Baixe Serway Physics 6th Edition Solutions e outras Notas de estudo em PDF para Cultura, somente na Docsity!
CHAPTER OUTLINE 1.1 Standards of Length, Mass, and Time 1.2 Matter and Model-Building 1.3 Density and Atomic Mass 1.4 Dimensional Analysis 1.5 Conversion of Units 1.6 Estimates and Order-of- Magnitude Calculations 1.7 Significant Figures
Physics and Measurement
ANSWERS TO QUESTIONS
Q1.1 Atomic clocks are based on electromagnetic waves which atoms emit. Also, pulsars are highly regular astronomical clocks.
Q1.2 Density varies with temperature and pressure. It would be necessary to measure both mass and volume very accurately in order to use the density of water as a standard.
Q1.3 People have different size hands. Defining the unit precisely would be cumbersome.
Q1.4 (a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilograms
Q1.5 (b) and (d). You cannot add or subtract quantities of different dimension.
Q1.6 A dimensionally correct equation need not be true. Example: 1 chimpanzee = 2 chimpanzee is dimensionally correct. If an equation is not dimensionally correct, it cannot be correct.
Q1.7 If I were a runner, I might walk or run 10 1 miles per day. Since I am a college professor, I walk about 10 0 miles per day. I drive about 40 miles per day on workdays and up to 200 miles per day on vacation.
Q1.8 On February 7, 2001, I am 55 years and 39 days old.
55 yr 365 25 1 .yr^ d 39 d 20 128 d 86 400 1 d s 1 74. 10 9 s ~ 109 s
F
HG^
I
KJ^
+ = FHG IKJ = ×.
Many college students are just approaching 1 Gs.
Q1.9 Zero digits. An order-of-magnitude calculation is accurate only within a factor of 10.
Q1.10 The mass of the forty-six chapter textbook is on the order of 10 0 kg.
Q1.11 With one datum known to one significant digit, we have 80 million yr + 24 yr = 80 million yr.
2 Physics and Measurement
SOLUTIONS TO PROBLEMS
Section 1.1 Standards of Length, Mass, and Time
No problems in this section
Section 1.2 Matter and Model-Building
P1.1 From the figure, we may see that the spacing between diagonal planes is half the distance between diagonally adjacent atoms on a flat plane. This diagonal distance may be obtained from the Pythagorean theorem, L (^) diag = L^2 + L^2. Thus, since the atoms are separated by a distance
L = 0 200. nm , the diagonal planes are separated by^1 2
L^2^ + L^2 = 0 141. nm.
Section 1.3 Density and Atomic Mass
*P1.2 Modeling the Earth as a sphere, we find its volume as^433 34 6 37 10 6 1 08 10
(^3 21 ) π r = πe. × m (^) j =. × m. Its
density is then ρ = = × ×
m = × V
24 21 3
kg. m
kg m. This value is intermediate between the tabulated densities of aluminum and iron. Typical rocks have densities around 2 000 to 3 000 kg m^3. The average density of the Earth is significantly higher, so higher-density material must be down below the surface.
P1.3 With V = abase area fbheight g V = eπ r^2 j h and ρ = mV , we have
ρ π (^) π ρ
F HG^
I KJ = ×
m r h^2
9
4 3
kg mm mm
mm m kg m
3
..^3 ..
a f a f
*P1.4 Let V represent the volume of the model, the same in ρ = m V
for both. Then ρ (^) iron = 9 35. kg V and
ρ (^) gold = gold
m V
. Next,
ρ ρ
gold iron
gold kg
m 9 35.
and m gold
3 3 kg (^3) 19.3 10 kg / m kg / m
= × kg ×
F HG^
I KJ^
P1.5 V = Vo − V (^) i = 4 r − r 3 2
π (^) e 3 13 j
ρ = mV , so m V r r
r r = = FHG IKJ − =
ρ ρ π 4 π ρ 3
3 13 2
3 13 e j
e j .
