

and the

q

Part A Find the powers

p

and q required for dimensional consistency.

Enter your answers numerically separated by a comma. ANSWER: ,

p q

= 0.5,-0.5

Correct

Passage Problem 1.53 Using a Cricket as a Thermometer All chemical reactions, whether organic or inorganic, proceed at a rate that depends on temperature-the higher the temperature, the higher the rate of reaction. This can be understood in terms of molecules moving with increased energy as the temperature is increased, and colliding with other molecules more frequently. In the case of organic reactions, the result is that metabolic processes speed up with increasing temperature. An increased or decreased metabolic rate can manifest itself in a number of ways. For example, a cricket trying to attract a mate chirps at a rate that depends on its overall rate of metabolism. As a result, the chirping rate of crickets depends directly on temperature. In fact, some people even use a pet cricket as a thermometer. The cricket that is most accurate as a thermometer is the snowy tree cricket (Oecanthus fultoni Walker). Its rate of chirping is described by the following formula: N

In this expression,

T

=

number of chirps per 13.0 seconds

=

T − 40.0

is the temperature in degrees Fahrenheit.

Part A

Which plot in the figure represents the chirping rate of the snowy tree cricket? ANSWER: A B C D E

Correct

Passage Problem 1.54 session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2469924

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Using a Cricket as a Thermometer All chemical reactions, whether organic or inorganic, proceed at a rate that depends on temperature-the higher the temperature, the higher the rate of reaction. This can be understood in terms of molecules moving with increased energy as the temperature is increased, and colliding with other molecules more frequently. In the case of organic reactions, the result is that metabolic processes speed up with increasing temperature. An increased or decreased metabolic rate can manifest itself in a number of ways. For example, a cricket trying to attract a mate chirps at a rate that depends on its overall rate of metabolism. As a result, the chirping rate of crickets depends directly on temperature. In fact, some people even use a pet cricket as a thermometer. The cricket that is most accurate as a thermometer is the snowy tree cricket (Oecanthus fultoni Walker). Its rate of chirping is described by the following formula: N

In this expression,

T

=

number of chirps per 13.0 seconds

=

T − 40.0

is the temperature in degrees Fahrenheit.

Part A If the temperature is 43 degrees Fahrenheit, how long does it take for the cricket to chirp 12 times? ANSWER: 12s 24s 52s 43s

Correct

Passage Problem 1.55 Using a Cricket as a Thermometer All chemical reactions, whether organic or inorganic, proceed at a rate that depends on temperature-the higher the temperature, the higher the rate of reaction. This can be understood in terms of molecules moving with increased energy as the temperature is increased, and colliding with other molecules more frequently. In the case of organic reactions, the result is that metabolic processes speed up with increasing temperature. An increased or decreased metabolic rate can manifest itself in a number of ways. For example, a cricket trying to attract a mate chirps at a rate that depends on its overall rate of metabolism. As a result, the chirping rate of crickets depends directly on temperature. In fact, some people even use a pet cricket as a thermometer. The cricket that is most accurate as a thermometer is the snowy tree cricket (Oecanthus fultoni Walker). Its rate of chirping is described by the following formula: N

In this expression,

T

=

number of chirps per 13.0 seconds

=

T − 40.0

is the temperature in degrees Fahrenheit.

Part A Your pet cricket chirps 112 times in one minute (60.0 s). What is the temperature in degrees Fahrenheit? ANSWER: 41.9 47.0 64.3 80.0

Correct

Passage Problem 1.56 Using a Cricket as a Thermometer All chemical reactions, whether organic or inorganic, proceed at a rate that depends on temperature-the higher the temperature, the higher the rate of session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2469924

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reaction. This can be understood in terms of molecules moving with increased energy as the temperature is increased, and colliding with other molecules more frequently. In the case of organic reactions, the result is that metabolic processes speed up with increasing temperature. An increased or decreased metabolic rate can manifest itself in a number of ways. For example, a cricket trying to attract a mate chirps at a rate that depends on its overall rate of metabolism. As a result, the chirping rate of crickets depends directly on temperature. In fact, some people even use a pet cricket as a thermometer. The cricket that is most accurate as a thermometer is the snowy tree cricket (Oecanthus fultoni Walker). Its rate of chirping is described by the following formula: N

In this expression,

T

=

number of chirps per 13.0 seconds

=

T − 40.0

is the temperature in degrees Fahrenheit.

Part A Suppose a snowy cricket is chirping when the temperature is 45.8 degrees Fahrenheit. How many oscillations does the radiation from a cesium-133 atom complete between successive chirps? ANSWER: 2.80×108 2.91×109 2.06×1010 3.09×1010

Correct

± Average Velocity from a Position vs. Time Graph Learning Goal: To learn to read a graph of position versus time and to calculate average velocity. In this problem you will determine the average velocity of a moving object from the graph of its position x(t) as a function of time t. A traveling object might move at different speeds and in different directions during an interval of time, but if we ask at what constant velocity the object would have to travel to achieve the same displacement over the given time interval, that is what we call the object's average velocity. We will use the notation vave [t1 , t2 ] to indicate average velocity over the time interval from t1 to t2 . For instance, vave [1, 3] is the average velocity over the time interval from t = 1 to t = 3.

Part A Consulting the graph shown in the figure, find the object's average velocity over the time interval from 0 to 1 second. Answer to the nearest integer.

Hint 1. Definition of average velocity Average velocity is defined as the constant velocity at which an object would have to travel to achieve a given displacement (difference between final and initial positions, which can be negative) over a given time interval, from the initial time ti to the final time tf . The average velocity is therefore equal to the displacement divided by the given time interval. In symbolic form, average velocity is given by vave [ti , tf ] =

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x(tf )−x(ti ) tf − ti

.

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ANSWER: vave [0, 1]

= 0

m/s

Correct

Part B Find the average velocity over the time interval from 1 to 3 seconds. Express your answer in meters per second to the nearest integer.

