Solutions manual for fundamentals of physic 9th by halliday

Solutions manual for fundamentals of physic 9th by halliday Solutions manual for fundamentals of physic 9th by halliday Solutions manual for fundamentals of physic 9th by halliday Solutions manual for fundamentals of physic 9th by halliday Solutions manual for fundamentals of physic 9th by halliday Solutions manual for fundamentals of physic 9th by halliday Solutions manual for fundamentals of physic 9th by halliday Solutions manual for fundamentals of physic 9th by halliday Solutions manual for fundamentals of physic 9th by halliday

c2011

PART 1

1 Measurement

2 Motion Along a Straight Line

3 Vectors

4 Motion in Two and Three Dimensions

5 Force and Motion — I

6 Force and Motion — II

7 Kinetic Energy and Work

8 Potential Energy and Conservation of Energy

9 Center of Mass and Linear Momentum

18 Temperature, Heat, and the First Law of Thermodynamics

19 The Kinetic Theory of Gases

20 Entropy and the Second Law of Thermodynamics

29 Magnetic Fields Due to Currents

30 Induction and Inductance

31 Electromagnetic Oscillations and Alternating Current

32 Maxwell’s Equations; Magnetism of Matter

38 Photons and Matter Waves

39 More About Matter Waves

40 All About Atoms

41 Conduction of Electricity in Solids

42 Nuclear Physics

43 Energy from the Nucleus

44 Quarks, Leptons, and the Big Bang

Chapter 1

1 Various geometric formulas are given in Appendix E

(a) Expressing the radius of the Earth as

1 gry = (1/10)(1/12)(72 points) = 0.60 point

Thus, 1 gry2 = (0.60 point)2 = 0.36 point2, which means that 0.50 gry = 0.18 point 2 2

3 The metric prefixes (micro, pico, nano, …) are given for ready reference on the inside front cover of the textbook (see also Table 1–2)

(a) Since 1 km = 1 × 103 m and 1 m = 1 × 106μm,

( ) ( 6 ) 51.0 yd = 0.91m 10 μm m =9.1×10 μm.

4 (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we obtain

(b) With 12 points = 1 pica, we have

( ) 1 inch 6 picas 12 points

5.0292 m

and

1 chain1.0 furlong 201.168 m (201.168 m ) 10 chains

6 We make use of Table 1-6

(a) We look at the first (“cahiz”) column: 1 fanega is equivalent to what amount of cahiz?

We note from the already completed part of the table that 1 cahiz equals a dozen fanega Thus, 1 fanega = 121 cahiz, or 8.33 × 10−2

cahiz Similarly, “1 cahiz = 48 cuartilla” (in the already completed part) implies that 1 cuartilla = 481 cahiz, or 2.08 × 10−2

cahiz Continuing in this way, the remaining entries in the first column are 6.94 × 10−3

and 3

3.47 10× −

(b) In the second (“fanega”) column, we find 0.250, 8.33 × 10−2, and 4.17 × 10−2

for the last three entries

(c) In the third (“cuartilla”) column, we obtain 0.333 and 0.167 for the last two entries (d) Finally, in the fourth (“almude”) column, we get 12 = 0.500 for the last entry

(e) Since the conversion table indicates that 1 almude is equivalent to 2 medios, our amount of 7.00 almudes must be equal to 14.0 medios

(f) Using the value (1 almude = 6.94 × 10−3

cahiz) found in part (a), we conclude that 7.00 almudes is equivalent to 4.86 × 10−2

cahiz

(g) Since each decimeter is 0.1 meter, then 55.501 cubic decimeters is equal to 0.055501

m3 or 55501 cm3 Thus, 7.00 almudes = 7.0012 fanega = 7.0012 (55501 cm3) = 3.24 × 104

cm3

7 We use the conversion factors found in Appendix D

1 acre ft = (43,560 ft ) ft = 43,560 ft⋅ ⋅Since 2 in = (1/6) ft, the volume of water that fell during the storm is

3

ft

3 3

8 From Fig 1-4, we see that 212 S is equivalent to 258 W and 212 – 32 = 180 S is equivalent to 216 – 60 = 156 Z The information allows us to convert S to W or Z