4 Physics and Measurement
P1.9 Mass of gold abraded: ∆ m = − = =
F HG^
I KJ^
3 80. g 3 35. g 0 45. g 0 45. g 1 kg = 4 5. × 10 −^4 10 g
b g 3 kg.
Each atom has mass m 0
(^2725) 197 197 1 66^10 1
= = × 3 27 10
F HG^
I KJ^
= ×
− − u u kg u
. (^). kg.
Now, ∆ m = ∆ N m 0 , and the number of atoms missing is
N
m m
= = ×
×
= ×
− 0 −
4 25
kg. kg
atoms.
The rate of loss is
∆ ∆ ∆ ∆
N
t N t
= ×^ FHG IKJFHG IKJFHG IKJFHG IKJ
= ×
21
11
atoms yr
yr 365.25 d
d 24 h
h 60 min
min 60 s
atoms s
P1.10 (a) m = ρ L^3 = 3 × −^6 = × −^ = × −
(^3 16 ) e7 86.^ g cm^ je5 00.^10 cm^ j 9 83.^10 g^ 9 83.^10 kg
(b) N m m
= = ×
×
= ×
− 0 −
19 27
kg. u kg 1 u
atoms e j
P1.11 (a) The cross-sectional area is
A = + = × −
m m m m m 2
a (^) fa (^) f a (^) fa (^) f .
The volume of the beam is
V = AL = (^) e 6 40. × 10 −^3 m (^2) ja1 50. m f = 9 60. × 10 −^3 m^3.
Thus, its mass is
m = ρ V = (^) e7 56. × 10 3 kg / m (^3) je9 60. × 10 −^3 m (^3) j = 72 6. kg.
FIG. P1.
(b) The mass of one typical atom is m 0
(^2726) 55 9 1 66^10 1
= × 9 28 10
F HG^
I KJ^
= ×
− −
. u.^ kg. u
a f kg. Now
m = Nm 0 and the number of atoms is N m m
×
− =^ ×
0 26
kg. kg
atoms.
Chapter 1 5
P1.12 (a) The mass of one molecule is m 0
(^2726) = 18 0 1 66^ ×^10 2 99 10
F HG^
I KJ^
= ×
− −
. u.^ kg. 1 u
kg. The number of
molecules in the pail is
N m pail m
kg 2.99 kg
= = molecules ×
− =^ ×
0 26
(b) Suppose that enough time has elapsed for thorough mixing of the hydrosphere.
N N
m both pail M
pail total
(4.02 10 25 molecules) kg kg
F HG^
I KJ^
= ×
×
F HG^
I KJ
or
N both = 3 65. × 10 4 molecules.
Section 1.4 Dimensional Analysis
P1.13 The term x has dimensions of L, a has dimensions of LT−^2 , and t has dimensions of T. Therefore, the equation x = ka mt^ n has dimensions of
L = (^) e LT −^2 j a fT
m (^) n or L T^1 0 = L T m^ n^ −^2 m^.
The powers of L and T must be the same on each side of the equation. Therefore,
L^1 = L m^ and m = 1.
Likewise, equating terms in T, we see that n − 2 m must equal 0. Thus, n = 2. The value of k , a dimensionless constant, cannot be obtained by dimensional analysis.
*P1.14 (a) Circumference has dimensions of L.
(b) Volume has dimensions of L^3.
(c) Area has dimensions of L^2.
Expression (i) has dimension L L 2 L
1 2 (^2) e j
/ = , so this must be area (c). Expression (ii) has dimension L, so it is (a). Expression (iii) has dimension L L e j 2 = L^3 , so it is (b). Thus, (a)^ = ii;^ (b)^ = iii, (c)^ = i.
Chapter 1 7
P1.21 Conceptualize : We must calculate the area and convert units. Since a meter is about 3 feet, we should expect the area to be about A ≈ (^) a 30 m (^) fa 50 m (^) f =1 500m 2. Categorize : We model the lot as a perfect rectangle to use Area = Length × Width. Use the conversion: 1 m = 3.281 ft.