Hint 1. Find the change in position The final and initial positions can be read off the y axis of the graph. What is the displacement during the time interval from 1 to 3 seconds? Express your answer numerically, in meters ANSWER: xf − xi

= 40

m

Hint 2. Definition of average velocity Average velocity is defined as the constant velocity at which an object would have to travel to achieve a given displacement (difference between final and initial positions, which can be negative) over a given time interval, from the initial time ti to the final time tf . The average velocity is therefore equal to the displacement divided by the given time interval. In symbolic form, average velocity is given by vave [ti , tf ] =

x(tf )−x(ti ) tf − ti

.

ANSWER: vave [1, 3]

= 20

m/s

Correct A note about instantaneous velocity. The instantaneous velocity at a certain moment in time is represented by the slope of the graph at that moment. For straight-line graphs, the (instantaneous) velocity remains constant over the interval, so the instantaneous velocity at any time during an interval is the same as the average velocity over that interval. For instance, in this case, the instantaneous velocity at any time from 1 to 3 seconds is the same as the average velocity of 20 m/s.

Part C Now find vave [0, 3]. Give your answer to three significant figures.

Hint 1. A note on the displacement Since the object's position remains constant from time 0 to time 1, the object's displacement from 0 to 3 is the same as in Part B. However, the time interval has changed. ANSWER: vave [0, 3]

= 13.3

m/s

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Correct Note that

v ave [0, 3]

is not equal to the simple arithmetic average of vave [0, 1] and vave [1, 3], i.e.,

v ave [0,1]+ vave [1,3] 2

, because they are averages

for time intervals of different lengths.

Part D Find the average velocity over the time interval from 3 to 6 seconds. Express your answer to three significant figures.

Hint 1. Determine the displacement What is the displacement? Answer to the nearest integer. ANSWER: x(6.0) − x(3.0)

= -40

m

Hint 2. Determine the time interval What is the time interval? Answer to two significant figures. ANSWER: tf − ti

= 3.0

s

ANSWER: vave [3.0, 6.0]

= -13.3

m/s

Correct

Part E Finally, find the average velocity over the whole time interval shown in the graph. Express your answer to three significant figures.

Hint 1. Determine the displacement What is the displacement? Answer to the nearest integer. ANSWER: x(6.0) − x(0.0)

= 0

m

ANSWER: vave [0.0, 6.0]

= 0

m/s

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Correct Note that though the average velocity is zero for this time interval, the instantaneous velocity (i.e., the slope of the graph) has several different values (positive, negative, zero) during this time interval. Note as well that since average velocity over a time interval is defined as the change in position (displacement) in the given interval divided by the time, the object can travel a great distance (here 80 meters) and still have zero average velocity, since it ended up exactly where it started. Therefore, zero average velocity does not necessarily mean that the object was standing still the entire time!

Running and Walking Ilya and Anya each can run at a speed of 7.00mph and walk at a speed of 4.00mph . They set off together on a route of length 5.00miles . Anya walks half of the distance and runs the other half, while Ilya walks half of the time and runs the other half.

Part A How long does it take Anya to cover the distance of 5.00miles ? Express your answer numerically, in minutes.

Hint 1. Compute the time that Anya walks Find the time tAnya,walk that it takes Anya to walk the first half of the distance, that is, to travel a distance 2.50miles at a speed of 4.00mph . Express your answer numerically, in minutes. ANSWER: tAnya,walk

= 37.5 minutes

Correct Hint 2. Compute the time that Anya runs Now find the time tAnya,run that Anya spends running. Express your answer numerically, in minutes. ANSWER: tAnya,run

= 21.4 minutes

Correct Hint 3. What equation to use Now just add the two times and obtain Anya's total time tAnya .

ANSWER: tAnya

= 58.9 minutes

Correct

Part B Find Anya's average speed. Express Anya's average speed

s ave,Anya

numerically, in miles per hour.

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Hint 1. Calculate Anya's average speed using the total distance and time You were given the total distance and have calculated the total time. Recall that average speed is equal to total distance traveled divided by the time it took to travel that distance. ANSWER: s ave,Anya

= 5.09 mph

Correct

Part C How long does it take Ilya to cover the distance? Express the time

tIlya

taken by Ilya numerically, in minutes.

Hint 1. Solve a related problem Consider a related problem. Ilya walks for a time tIlya /2 at speed 4.00mph and runs for a time tIlya /2 at speed 7.00mph . Find the total distance dIlya that he travels in time tIlya in terms of given quantities and tIlya . In the answer box, write the factor that multiplies tIlya in your expression for the distance (in units of miles per hour). ANSWER: = 5.50

dIlya

tIlya

ANSWER: tIlya

= 54.5 minutes

Correct

Part D Now find Ilya's average speed. Express Ilya's average speed

s ave,Ilya

numerically, in miles per hour.

Hint 1. Calculate Ilya's average speed using the total distance and total time You were given the total distance and have calculated the total time. Recall that average speed is equal to total distance traveled divided by the amount of time it took to travel that distance. ANSWER: s ave,Ilya

= 5.50

Correct From this result, you can see that when a journey consists of equal amounts of time walking and running, the average speed will be equal to the simple average of the walking and running speeds. This is not the case when the journey is broken into walking and running segments in some other manner.

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To describe the motion of a particle along a straight line, it is often convenient to draw a graph representing the position of the particle at different times. This type of graph is usually referred to as an x vs. t graph. To draw such a graph, choose an axis system in which time t is plotted on the horizontal axis and position x on the vertical axis. Then, indicate the values of x at various times t. Mathematically, this corresponds to plotting the variable x as a function of t. An example of a graph of position as a function of time for a particle traveling along a straight line is shown below. Note that an x vs. t graph like this does not represent the path of the particle in space. Now let's study the graph shown in the figure in more detail. Refer to this graph to answer Parts A, B, and C.

Part A What is the total distance Δx traveled by the particle? Express your answer in meters.

Hint 1. Total distance The total distance Δx traveled by the particle is given by the difference between the initial position x0 at t = 50.0 s

t = 0.0 s

and the position x at

. In symbols, Δx = x − x0

.

Hint 2. How to read an x vs. t graph Remember that in an x vs. t graph, time t is plotted on the horizontal axis and position x on the vertical axis. For example, in the plot shown in the figure, x = 16.0 m at t = 10.0 s.

ANSWER: Δx

= 30

m

Correct

Part B What is the average velocity

v av

of the particle over the time interval

Δt = 50.0 s

?