(a) In units of W, we have

9 The volume of ice is given by the product of the semicircular surface area and the

thickness The area of the semicircle is A = πr2

/2, where r is the radius Therefore, the

volume is

11 (a) Presuming that a French decimal day is equivalent to a regular day, then the ratio

of weeks is simply 10/7 or (to 3 significant figures) 1.43

(b) In a regular day, there are 86400 seconds, but in the French system described in the problem, there would be 105 seconds The ratio is therefore 0.864

12 A day is equivalent to 86400 seconds and a meter is equivalent to a million micrometers, so

3 7 10

6

when t' A − tA = 600 s

16 We denote the pulsar rotation rate f (for frequency)

3

1 rotation1.55780644887275 10 s

×

(a) Multiplying f by the time-interval t = 7.00 days (which is equivalent to 604800 s, if

we ignore significant figure considerations for a moment), we obtain the number of

×

which should now be rounded to 3.88 × 108

rotations since the time-interval was specified in the problem to three significant figures

(b) We note that the problem specifies the exact number of pulsar revolutions (one million) In this case, our unknown is t, and an equation similar to the one we set up in part (a) takes the form N = ft, or

which yields the result t = 1557.80644887275 s (though students who do this calculation

on their calculator might not obtain those last several digits)

(c) Careful reading of the problem shows that the time-uncertainty per revolution is

to tell what the correction should be The following gives the corrections (in seconds) that must be applied to the reading on each clock for each 24-h period The entries were determined by subtracting the clock reading at the end of the interval from the clock reading at the beginning

so we judge clock C to be the best and clock D to be the next best The correction that must be applied to clock A is in the range from 15 s to 17s For clock B it is the range from -5 s to +10 s, for clock E it is in the range from -70 s to -2 s After C and D, A has the smallest range of correction, B has the next smallest range, and E has the greatest range From best to worst, the ranking of the clocks is C, D, A, B, E

18 The last day of the 20 centuries is longer than the first day by

(20 century) (0.001 s century)= 0.02 s

The average day during the 20 centuries is (0 + 0.02)/2 = 0.01 s longer than the first day

Since the increase occurs uniformly, the cumulative effect T is

or roughly two hours

19 When the Sun first disappears while lying down, your line of sight to the top of the

Sun is tangent to the Earth’s surface at point A shown in the figure As you stand, elevating your eyes by a height h, the line of sight to the Sun is tangent to the Earth’s surface at point B

Let d be the distance from point B to your eyes From the Pythagorean theorem, we have

points A and B is θ, which is also the angle through which the Sun moves about Earth

during the time interval t = 11.1 s The value of θ can be obtained by using

Using d =rtanθ , we have d2 =r2tan2θ =2rh, or

2

2tan

h r

θ

=

Using the above value for θ and h = 1.7 m, we have r=5.2 10 m.× 6

20 (a) We find the volume in cubic centimeters

(b) The volume found in part (a) is converted (by dividing by (100 cm/m)3) to 0.731 m3, which corresponds to a mass of

4.06 10 min = 0.77 y0.0018 kg min = ×

after dividing by the number of minutes in a year (365 days)(24 h/day) (60 min/h)

21 If M E is the mass of Earth, m is the average mass of an atom in Earth, and N is the

number of atoms, then M E = Nm or N = M E /m We convert mass m to kilograms using

49

19.32 g

19.32 g/cm

1 cm

m V

(a) We take the volume of the leaf to be its area A multiplied by its thickness z With

density ρ = 19.32 g/cm3 and mass m = 27.63 g, the volume of the leaf is found to be

(b) The volume of a cylinder of length A is V = AA where the cross-section area is that of

a circle: A = πr2 Therefore, with r = 2.500 × 10−6 m and V = 1.430 × 10−6

m3, we obtain

4

2 7.284 10 m 72.84 km

V r

Thus, the mass of a cubic meter of water is 1000 kg

(b) We divide the mass of the water by the time taken to drain it The mass is found from

M = ρV (the product of the volume of water and its density):

5.70 10 kg

158 kg s

3.6 10 s

M R t

×

×

24 The metric prefixes (micro (μ), pico, nano, …) are given for ready reference on the

inside front cover of the textbook (see also Table 1–2) The surface area A of each grain

density, ρ =m V/ , so that the mass can be found from m = ρV, where ρ = 2600 kg/m3

Thus, using V = 4πr3/3, the mass of each grain is

( 6 )3 3

9 3

We observe that (because a cube has six equal faces) the indicated surface area is 6 m2

The number of spheres (the grains of sand) N that have a total surface area of 6 m2 is given by