Analyze : A = LW = 100 FHG 1 IKJ FHG IKJ = × 3 281
ft m 1 39 10 3 ft
ft m ft
a f a f = 1 390 m 2 m^2
..
Finalize : Our calculated result agrees reasonably well with our initial estimate and has the proper units of m 2. Unit conversion is a common technique that is applied to many problems.
P1.22 (a) V = a40.0 m fa20.0 m fa 12.0 m f = 9 60. × 10 3 m 3
V = 9 60. × 10 3 m 3 b3.28 ft 1 m g 3 = 3 39. × 105 ft^3
(b) The mass of the air is
m = ρair V = (^) e1 20. kg m (^3) je9.60 × 10 3 m (^3) j = 1 15. × 104 kg.
The student must look up weight in the index to find
F (^) g = mg = (^) e1.15 × 10 4 kg (^) je9.80 m s 2 j = 1.13 × 10 5 N.
Converting to pounds,
Fg = (^) e1 13. × 10 5 N (^) jb1 lb 4.45 N g = 2 54. × 104 lb.
P1.23 (a) Seven minutes is 420 seconds, so the rate is
r = 30 0 = × − 420
. gal 7 14. 10 2 s
gal s.
(b) Converting gallons first to liters, then to m 3 ,
r
r
= × F HG^
I KJ
F HG^
I KJ = ×
− −
−
7 14 10 3 786^10
2 3
4
gal s L 1 gal
m 1 L m s
3
3
e j
(c) At that rate, to fill a 1-m 3 tank would take
t = ×
F HG^
I KJ
F HG^
I KJ^
4
m m s
h 3 600
h
3
. 3.^.
8 Physics and Measurement
*P1.24 (a) Length of Mammoth Cave = 348 FHG 1 609 IKJ = = × = × 1
mi km 560 5 60 10 5 5 60 107 mi
. (^) km. m. cm.
(b) Height of Ribbon Falls = 1 612 FHG 0 IKJ = = = × 1
ft .304 8 m 491 m 0 491 4 91 10 4 ft
. km. cm.
(c) Height of Denali = 20 320 ft FHG 0 .304 8 m 1 ft IKJ = 6 .19 km = 6 19. × 10 3 m = 6 19. × 105 cm.
(d) Depth of King’s Canyon = 8 200 FHG 0 IKJ = = × = × 1
ft .304 8 m 2 2 50 10 3 2 50 105 ft
.50 km. m. cm.
P1.25 From Table 1.5, the density of lead is 1 13. × 10 4 kg m 3 , so we should expect our calculated value to be close to this number. This density value tells us that lead is about 11 times denser than water, which agrees with our experience that lead sinks.
Density is defined as mass per volume, in ρ = m V
. We must convert to SI units in the calculation.
ρ =
F HG^
I KJ
F HG^
I KJ^ =^ ×
3 1 1 14^10
.^34
g cm
kg g
cm m.^ kg m^
3
At one step in the calculation, we note that one million cubic centimeters make one cubic meter. Our result is indeed close to the expected value. Since the last reported significant digit is not certain, the difference in the two values is probably due to measurement uncertainty and should not be a concern. One important common-sense check on density values is that objects which sink in water must have a density greater than 1 g cm 3 , and objects that float must be less dense than water.
P1.26 It is often useful to remember that the 1 600-m race at track and field events is approximately 1 mile in length. To be precise, there are 1 609 meters in a mile. Thus, 1 acre is equal in area to
1 6401 1 609^ 4 05 10
2 acre mi^3 acres
m mi m
(^2 ) a f
F HG^
I KJ
F HG^
I KJ^ =^.^ ×.
*P1.27 The weight flow rate is 1 200 ton 2 000^1 1 h
lb ton
h 60 min
min 60 s
FHG IKJFHG IKJFHG IKJ = lb s.