Express your answer in meters per second.

Hint 1. Definition and graphical interpretation of average velocity The average velocity

vav

of a particle that travels a distance Δx along a straight line in a time interval v av =

Δx Δt

Δt

is defined as

.

In an x vs. t graph, then, the average velocity equals the slope of the line connecting the initial and final positions.

Hint 2. Slope of a line The slope m of a line from point A, with coordinates

(xA , y

A

)

, to point B, with coordinates

m= session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2469924

xB − xA

(xB , y

B

)

, is equal to the "rise" over the "run," or

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Homework 1 m=

xB − xA y B −y A

.

ANSWER: vav

= 0.600

m/s

Correct The average velocity of a particle between two positions is equal to the slope of the line connecting the two corresponding points in an x vs. t graph.

Part C What is the instantaneous velocity

v

of the particle at

t = 10.0 s

?

Express your answer in meters per second.

Hint 1. Graphical interpretation of instantaneous velocity The velocity of a particle at any given instant of time or at any point in its path is called instantaneous velocity. In an x vs. t graph of the particle's motion, you can determine the instantaneous velocity of the particle at any point in the curve. The instantaneous velocity at any point is equal to the slope of the line tangent to the curve at that point. ANSWER: v

= 0.600

m/s

Answer Requested The instantaneous velocity of a particle at any point on its x vs. t graph is the slope of the line tangent to the curve at that point. Since in the case at hand the curve is a straight line, the tangent line is the curve itself. Physically, this means that the instantaneous velocity of the particle is constant over the entire time interval of motion. This is true for any motion where distance increases linearly with time.

Another common graphical representation of motion along a straight line is the v vs. t graph, that is, the graph of (instantaneous) velocity as a function of time. In this graph, time t is plotted on the horizontal axis and velocity v on the vertical axis. Note that by definition, velocity and acceleration are vector quantities. In straight-line motion, however, these vectors have only one nonzero component in the direction of motion. Thus, in this problem, we will call v the velocity and a the acceleration, even though they are really the components of the velocity and acceleration vectors in the direction of motion.

Part D Which of the graphs shown is the correct v vs. t plot for the motion described in the previous parts?

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Hint 1. How to approach the problem Recall your results found in the previous parts, namely the fact that the instantaneous velocity of the particle is constant. Which graph represents a variable that always has the same constant value at any time? ANSWER: Graph A Graph B Graph C Graph D

Correct Whenever a particle moves with constant nonzero velocity, its x vs. t graph is a straight line with a nonzero slope, and its v vs. t curve is a horizontal line.

Part E Shown in the figure is the v vs. t curve selected in the previous part. What is the area A of the shaded region under the curve? Express your answer in meters.

Hint 1. How to approach the problem The shaded region under the v vs. t curve is a rectangle whose horizontal and vertical sides lie on the t axis and the v axis, respectively. Since session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2469924

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the area of a rectangle is the product of its sides, in this case the area of the shaded region is the product of a certain quantity expressed in seconds and another quantity expressed in meters per second. The area itself, then, will be in meters. ANSWER: A

= 30

m

Correct Compare this result with what you found in Part A. As you can see, the area of the region under the v vs. t curve equals the total distance traveled by the particle. This is true for any velocity curve and any time interval: The area of the region that extends over a time interval Δt under the v vs. t curve is always equal to the distance traveled in Δt.

Analyzing Position versus Time Graphs: Conceptual Question Two cars travel on the parallel lanes of a two-lane road. The cars’ motions are represented by the position versus time graph shown in the figure. Answer the questions using the times from the graph indicated by letters.

Part A At which of the times do the two cars pass each other?

Hint 1. Two cars passing Two objects can pass each other only if they have the same position at the same time. ANSWER: A B C D E None Cannot be determined

Correct

Part B Are the two cars traveling in the same direction when they pass each other? ANSWER:

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yes no

Correct

Part C At which of the lettered times, if any, does car #1 momentarily stop?

Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the "rise" (change in position) over the "run" (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: A B C D E none cannot be determined

Correct

Part D At which of the lettered times, if any, does car #2 momentarily stop?

Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the "rise" (change in position) over the "run" (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: A B C D E none cannot be determined

Correct

Part E At which of the lettered times are the cars moving with nearly identical velocity?

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Hint 1. Determining Velocity from a Position versus Time Graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: A B C D E None Cannot be determined

Correct

Velocity from Graphs of Position versus Time An object moves along the x axis during four separate trials. Graphs of position versus time for each trial are shown in the figure.

Part A During which trial or trials is the object's velocity not constant? Check all that apply.

Hint 1. Finding velocity from a position versus time graph On a graph of coordinate x as a function of time t, the instantaneous speed at any point is equal to the slope of the curve at that point.

Hint 2. Equation for slope The slope of a line is its rise divided by the run: slope =

Δx Δt

.

ANSWER:

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Trial A Trial B Trial C Trial D

Correct The graph of the motion during Trial B has a changing slope and therefore is not constant. The other trials all have graphs with constant slope and thus correspond to motion with constant velocity.

Part B During which trial or trials is the magnitude of the average velocity the largest? Check all that apply.

Hint 1. Definition of average velocity Recall that average velocity =

Δ(position) Δ(time)

=

Δx Δt

.

Then note that the question asks only about the magnitude of the velocity. ANSWER: Trial A Trial B Trial C Trial D

All attempts used; correct answer displayed While Trial B and Trial D do not have the same average velocity, the only difference is the direction! The magnitudes are the same. Neither one is "larger" than the other, and it is only because of how we chose our axes that Trial B has a positive average velocity while Trial D has a negative average velocity. In Trial C the object does not move, so it has an average velocity of zero. During Trial A the object has a positive average velocity but its magnitude is less than that in Trial B and Trial D.

Given Positions, Find Velocity and Acceleration Learning Goal: To graph position, velocity, and acceleration of an object given a table of positions vs. time. The table below shows the x coordinate of a moving object. The position is tabulated at one-second intervals. The x coordinate is indicated below each time. time (s)

0

1

2

3

4

5

6

7

8

9

x (m)

0

1

4

9

16

24

32

40

46

48

Part A Which graph best represents the function x(t), describing the object's position vs. time?