25 The volume of the section is (2500 m)(800 m)(2.0 m) = 4.0 × 106 m3 Letting “d”

stand for the thickness of the mud after it has (uniformly) distributed in the valley, then

its volume there would be (400 m)(400 m)d Requiring these two volumes to be equal,

we can solve for d Thus, d = 25 m The volume of a small part of the mud over a patch

of area of 4.0 m2 is (4.0)d = 100 m3 Since each cubic meter corresponds to a mass of

1900 kg (stated in the problem), then the mass of that small part of the mud is

5

1.9 10 kg×

26 (a) The volume of the cloud is (3000 m)π(1000 m)2 = 9.4 ×109 m3 Since each cubic meter of the cloud contains from 50 × 106 to 500 × 106 water drops, then we conclude that the entire cloud contains from 4.7 × 1018 to 4.7 × 1019 drops Since the volume of each drop is 43π(10×10− 6 m)3 = 4.2×10−15 m3, then the total volume of water in a cloud

If we ignore the empty spaces between the close-packed spheres, then the density of an

individual iron atom will be the same as the density of any iron sample That is, if M is the mass and V is the volume of an atom, then

M V

chee1tahil

hoon

1 chee

g1hoon

(b) At t=4.21 s, the water mass is

0.8( 4.21 s) 5.00(4.21) 3.00(4.21) 20.00 23.2 g

(c) The rate of mass change at t=2.00 s is

0.2 2.00 s

2

g 1 kg 60 s4.00(2.00) 3.00 g/s 0.48 g/s 0.48

s 1000 g 1 min2.89 10 kg/min

3

g 1 kg 60 s4.00(5.00) 3.00 g/s 0.101 g/s 0.101

s 1000 g 1 min6.05 10 kg/min

m V

If we neglect the volume of the empty spaces between the candies, then the total mass of

the candies in the container when filled to height h is M =ρAh, where

2(14.0 cm)(17.0 cm) 238 cm

unchanged Thus, the rate of mass change is given by

(4.00 10 kg/cm )(238 cm )(0.250 cm/s)0.0238 kg/s 1.43 kg/min

) in addition to a rectangular box (height h´ = 6.0 m

and same base) Therefore,

31

(b) In this case, each dimension (relative to the real house) is reduced by a factor of 1/144 Therefore,

33 In this problem we are asked to differentiate between three types of tons:

displacement ton, freight ton and register ton, all of which are units of volume The three

different tons are given in terms of barrel bulk, with

(a) The difference between 73 “freight” tons and 73 “displacement” tons is

73(freight tons displacement tons) 73(32.124 U.S bushels 28.108 U.S bushels)293.168 U.S bushels 293 U.S bushels

34 The customer expects a volume V1 = 20 × 7056 in3

and receives V2 = 20 × 5826 in.3

, the difference being Δ = −V V1 V2 =24600 in.3, or

where Appendix D has been used

35 The first two conversions are easy enough that a formal conversion is not especially called for, but in the interest of practice makes perfect we go ahead and proceed formally:

36 Table 7 can be completed as follows:

(a) It should be clear that the first column (under “wey”) is the reciprocal of the first row – so that 109 = 0.900, 403 = 7.50 × 10−2, and so forth Thus, 1 pottle = 1.56 × 10−3

wey and 1 gill = 8.32 × 10−6

wey are the last two entries in the first column

(b) In the second column (under “chaldron”), clearly we have 1 chaldron = 1 chaldron (that is, the entries along the “diagonal” in the table must be 1’s) To find out how many chaldron are equal to one bag, we note that 1 wey = 10/9 chaldron = 40/3 bag so that 121chaldron = 1 bag Thus, the next entry in that second column is 121 = 8.33 × 10−2

Similarly, 1 pottle = 1.74 × 10−3 chaldron and 1 gill = 9.24 × 10−6

gill, 1 pottle = 188 gill, and, of course, 1 gill = 1 gill

(f) Using the information from part (c), 1.5 chaldron = (1.5)(12.0) = 18.0 bag And since each bag is 0.1091 m3 we conclude 1.5 chaldron = (18.0)(0.1091) = 1.96 m3

37 The volume of one unit is 1 cm3 = 1 ×10−6 m3, so the volume of a mole of them is 6.02 × 1023 cm3 = 6.02 × 1017 m3 The cube root of this number gives the edge length:

( ) ( )( )

total

2 2

3.00 acre 25.0 perch 4.00 perch

answer: Atotal = 14.5 roods

(b) We convert our intermediate result in part (a):

If the fuel consumption rate is R (in miles/gallon), then the amount of gasoline (in gallons) needed for a trip of distance d (in miles) would be

(miles)(gallon)

(miles/gallon)

d V

R

=

Since the car was manufactured in the U.K., the fuel consumption rate is calibrated based

on U.K gallon, and the correct interpretation should be “40 miles per U.K gallon.” In U.K., one would think of gallon as U.K gallon; however, in the U.S., the word “gallon” would naturally be interpreted as U.S gallon Note also that since

1 U.K gallon=4.5460900 L and 1 U.S gallon =3.7854118 L, the relationship between the two is

This means that the driver mistakenly believes that the car should need 18.8 U.S gallons (b) Using the conversion factor found above, the actual amount required is equivalent to

1.4 10

5 10 3.0 10

×

43 A million milligrams comprise a kilogram, so 2.3 kg/week is 2.3 × 106 mg/week Figuring 7 days a week, 24 hours per day, 3600 second per hour, we find 604800 seconds are equivalent to one week Thus, (2.3×106 mg/week)/(604800 s/week) = 3.8 mg/s

44 The volume of the water that fell is

45 The number of seconds in a year is 3.156 × 107

This is listed in Appendix D and results from the product

(365.25 day/y) (24 h/day) (60 min/h) (60 s/min)

(a) The number of shakes in a second is 108; therefore, there are indeed more shakes per second than there are seconds per year

(b) Denoting the age of the universe as 1 u-day (or 86400 u-sec), then the time during which humans have existed is given by

10

6 10

Therefore, in atomic mass units, the average mass of one atom in the common Eastern mole is

23 23

1 97

2

kenm

mm

2 2

2 2

(b) Similarly, we find

11

1 97

3

3ken

m

mm

50 According to Appendix D, a nautical mile is 1.852 km, so 24.5 nautical miles would

be 45.374 km Also, according to Appendix D, a mile is 1.609 km, so 24.5 miles is 39.4205 km The difference is 5.95 km

51 (a) For the minimum (43 cm) case, 9 cubits converts as follows:

mm, respectively

(c) We can convert length and diameter first and then compute the volume, or first

compute the volume and then convert We proceed using the latter approach (where d is

diameter and A is length)

Similarly, with 0.43 m replaced by 0.53 m, we obtain Vcylinder, max = 4.2 m3

52 Abbreviating wapentake as “wp” and assuming a hide to be 110 acres, we set up the ratio 25 wp/11 barn along with appropriate conversion factors:

equal to the arc length s divided by the radius R For a very large radius circle and small

value of θ, the arc may be approximated as the straight line-segment of length 1 AU Thus,

×Next, we relate AU to light-year (ly) Since a year is about 3.16 × 107

Our results can be further combined to give 1 pc=3.2 ly

54 (a) Using Appendix D, we have 1 ft = 0.3048 m, 1 gal = 231 in.3, and 1 in.3 = 1.639 ×

10−2 L From the latter two items, we find that 1 gal = 3.79 L Thus, the quantity 460

ft2/gal becomes

2 2

Chapter 2

1 The speed (assumed constant) is v = (90 km/h)(1000 m/km) ⁄ (3600 s/h) = 25 m/s

Thus, in 0.50 s, the car travels a distance d = vt = (25 m/s)(0.50 s) ≈ 13 m

2 (a) Using the fact that time = distance/velocity while the velocity is constant, we find

avg 73.2 m 73.2 m

3.05 m 1.22 m/s

3 Since the trip consists of two parts, let the displacements during first and second parts of the motion be Δx 1 and Δx 2, and the corresponding time intervals be Δt 1 and Δt 2 , respectively Now, because the problem is one-dimensional and both displacements are in the same direction, the total displacement is Δx = Δx 1 + Δx 2 , and the total time for the trip is Δt = Δt 1 + Δt 2 Using the definition of average velocity given in Eq 2-2,

Δ + Δ Δ

To find the average speed, we note that during a time Δt if the velocity remains a positive constant, then the speed is equal to the magnitude of velocity, and the distance is equal to the magnitude of displacement, with d = Δ = Δ | x| v t

(a) During the first part of the motion, the displacement is Δx 1 = 40 km and the time interval is

t

Δ

Δ (b) In this case, the average speed is the same as the magnitude of the average velocity: savg = 40 km/h.