P1.28 1 mi = 1 609 m = 1 609. km; thus, to go from mph to km h , multiply by 1.609.
(a) 1 mi h = 1 609. km h
(b) 55 mi h = 88 5. km h
(c) 65 mi h = 104 6. km h. Thus, ∆ v = 16 1. km h.
10 Physics and Measurement
P1.35 (a) d d
d nucleus, scale nucleus, real d
atom, scale atom, real
m ft 1.06 10 m
= ft
F HG^
I KJ^
= ×
×
F HG^
I KJ^ =^ ×
− −
10 e.^ j.^3 , or
d nucleus, scale = (^) e 6 79. × 10 −^3 ft (^) jb304 8. mm 1 ft (^) g= 2 07. mm
(b)
V
V
r r
d d
r r
atom nucleus
atom nucleus
atom nucleus
atom nucleus
m m times as large
= = F HG^
I KJ^
= F HG^
I KJ^
= ×
×
F HG^
I KJ = ×
− −
4 3 4 3
(^3 3 ) 15
3
13
3 3
π π
*P1.36 scale distance between
real distance
scale factor km^
m m
= F km HG^
I KJ
F HG^
I KJ^
= × ×
×
F HG^
I KJ^
− 4 0 10 7 0^10 1 4 10
3
. (^9)
e j.
P1.37 The scale factor used in the “dinner plate” model is
S =
×
0 25 = × −
5.^2
m lightyears
.5 m lightyears.
The distance to Andromeda in the scale model will be
D (^) scale = D (^) actual S = (^) e2.0 × 10 6 lightyears (^) je 2.5 × 10 −^6 m lightyears (^) j = 5 0. m.
P1.38 (a) A A
r r
r r
Earth Moon
Earth Moon^2
Earth Moon
m cm m cm
= = F HG^
I KJ^
×
×
F
H
GG
I
K
(^4) JJ = 4
2 2 6 8
2 π π
e jb. g
(b) (^) VV rr
r r
Earth Moon
Earth Moon
Earth^3 Moon
m cm m cm
F HG^
I KJ^
×
×
F
H
GG
I
K
JJ =
4 3 4 3
6 8
3 3
π π
e jb^ g
P1.39 To balance, m (^) Fe = m Alor ρFe V Fe (^) =ρAl V Al
ρ π ρ π
ρ ρ
Fe Fe Al Al
Al Fe FeAl cm^ cm
3 3
1 3 (^) 1 3
F HG
I KJ^ =^
F HG
I KJ
= F HG^
I KJ^
= FHG IKJ =
r r
r r
/ (^) /
.. .
a f..
Chapter 1 11
P1.40 The mass of each sphere is
m (^) Al = ρAl V Al (^) =^4 π ρAl r^ Al 3
3
and
m (^) Fe = ρFe V Fe (^) =^4 π ρFe Fe r 3
3 .
Setting these masses equal,
4 3
π ρAl r Al^3 π ρFe Fe r^3 = and r (^) Al r Fe Fe Al
= ρ ρ
Section 1.6 Estimates and Order-of-Magnitude Calculations
P1.41 Model the room as a rectangular solid with dimensions 4 m by 4 m by 3 m, and each ping-pong ball as a sphere of diameter 0.038 m. The volume of the room is 4 × 4 × 3 = 48 m 3 , while the volume of one ball is
4 3
π 0 038. m^3 2 87. 105 2
FHG IKJ = × − (^) m 3.
Therefore, one can fit about 48 2 87 10
× −^
ping-pong balls in the room. As an aside, the actual number is smaller than this because there will be a lot of space in the room that cannot be covered by balls. In fact, even in the best arrangement, the so-called “best packing fraction” is^1 6
π 2 = 0 74. so that at least 26% of the space will be empty. Therefore, the above estimate reduces to 1 67. × 10 6 × 0 740. ~ 106.