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ANSWER: 1 2 3 4

Correct

Part B Which of the following graphs best represents the function v(t), describing the object's velocity as a function of time?

Hint 1. Find the velocity toward the end of the motion The velocity is given by the slope of the displacement (position) curve. Using this fact, the velocity toward the end of the motion is __________.

Hint 1. Finding the slope The slope of a displacement

x

versus time t curve is given by the equation slope =

x2 −x1 t2 −t1

.

ANSWER: positive and increasing positive and decreasing negative and increasing negative and decreasing

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Hint 2. Identify the implications of zero velocity Two of the possible velocity vs. time graphs indicate zero velocity between t graph look like in this region?

= 4

and t

= 7 s

. What would the corresponding position vs. time

ANSWER: a horizontal line straight but sloping up to the right straight but sloping down to the right curved upward curved downward

ANSWER: 1 2 3 4

Correct In principle, you could also compute and plot the average velocity. The expression for the average velocity is vavg [t1 , t2 ] =

x(t2 )−x(t1 ) t2 −t1

.

The notation vavg [t1 , t2 ] emphasizes that this is not an instantaneous velocity, but rather an average over an interval from

t1

to t2 . After you

compute this, you must put a single point on the graph of velocity vs. time. The most accurate place to plot the average velocity is at the middle of the time interval over which the average was computed.

Part C Which of the following graphs best represents the function a(t), describing the acceleration of this object?

Hint 1. Find the acceleration toward the end of the motion The acceleration is given by the slope of the velocity curve. Toward the end of the motion, when the velocity is decreasing, the acceleration must be __________.

Hint 1. Finding the slope The slope of a displacement

x

versus time t curve is given by the equation \large{{\rm slope} = \frac {v_2 - v_1}{t_2 - t_1}}.

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zero positive negative

Hint 2. Calculate the acceleration in the region of constant velocity What is the acceleration \texttip{a}{a} over the interval during which the object travels at constant speed? Express your answer numerically in meters per second per second. ANSWER: \texttip{a}{a} = 0 \rm m/s^2

ANSWER: 1 2 3 4

Correct In one dimension, a linear increase or decrease in the velocity of an object over a given time interval implies constant acceleration over that particular time interval. You can find the magnitude of the acceleration using the formula for average acceleration over a time interval: \large{a_{\rm avg}[t_1,t_2] = \frac{v(t_2)-v(t_1)}{t_2-t_1}}. When the acceleration is constant over an extended interval, you can choose any value of \texttip{t_{\rm 1}}{t_1} and \texttip{t_{\rm 2}}{t_2} within the interval to compute the average.

What Velocity vs. Time Graphs Can Tell You A common graphical representation of motion along a straight line is the v vs. t graph, that is, the graph of (instantaneous) velocity as a function of time. In this graph, time \texttip{t}{t} is plotted on the horizontal axis and velocity \texttip{v}{v} on the vertical axis. Note that by definition, velocity and acceleration are vector quantities. In straight-line motion, however, these vectors have only a single nonzero component in the direction of motion. Thus, in this problem, we will call \texttip{v}{v} the velocity and \texttip{a}{a} the acceleration, even though they are really the components of the velocity and acceleration vectors in the direction of motion, respectively. Here is a plot of velocity versus time for a particle that travels along a straight line with a varying velocity. Refer to this plot to answer the following questions.

Part A What is the initial velocity of the particle, \texttip{v_{\rm 0}}{v_0}? Express your answer in meters per second.

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Hint 1. Initial velocity The initial velocity is the velocity at t=0\;\rm s.

Hint 2. How to read a v vs. t graph Recall that in a graph of velocity versus time, time is plotted on the horizontal axis and velocity on the vertical axis. For example, in the plot shown in the figure, v=2.00\;\rm m/s at t=30.0\;\rm s.

ANSWER: \texttip{v_{\rm 0}}{v_0} = 0.5 \rm m/s

Correct

Part B What is the total distance \texttip{\Delta x}{Deltax} traveled by the particle? Express your answer in meters.

Hint 1. How to approach the problem Recall that the area of the region that extends over a time interval \texttip{\Delta t}{Deltat} under the v vs. t curve is always equal to the distance traveled in \texttip{\Delta t}{Deltat}. Thus, to calculate the total distance, you need to find the area of the entire region under the v vs. t curve. In the case at hand, the entire region under the v vs. t curve is not an elementary geometrical figure, but rather a combination of triangles and rectangles.

Hint 2. Find the distance traveled in the first 20.0 seconds What is the distance \texttip{\Delta x_{\rm 1}}{Deltax_1} traveled in the first 20 seconds of motion, between t=0.0\;\rm s and t=20.0\;\rm s? Express your answer in meters.

Hint 1. Area of the region under the v vs. t curve The region under the v vs. t curve between t=0.0\;\rm s and t=20.0\;\rm s can be divided into a rectangle of dimensions 20.0\;\rm s by 0.50\;\rm m/s, and a triangle of base 20.0\;\rm s and height 1.50\;\rm m/s, as shown in the figure.

ANSWER: \texttip{\Delta x_{\rm 1}}{Deltax_1} = 25 \rm m

Hint 3. Find the distance traveled in the second 20.0 seconds What is the distance \texttip{\Delta x_{\rm 2}}{Deltax_2} traveled in the second 20 seconds of motion, from t=20.0\;\rm s to t=40.0\;\rm s?

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Express your answer in meters.

Hint 1. Area of the region under the v vs. t curve The region under the v vs. t curve between t=20.0\;\rm s and t=40.0\;\rm s is a rectangle of dimensions 20.0\;\rm s by 2.00\;\rm m/s, as shown in the figure.

ANSWER: \texttip{\Delta x_{\rm 2}}{Deltax_2} = 40 \rm m

Hint 4. Find the distance traveled in the last 10.0 seconds What is the distance \texttip{\Delta x_{\rm 3}}{Deltax_3} traveled in the last 10 seconds of motion, from t=40.0\;\rm s to t=50.0\;\rm s? Express your answer in meters.

Hint 1. Area of the region under the v vs. t curve The region under the v vs. t curve between t=40.0\;\rm s and t=50.0\;\rm s is a triangle of base 10.0\;\rm s and height 2.00\;\rm m/s, as shown in the figure.