P1.42 A reasonable guess for the diameter of a tire might be 2.5 ft, with a circumference of about 8 ft. Thus, the tire would make (^) b50 000 mi (^) gb5 280 ft mi (^) gb 1 rev 8 ft (^) g= 3 × 10 7 rev ~ 10 7 rev.
P1.43 In order to reasonably carry on photosynthesis, we might expect a blade of grass to require at least 1 16 in^43 10 ft
(^2) = × − (^5 2). Since 1 acre = 43 560 ft 2 , the number of blades of grass to be expected on a
quarter-acre plot of land is about
n = = ×
total area − = × area per blade
acre ft acre ft blade
2.5 10 blades blades
2 2
a fe j .
Chapter 1 13 Section 1.7 Significant Figures
*P1.48 METHOD ONE We treat the best value with its uncertainty as a binomial (^) a 21 3. ± 0 2. (^) f cm (^) a 9 8. ±0 1.fcm ,
A = 21 3 9 8. a. f ± 21 3 0 1. a f. ± 0 2 9 8. a. f a±0 2. fa f0 1. cm 2.
The first term gives the best value of the area. The cross terms add together to give the uncertainty and the fourth term is negligible.
A = 209 cm 2 ± 4 cm^2.
METHOD TWO We add the fractional uncertainties in the data.
A = a 21 3. cm fa9 8. cm f± FHG (^) 21 30 2.. +9 80 1.. IKJ = 209 cm 2 ± 2% = 209 cm^2 ± 4 cm^2
P1.49 (a)^ π^ π π
r^2 2 2 2 2
. m. m m m m m m m
a f (. ) (. )(. ) (. )
(b) 2 π r = 2 πa10 5. m ± 0 2. m f = 66 0. m ±1 3. m
P1.50 (a) 3 (b) 4 (c) 3 (d) 2
P1.51 r m m r
= ± = ± ×
2
43 3
.. cm.. m .. kg
a f a f a f
c h
ρ π
also, δ ρ ρ
= δ^^ m + δ m
r r
In other words, the percentages of uncertainty are cumulative. Therefore,
δ ρ ρ =^ +^ =
a f (^) ,
ρ π
×
1 85− = ×
43 2 3
c h e mj
kg m
and ρ ± δ ρ= a1 61. ± 0 17. f × 10 3 kg m 3 = a1 6. ± 0 2. f× 103 kg m^3.
14 Physics and Measurement P1.52 (a) 756.?? 37.2?
- /5 / 3 = 797
(b) 0 003 2. a 2 s.f. f × 356 3. a4 s.f. f = 1 140 16. = a2 s.f. f 1 1.
(c) 5.620 4 s.f.a f × πa> 4 s.f. f =17.656 = 4 s.f.a f 17.
*P1.53 We work to nine significant digits:
1 yr = 1 yr 365 242 1991 yr^ d^24 1 d^ h^60 1 h^ min 1 min^60 s 31 556 926 0s
F HG^
I KJ
F HG^
I KJ
F HG^
I KJ
F HG^
I KJ^ =
P1.54 The distance around is 38.44 m + 19.5 m + 38.44 m + 19.5 m = 115.88 m, but this answer must be rounded to 115.9 m because the distance 19.5 m carries information to only one place past the decimal. 115 9. m
P1.55 (^) V V V V V V V V
1 2 1 2 1 2 3
b g a fa fa f a fa fa f e j
m m m m m m m m m m m m m
3 3 3 3
δ
δ
δ
δ
A
A
1 1 1 1 1 1
U
V
| | |
W
| | |
m 19.0 m m 1.0 m cm 9.0 cm
w w t t
V
V
FIG. P1.
Additional Problems
P1.56 It is desired to find the distance x such that
x 100 x
m
= m
(i.e., such that x is the same multiple of 100 m as the multiple that 1 000 m is of x ). Thus, it is seen that
x^2 = a 100 m fb1 000 mg = 1 00. × 105 m 2
and therefore
x = 1 00. × 10 5 m 2 = 316 m.
16 Physics and Measurement
P1.60 (^) α ′(deg) α(rad) (^) tana f α sina f α difference 15.0 0.262 0.268 0.259 3.47% 20.0 0.349 0.364 0.342 6.43% 25.0 0.436 0.466 0.423 10.2% 24.0 0.419 0.445 0.407 9.34% 24.4 0.426 0.454 0.413 9.81% 24.5 0.428 0.456 0.415 9.87% 24.6 0.429 0.458 0.416 9.98% 24.6° 24.7 0.431 0.460 0.418 10.1%
P1.61 2 15 0 2 39
π r r h r h
. m . m tan 55.
a.^ m^ ftan(^.^ )^. m
5555 °°
h
rr
h
FIG. P1.
- P1.62 Let d represent the diameter of the coin and h its thickness. The mass of the gold is
m = V = At = d^ + dh t
F HG^
I KJ
ρ ρ ρ^2 π^ π 4
2
where t is the thickness of the plating.
m = +
L N
M M
O Q
P P
×
= × = =
2
..^...^4
.
. $10 $0..
π^ a^ f^ πa fa f e j
grams cost grams gram cents
This is negligible compared to $4.98.
P1.63 The actual number of seconds in a year is
b86 400 s day gb365.25 day yr g=^ 31 557 600 s yr.
The percent error in the approximation is
π × − × =
(^7) s yr s yr
s yr
e j b^ g^
..
Chapter 1 17
P1.64 (a) V = L^3 , A = L^2 , h = L
V = A h
L^3 = L L^2 = L^3. Thus, the equation is dimensionally correct.
(b) V (^) cylinder = π R h^2 = (^) eπ R^2 j h = Ah , where A =π R^2 V (^) rectangular object = A wh = a fA w h = Ah , where A = A w
P1.65 (a) The speed of rise may be found from
v = Vol rate of flow D = = (Area:
cm 3 s (^) cm s cm
a f π^2 πa f^2 4
6 30 4
(b) Likewise, at a 1.35 cm diameter,
v = 16 51.35^ = 11 5 4
2
. cm 3 s (^). cm s π (^) a cmf.
P1.66 (a) 1 cubic meter of water has a mass
m = ρ V = 1 00 × 10 −^3 3 1 00 3 10 2 = 1 000
3 e.^ kg cm^ je.^ m^ je cm m^ j kg
(b) As a rough calculation, we treat each item as if it were 100% water.
cell: kg m m
kg
kidney:. kg cm cm
kg
fly: kg cm mm mm cm mm
3
3
3 2
m V R D
m V R
m D h
= = FHG IKJ = FHG IKJ = FHG IKJ ×
= ×
= = FHG IKJ = × FHG IKJ
=
= FHG IKJ = × FHG IKJ
−
−
−
− −
ρ ρ π ρ π π
ρ ρ π π
ρ π π
6 1 000^
6 1 0^10
3 3 6 3
16
3 3 3
2 3 1
e j e j
e j
e j a^ f a^ fe
.. (^) j
3
= 1 3. × 10 −^5 kg
P1.67 V 20 mpg= (^10 cars 20 )(mi gal^10 mi yr )= 5.0 × 10 10 gal yr
8 4
V 25 mpg cars^ mi yr^10 mi gal
= (^10 )(^10 )= 4.0 × 10 gal yr 25
8 4
Fuel saved = V 25 mpg (^) − V 20 mpg = 1 0. × 10 10 gal yr
Chapter 1 19
ANSWERS TO EVEN PROBLEMS
P1.2 5 52. × 10 3 kg m 3 , between the densities of aluminum and iron, and greater than the densities of surface rocks.
P1.34 1 3. × 10 21 kg
P1.36 200 km
P1.4 23.0 kg^ P1.38^ (a) 13.4; (b) 49.