ANSWER: \texttip{\Delta x_{\rm 3}}{Deltax_3} = 10 \rm m

ANSWER: \texttip{\Delta x}{Deltax} = 75 \rm m

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Answer Requested

Part C What is the average acceleration \texttip{a_{\rm av}}{a_av} of the particle over the first 20.0 seconds? Express your answer in meters per second per second.

Hint 1. Definition and graphical interpretation of average acceleration The average acceleration \texttip{a_{\rm av}}{a_av} of a particle that travels along a straight line in a time interval \texttip{\Delta t}{Deltat} is the ratio of the change in velocity \texttip{\Delta v}{Deltav} experienced by the particle to the time interval \texttip{\Delta t}{Deltat}, or \large{a_{\rm av}=\frac{\Delta v}{\Delta t}}. In a v vs. t graph, then, the average acceleration equals the slope of the line connecting the two points representing the initial and final velocities.

Hint 2. Slope of a line The slope \texttip{m}{m} of a line from point A, of coordinates (x_{\rm A}, y_{\rm A}), to point B, of coordinates (x_{\rm B}, y_{\rm B}), is equal to the "rise" over the "run," or \large{m=\frac{y_{\rm B}-y_{\rm A}}{x_{\rm B}-x_{\rm A}}}.

ANSWER: \texttip{a_{\rm av}}{a_av} = 0.075 \rm m/s^2

Answer Requested The average acceleration of a particle between two instants of time is the slope of the line connecting the two corresponding points in a v vs. t graph.

Part D What is the instantaneous acceleration \texttip{a}{a} of the particle at t=45.0\;\rm s?

Hint 1. Graphical interpretation of instantaneous acceleration The acceleration of a particle at any given instant of time or at any point in its path is called the instantaneous acceleration. If the v vs. t graph of the particle's motion is known, you can directly determine the instantaneous acceleration at any point on the curve. The instantaneous acceleration at any point is equal to the slope of the line tangent to the curve at that point.

Hint 2. Slope of a line The slope \texttip{m}{m} of a line from point A, of coordinates (x_{\rm A}, y_{\rm A}), to point B, of coordinates (x_{\rm B}, y_{\rm B}), is equal to the "rise" over the "run," or \large{m=\frac{y_{\rm B}-y_{\rm A}}{x_{\rm B}-x_{\rm A}}}.

ANSWER: 1 \rm m/s^2 0.20 \rm m/s^2 \texttip{a}{a} =

-0.20 \rm m/s^2 0.022 \rm m/s^2 -0.022 \rm m/s^2

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Correct The instantaneous acceleration of a particle at any point on a v vs. t graph is the slope of the line tangent to the curve at that point. Since in the last 10 seconds of motion, between t=40.0\;\rm s and t=50.0\;\rm s, the curve is a straight line, the tangent line is the curve itself. Physically, this means that the instantaneous acceleration of the particle is constant over that time interval. This is true for any motion where velocity increases linearly with time. In the case at hand, can you think of another time interval in which the acceleration of the particle is constant?

Now that you have reviewed how to plot variables as a function of time, you can use the same technique and draw an acceleration vs. time graph, that is, the graph of (instantaneous) acceleration as a function of time. As usual in these types of graphs, time \texttip{t}{t} is plotted on the horizontal axis, while the vertical axis is used to indicate acceleration \texttip{a}{a}.

Part E Which of the graphs shown below is the correct acceleration vs. time plot for the motion described in the previous parts?

Hint 1. How to approach the problem Recall that whenever velocity increases linearly with time, acceleration is constant. In the example here, the particle's velocity increases linearly with time in the first 20.0 \rm s of motion. In the second 20.0 \rm s , the particle's velocity is constant, and then it decreases linearly with time in the last 10 \rm s. This means that the particle's acceleration is constant over each time interval, but its value is different in each interval.

Hint 2. Find the acceleration in the first 20 \rm s What is \texttip{a_{\rm 1}}{a_1}, the particle's acceleration in the first 20 \rm s of motion, between t=0.0\;\rm s and t=20.0\;\rm s? Express your answer in meters per second per second.

Hint 1. Constant acceleration Since we have already determined that in the first 20 \rm s of motion the particle's acceleration is constant, its constant value will be equal to the average acceleration that you calculated in Part C. ANSWER: \texttip{a_{\rm 1}}{a_1} = 0.075 \rm m/s^2

Hint 3. Find the acceleration in the second 20 \rm s What is \texttip{a_{\rm 2}}{a_2}, the particle's acceleration in the second 20 \rm s of motion, between t=20.0\;\rm s and t=40.0\;\rm s? session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2469924

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Express your answer in meters per second per second.

Hint 1. Constant velocity In the second 20 \rm s of motion, the particle's velocity remains unchanged. This means that in this time interval, the particle does not accelerate. ANSWER: \texttip{a_{\rm 2}}{a_2} = 0 \rm m/s^2

Hint 4. Find the acceleration in the last 10 \rm s What is \texttip{a_{\rm 3}}{a_3}, the particle's acceleration in the last 10 \rm s of motion, between t=40.0\;\rm s and t=50.0\;\rm s? Express your answer in meters per second per second.

Hint 1. Constant acceleration Since we have already determined that in the last 10 \rm s of motion the particle's acceleration is constant, its constant value will be equal to the instantaneous acceleration that you calculated in Part D. ANSWER: \texttip{a_{\rm 3}}{a_3} = -0.20 \rm m/s^2

ANSWER: Graph A Graph B Graph C Graph D

Correct In conclusion, graphs of velocity as a function of time are a useful representation of straight-line motion. If read correctly, they can provide you with all the information you need to study the motion.

Direction of Velocity and Acceleration Vector Quantities Conceptual Question For each of the motions described below, determine the algebraic sign (+, -, or 0) of the velocity and acceleration of the object at the time specified. For all of the motions, the positive y axis is upward.

Part A An elevator is moving downward when someone presses the emergency stop button. The elevator comes to rest a short time later. Give the signs for the velocity and the acceleration of the elevator after the button has been pressed but before the elevator has stopped. Enter the correct sign for the elevator's velocity and the correct sign for the elevator's acceleration, separated by a comma. For example, if you think that the velocity is positive and the acceleration is negative, then you would enter +,-. If you think that both are zero, then you would enter 0,0.

Hint 1. Algebraic sign of velocity The algebraic sign of velocity is determined solely by comparing the direction in which the object is moving with the direction that is defined to be positive. In this example, upward is defined to be positive. Therefore, any object moving upward, whether speeding up, slowing down, or traveling at constant speed, has positive velocity.

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Hint 2. Algebraic sign of acceleration The algebraic sign of acceleration is more difficult to determine than the algebraic sign of velocity. The acceleration of an object points in the same direction as the change in the velocity of an object. If an object is speeding up, the change in the velocity points in the same direction as the velocity:

If an object is slowing down, the change in velocity points in the opposite direction to that of the velocity:

Once you know the direction of the acceleration, you can determine its sign by comparing it to the defined positive direction, in this case, upward. ANSWER: -,+

Correct

Part B A child throws a baseball directly upward. What are the signs of the velocity and acceleration of the ball immediately after the ball leaves the child's hand? Enter the correct sign for the baseball's velocity and the correct sign for the baseball's acceleration, separated by a comma. For example, if you think that the velocity is positive and the acceleration is negative, then you would enter +,-. If you think that both are zero, then you would enter 0,0.

Hint 1. Algebraic sign of velocity The algebraic sign of velocity is determined solely by comparing the direction in which the object is moving with the direction that is defined to be positive. In this example, upward is defined to be positive. Therefore, any object moving upward, whether speeding up, slowing down, or traveling at constant speed, has positive velocity.

Hint 2. Algebraic sign of acceleration The algebraic sign of acceleration is more difficult to determine than the algebraic sign of velocity. The acceleration of an object points in the same direction as the change in the velocity of an object. If an object is speeding up, the change in the velocity points in the same direction as the velocity: session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2469924

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If an object is slowing down, the change in velocity points in the opposite direction to that of the velocity:

Once you know the direction of the acceleration, you can determine its sign by comparing it to the defined positive direction, in this case, upward. ANSWER: +,-

Correct

Part C A child throws a baseball directly upward. What are the signs of the velocity and acceleration of the ball at the very top of the ball's motion (i.e., the point of maximum height)? Enter the correct sign for the baseball's velocity and the correct sign for the baseball's acceleration, separated by a comma. For example, if you think that the velocity is positive and the acceleration is negative, then you would enter +,-. If you think that both are zero, then you would enter 0,0.

Hint 1. Algebraic sign of velocity The algebraic sign of velocity is determined solely by comparing the direction in which the object is moving with the direction that is defined to be positive. In this example, upward is defined to be positive. Therefore, any object moving upward, whether speeding up, slowing down, or traveling at constant speed, has positive velocity.

Hint 2. Algebraic sign of acceleration The algebraic sign of acceleration is more difficult to determine than the algebraic sign of velocity. The acceleration of an object points in the same direction as the change in the velocity of an object. If an object is speeding up, the change in the velocity points in the same direction as the velocity:

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If an object is slowing down, the change in velocity points in the opposite direction as the velocity:

Once you know the direction of the acceleration, you can determine its sign by comparing it to the defined positive direction, in this case, upward. ANSWER: 0,-

Correct

One-Dimensional Kinematics with Constant Acceleration Learning Goal: To understand the meaning of the variables that appear in the equations for one-dimensional kinematics with constant acceleration. Motion with a constant nonzero acceleration is not uncommon in the world around us. Falling (or thrown) objects and cars starting and stopping approximate this type of motion. It is also the type of motion most frequently involved in introductory kinematics problems. The kinematic equations for such motion can be written as \large{x(t) = x_{\rm i} +v_{\rm i}t + \frac{1}{2}at^2}, v(t) = v_{\rm i} + at, where the symbols are defined as follows: \texttip{x\left(t\right)}{x(t)} is the position of the particle at time \texttip{t}{t}; \texttip{x_{\rm i}}{x_i} is the initial position of the particle; \texttip{v\left(t\right)}{v(t)} is the velocity of the particle at time \texttip{t}{t}; \texttip{v_{\rm i}}{v_i} is the initial velocity of the particle; \texttip{a}{a} is the acceleration of the particle. In answering the following questions, assume that the acceleration is constant and nonzero: a \neq 0.

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Part A The quantity represented by \texttip{x}{x} is a function that changes over time (i.e., is not constant). ANSWER: true false

Correct

Part B The quantity represented by \texttip{x_{\rm i}}{x_i} is a function of time (i.e., is not constant). ANSWER: true false

Correct Recall that \texttip{x_{\rm i}}{x_i} represents an initial value, not a variable. It refers to the position of an object at some initial moment.

Part C The quantity represented by \texttip{v_{\rm i}}{v_i} is a function of time (i.e., is not constant). ANSWER: true false

Correct

Part D The quantity represented by \texttip{v\left(t\right)}{v(t)} is a function of time (i.e., is not constant). ANSWER: true false

Correct The velocity \texttip{v\left(t\right)}{v(t)} always varies with time when the linear acceleration is nonzero.

Part E A particle moves with constant acceleration \texttip{a}{a}. The expression v_{\rm i}+at represents the particle's velocity at what instant in time? ANSWER:

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at time t=0 at the "initial" time when a time \texttip{t}{t} has passed since the beginning of the particle's motion, when its velocity was \texttip{v_{\rm i}}{v_i}

Correct More generally, the equations of motion can be written as \large{x(t) = x_{\rm i} +v_{\rm i}\;\Delta t + \frac{1}{2}a \;(\Delta t)^2} and v(t) = v_{\rm i} + a \;\Delta t. Here \texttip{\Delta t}{Deltat} is the time that has elapsed since the beginning of the particle's motion, that is, \Delta t = t - t_{\rm i}, where \texttip{t}{t} is the current time and \texttip{t_{\rm i}}{t_i} is the time at which we start measuring the particle's motion. The terms \texttip{x_{\rm i}}{x_i} and \texttip{v_{\rm i}} {v_i} are, respectively, the position and velocity at t=t_{\rm i}. As you can now see, the equations given at the beginning of this problem correspond to the case t_{\rm i} = 0, which is a convenient choice if there is only one particle of interest.