P1.40 r (^) Al r Fe Fe Al
= F HG^
I KJ
ρ ρ
P1.6 7.69 cm^ 1 3
P1.8 (a) and (b) see the solution, N A = 6 022 137. × 10 23 ; (c) 18.0 g; (d) 44.0 g P1.42^ ~ 10^7 rev
P1.10 (a) 9 83. × 10 −^16 g ; (b) 1 06. × 10 7 atoms^ P1.44^ ~ 10^9 raindrops
P1.46 ~ 10 11 cans; ~ 10 5 tons P1.12 (a) 4 02.^ × 10 25 molecules; (b) 3 65. × 10 4 molecules (^) P1.48 a 209 ± 4 f cm^2
P1.14 (a) ii; (b) iii; (c) i (^) P1.50 (a) 3; (b) 4; (c) 3; (d) 2
P1.16 (a) M L T 2
⋅ (^) ; (b) 1 newton = 1 kg m s⋅ 2 P1.52 (a) 797; (b) 1.1; (c) 17.
P1.54 115.9 m P1.18 35 7. m 2 P1.56 316 m P1.20 1 39. × 10 −^4 m 3 P1.58 4 50. m 2 P1.22 (a) 3 39. × 10 5 ft^3 ; (b) 2 54. × 10 4 lb P1.60 see the solution; 24.6° P1.24 (a) 560 km = 5 60. × 10 5 m = 5 60. × 107 cm; (b) 491 m = 0 491. km = 4 91. × 10 4 cm ; P1.62^ 3 64.^ cents^ ; no (c) 6 .19 km = 6 19. × 10 3 m = 6 19. × 105 cm; (^) P1.64 see the solution (d) 2 .50 km = 2 50. × 10 3 m = 2 50. × 105 cm P1.66 (a) 1 000 kg; (b) 5 2. × 10 −^16 kg ; 0 27. kg ; P1.26 4 05 (^10) 1 3. × 10 − (^5) kg
. × 3 m 2
P1.28 (a) 1 mi h = 1 609. km h; (b) 88 5. km h ; (c) 16 1. km h^ P1.68^ 8 32.^ ×^10 −^4 m s^ ; a snail
P1.30 1 19. × 10 57 atoms^ P1.70^ see the solution
P1.32 2 57. × 10 6 m^3
CHAPTER OUTLINE 2.1 Position, Velocity, and Speed 2.2 Instantaneous Velocity and Speed 2.3 Acceleration 2.4 Motion Diagrams 2.5 One-Dimensional Motion with Constant Acceleration 2.6 Freely Falling Objects 2.7 Kinematic Equations Derived from Calculus
Motion in One Dimension
ANSWERS TO QUESTIONS
Q2.1 If I count 5.0 s between lightning and thunder, the sound has
traveled b 331 m s ga5 0. s f= 1 7. km. The transit time for the light
is smaller by
3 00 10 331
× m s=. × m s
times,
so it is negligible in comparison.
Q2.2 Yes. Yes, if the particle winds up in the + x region at the end.
Q2.3 Zero.
Q2.4 Yes. Yes.
Q2.5 No. Consider a sprinter running a straight-line race. His average velocity would simply be the length of the race divided by the time it took for him to complete the race. If he stops along the way to tie his shoe, then his instantaneous velocity at that point would be zero.
Q2.6 We assume the object moves along a straight line. If its average velocity is zero, then the displacement must be zero over the time interval, according to Equation 2.2. The object might be stationary throughout the interval. If it is moving to the right at first, it must later move to the left to return to its starting point. Its velocity must be zero as it turns around. The graph of the motion shown to the right represents such motion, as the initial and final positions are the same. In an x vs. t graph, the instantaneous velocity at any time t is the slope of the curve at that point. At t 0 in the graph, the slope of the curve is zero, and thus the instantaneous velocity at that time is also zero.
x
t 0 t
FIG. Q2.
Q2.7 Yes. If the velocity of the particle is nonzero, the particle is in motion. If the acceleration is zero, the velocity of the particle is unchanging, or is a constant.