Displacement versus Time Graphs Conceptual Question The motions described in each of the questions take place at an intersection on a two-lane road with a stop sign in each direction. For each motion, select the correct position versus time graph. For all of the motions, the stop sign is at the position x=0, and east is the positive x direction.

Part A A driver ignores the stop sign and continues driving east at constant speed.

Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the rise (change in position) over the run (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed.

Hint 2. Driving east Since east is defined as the positive x direction, a car traveling east must have a positive velocity. A positive velocity is represented as a positive slope on a position versus time graph.

Hint 3. Constant speed Since velocity is represented by the slope on a position versus time graph, a car moving at constant speed must be represented by a line of constant slope. ANSWER:

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A B C D E F

Correct

Part B A driver ignores the stop sign and continues driving west at constant speed.

Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the rise (change in position) over the run (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed.

Hint 2. Driving west Since east is defined as the positive x direction, a car traveling west must have a negative velocity. A negative velocity is represented as a negative slope on a position versus time graph.

Hint 3. Constant speed Since velocity is represented by the slope on a position versus time graph, a car moving at constant speed must be represented by a line of constant slope. ANSWER: A B C D E F

Correct

Part C A driver, traveling west, slows and stops at the stop sign.

Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the rise (change in position) over the run (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed.

Hint 2. Driving west Since east is defined as the positive x direction, a car traveling west must have a negative velocity. A negative velocity is represented as a negative slope on a position versus time graph.

Hint 3. Acceleration Since velocity is represented by the slope on a position versus time graph, a car that accelerates must be represented as a curve with changing slope. If a car slows, then the slope of the graph must approach zero. If a car's speed increases, the slope must become more positive or more negative (depending upon which direction it is moving).

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ANSWER: A B C D E F

Correct

Part D A driver, after stopping at the stop sign, accelerates to the east.

Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the rise (change in position) over the run (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed.

Hint 2. Driving east Since east is defined as the positive x direction, a car traveling east must have a positive velocity. A positive velocity is represented as a positive slope on a position versus time graph.

Hint 3. Acceleration Since velocity is represented by the slope on a position versus time graph, a car that accelerates must be represented as a curve with changing slope. If a car slows, then the slope of the graph must approach zero. If a car's speed increases, the slope must become more positive or more negative (depending upon which direction it is moving). ANSWER: A B C D E F

Correct

Problem 2.1 Referring to the figure, you walk from your home to the library, then to the park.

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Part A What is the distance traveled? ANSWER: d = 1.95 {\rm mi}

Correct

Part B What is your displacement? ANSWER: \Delta x = 0.750 {\rm mi}

Correct

Problem 2.3 The golfer in the figure sinks the ball in two putts, as shown.

Part A What is the distance traveled by the ball? Express your answer using two significant figures. ANSWER:

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d = 15 {\rm m}

Correct

Part B What is the displacement of the ball? Express your answer using two significant figures. ANSWER: \Delta x = 10 {\rm m}

Correct

Problem 2.5 A jogger runs on the track shown in the figure.

Part A Neglecting the curvature of the corners, what is the distance traveled and the displacement in running from point A to point B? Enter your answers numerically separated by a comma. ANSWER: d, \Delta x = 130,100 {\rm m}

Correct

Part B Find the distance and displacement for a complete circuit of the track. Enter your answers numerically separated by a comma. ANSWER: d, \Delta x = 260,0 {\rm m}

Correct

CE Predict/Explain 2.7 You drive your car in a straight line at 15 {\rm m}/{\rm s} for 10 kilometers, then at 25 {\rm m}/{\rm s} for another 10 kilometers.

Part A Is your average speed for the entire trip more than, less than, or equal to 20 {\rm m}/{\rm s}? ANSWER: session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2469924

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more than 20 {\rm m}/{\rm s} less than 20 {\rm m}/{\rm s} equal to 20 {\rm m}/{\rm s}

Correct

Part B Choose the best explanation from among the following: ANSWER: More time is spent at 15 {\rm m}/{\rm s} than at 25 {\rm m}/{\rm s}. The average of 15 {\rm m}/{\rm s} and 25 {\rm m}/{\rm s} is 20 {\rm m}/{\rm s}. Less time is spent at 15 {\rm m}/{\rm s} than at 25 {\rm m}/{\rm s}.

Correct

Problem 2.9 Joseph DeLoach of the United States set an Olympic record in 1988 for the 200-meter dash with a time of 19.75 seconds.

Part A What was his average speed? Give your answer in meters per second and miles per hour. Express your answer using four significant figures. ANSWER: v_{\rm av} = 10.13 {\rm m/s}

Correct

Part B Express your answer using four significant figures. ANSWER: v_{\rm av} = 22.66 {\rm mi/h}

Correct

Problem 2.11 Kangaroos have been clocked at speeds of 65 {\rm km}/{\rm h}.

Part A How far can a kangaroo hop in 3.6 minutes at this speed? Express your answer using two significant figures. ANSWER: d = 3.9 {\rm km} session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2469924

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Correct

Part B How long will it take a kangaroo to hop 0.20{\rm km} at this speed? Express your answer using two significant figures. ANSWER: t = 11 {\rm s}

Correct

Problem 2.13 Radio waves travel at the speed of light, approximately 186,000 miles per second.

Part A How long does it take for a radio message to travel from Earth to the Moon and back? (See the inside back cover of the textbook for the necessary data.) ANSWER: t = 2.57 {\rm s}

Correct

Problem 2.17 A finch rides on the back of a Galapagos tortoise, which walks at the stately pace of 0.060 {\rm m}/{\rm s}. After 1.2 minutes the finch tires of the tortoise's slow pace, and takes flight in the same direction for another 1.2 minutes at 12 {\rm m}/{\rm s}.

Part A What was the average speed of the finch for this 2.4-minute interval? Express your answer using two significant figures. ANSWER: v_{\rm ave} = 6.0 {\rm m/s}

Correct

Problem 2.19 A dog runs back and forth between its two owners, who are walking toward one another. The dog starts running when the owners are 10.0 {\rm m} apart.

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Part A If the dog runs with a speed of 3.0 {\rm m}/{\rm s}, and the owners each walk with a speed of 1.3 {\rm m}/{\rm s}, how far has the dog traveled when the owners meet? Express your answer using two significant figures. ANSWER: d = 11 {\rm m}

Correct

Problem 2.21 In heavy rush-hour traffic you drive in a straight line at 12 {\rm m/s} for 1.5 minutes, then you have to stop for 3.5 minutes, and finally you drive at 15 {\rm m/s} for another 2.5 minutes.

Part A Plot a position-versus-time graph for this motion. Your plot should extend from t = 0 to t = 7.5 minutes. Assume x=0 and t=0 at the start of your motion. ANSWER:

Answer Requested

Part B Use your plot from part A to calculate the average velocity between t = 0 to t = 7.5 minutes. Express your answer using two significant figures. ANSWER: v_{\rm av} = 7.4 {\rm m/s}

Correct

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Problem 2.25 The position of a particle as a function of time is given by x = (6\;{\rm{m}}/{\rm{s}})t + ( - 2\;{\rm{m}}/{\rm{s}}^2 )t^2.

Part A Plot \texttip{x}{x} versus \texttip{t}{t} for t = 0 to t = 2\;{\rm{s}}. ANSWER:

Answer Requested

Part B Find the average velocity of the particle from t = 0 to t = 1\;{\rm{s}}. Express your answer using two significant figures. ANSWER: v_{\rm av} = 4.0 {\rm m/s}

Correct

Part C Find the average speed from t = 0 to t = 1\;{\rm{s}}. Express your answer using two significant figures. ANSWER: s_{\rm av} = 4.0 {\rm m/s}

Correct

Problem 2.27 On your wedding day you leave for the church 28.0 minutes before the ceremony is to begin, which should be plenty of time since the church is only 11.5 session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2469924

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miles away. On the way, however, you have to make an unanticipated stop for construction work on the road. As a result, your average speed for the first 13.0 minutes is only 5.0 {\rm mi}/{\rm h}.

Part A What average speed do you need for the rest of the trip to get you to the church on time? ANSWER: s = 41.7 {\rm mi/h}

Answer Requested

Problem 2.29 The position of a particle as a function of time is given by x = (2.0\;{\rm m/s})t + ( - 3.0\;{\rm m/s}^3 )t^3.

Part A Plot \texttip{x}{x} versus \texttip{t}{t} for time from t = 0 to t = 1.0\;{\rm{s}}. ANSWER:

Typesetting math: 76%

Answer Requested

Part B Find the average velocity of the particle from \texttip{t}{t_1} = 0.35{\rm s} to \texttip{t}{t_2} = 0.45{\rm s} . Express your answer using two decimal places. ANSWER: v_{\rm av} = 0.55 {\rm m/s}

Correct

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Homework 1

Find the average velocity of the particle from \texttip{t}{t_3} = 0.39{\rm s} to \texttip{t}{t_4} = 0.41{\rm s} . Express your answer using two decimal places. ANSWER: v_{\rm av} = 0.56 {\rm m/s}

Correct

Part D Do you expect the instantaneous velocity at \texttip{t}{t_0} = 0.40{\rm s} to be closer to 0.54{\rm m/s} , 0.56{\rm m/s} , or 0.58{\rm m/s} ? ANSWER: 0.54{\rm m/s} 0.56{\rm m/s} 0.58{\rm m/s}

Correct

Part E Explain. ANSWER: 3702 Character(s) remaining I would expect the velocity to be about 0.56 m/s because 0.4s is between t = 0.39s and t = 0.41s.

Graded, see 'My Answers' for details

CE Predict/Explain 2.31 Two bows shoot identical arrows with the same launch speed. To accomplish this, the string in bow 1 must be pulled back farther when shooting its arrow than the string in bow 2.

Part A Is the acceleration of the arrow shot by bow 1 greater than, less than, or equal to the acceleration of the arrow shot by bow 2? ANSWER: greater than the acceleration of the arrow shot by bow 2 less than the acceleration of the arrow shot by bow 2 equal to the acceleration of the arrow shot by bow 2

Correct

Part B Choose the best explanation from among the following: ANSWER: session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2469924

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Homework 1

The arrow in bow 1 accelerates for a greater time. Both arrows start from rest. The arrow in bow 2 accelerates for a greater time.

Correct

Problem 2.33 At the starting gun, a runner accelerates at 2.0{\rm m/s^2} for 4.8{\rm s} . The runner's acceleration is zero for the rest of the race.

Part A What is the speed of the runner at t = 2.0\;{\rm{s}}? Express your answer using two significant figures. ANSWER: v = 4.0 {\rm m/s}

Correct

Part B What is the speed of the runner at the end of the race? Express your answer using two significant figures. ANSWER: v = 9.6 {\rm m/s}

Correct

Problem 2.35 A car is traveling due north at 19.1{\rm m/s} .

Part A Find the velocity of the car after 5.28{\rm s} if its acceleration is 1.26{\rm m/s^2} due north. ANSWER: v = 25.8 {\rm m/s} north

Correct

Part B Find the velocity of the car after 5.28{\rm s} if its acceleration is 1.06{\rm m/s^2} due south. ANSWER: v = 13.5 {\rm m/s} north

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Homework 1

Answer Requested

Problem 2.37 A person on horseback moves according to the velocity-versus-time graph shown in the figure.

Part A Find the displacement of the person for segment A of the motion. ANSWER: \Delta x_A = 10 {\rm m}

Correct

Part B Find the displacement of the person for segment B of the motion. ANSWER: \Delta x_B = 20 {\rm m}

Correct

Part C Find the displacement of the person for segment C of the motion. ANSWER: \Delta x_C = 40 {\rm m}

Correct

Problem 2.41 As a train accelerates away from a station, it reaches a speed of 4.8{\rm m/s} in 4.9{\rm s} .

Part A If the train's acceleration remains constant, what is its speed after an additional 7.0{\rm s} has elapsed? session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2469924

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Homework 1

Express your answer using two significant figures. ANSWER: v = 12 {\rm m/s}

Answer Requested Score Summary: Your score on this assignment is 81.3%. You received 86.15 out of a possible total of 106 points.

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