(c) The graph of the entire trip is shown below; it consists of two contiguous line segments, the first having a slope of 30 km/h and connecting the origin to (Δt 1 , Δx 1 ) =

(1.33 h, 40 km) and the second having a slope of 60 km/h and connecting (Δt1 , Δx1 )

to (Δt, Δx) = (2.00 h, 80 km)

4 Average speed, as opposed to average velocity, relates to the total distance, as

opposed to the net displacement The distance D up the hill is, of course, the same as

the distance down the hill, and since the speed is constant (during each stage of the

motion) we have speed = D/t Thus, the average speed is

D D v

D v

will use), but if we wished to make the units explicit we would write

For later reference, we also note that the position at t = 0 is x = 0

(e) The position at t = 0 is subtracted from the position at t = 4 s to find the

displacement Δx = 12 m

(f) The position at t = 2 s is subtracted from the position at t = 4 s to give the

displacement Δx = 14 m Eq 2-2, then, leads to

t

Δ

Δ

(g) The position of the object for the interval 0 ≤ t ≤ 4 is plotted below The straight

line drawn from the point at (t, x) = (2 s , –2 m) to (4 s, 12 m) would represent the

average velocity, answer for part (f)

6 Huber’s speed is

v0 = (200 m)/(6.509 s) =30.72 m/s = 110.6 km/h, where we have used the conversion factor 1 m/s = 3.6 km/h Since Whittingham beat

Huber by 19.0 km/h, his speed is v1 = (110.6 km/h + 19.0 km/h) = 129.6 km/h, or 36 m/s (1 km/h = 0.2778 m/s) Thus, using Eq 2-2, the time through a distance of 200 m for Whittingham is

36 m/s

x t v

Δ

7 Recognizing that the gap between the trains is closing at a constant rate of 60 km/h,

the total time that elapses before they crash is t = (60 km)/(60 km/h) = 1.0 h During this time, the bird travels a distance of x = vt = (60 km/h)(1.0 h) = 60 km

8 The amount of time it takes for each person to move a distance L with speed v is s

/ s

t L v

Δ = With each additional person, the depth increases by one body depth d

(a) The rate of increase of the layer of people is

(0.25 m)(3.50 m/s)

0.50 m/s

s s

5.0 m

10 s 0.50 m/s

D t R

where we set L1 ≈ 1000 m in the last step Thus, if L1 and L2 are no different than

about 1.4 m, then runner 1 is indeed faster than runner 2 However, if L1 is shorter

than L2 by more than 1.4 m, then runner 2 would actually be faster

10 Let v be the speed of the wind and w v be the speed of the car c

(a) Suppose during time interval t , the car moves in the same direction as the wind 1

Then the effective speed of the car is given byv eff,1= + , and the distance traveled v c v w

is d v t= eff,1 1= (v c+v t w)1 On the other hand, for the return trip during time interval t2 , the car moves in the opposite direction of the wind and the effective speed would be ,2

v = − The distance traveled is v v d v= eff,2 2t = (v c−v t w)2 The two expressions can be rewritten as

d d v

t t

method 1 gives the car’s speed v a in windless situation c

(b) If method 2 is used, the result would be

160 km Thus, a speed v = (160 km)/(1.25 h) = 128 km/h is needed

12 (a) Let the fast and the slow cars be separated by a distance d at t = 0 If during the

time interval t L v= / s = (12.0 m) /(5.0 m/s) 2.40 s = in which the slow car has moved

a distance of L= 12.0 m, the fast car moves a distance of vt d L= + to join the line

of slow cars, then the shock wave would remain stationary The condition implies a separation of

(25 m/s)(2.4 s) 12.0 m 48.0 m.

(b) Let the initial separation at t= be 0 d = 96.0 m. At a later time t, the slow and

the fast cars have traveled x v t= s and the fast car joins the line by moving a distance

shock

4.80 s

x v

t

Δ

(c) Since x L> , the direction of the shock wave is downstream

13 (a) Denoting the travel time and distance from San Antonio to Houston as T and D,

respectively, the average speed is

which should be rounded to 73 km/h

(b) Using the fact that time = distance/speed while the speed is constant, we find

+ which should be rounded to 68 km/h

(c) The total distance traveled (2D) must not be confused with the net displacement

(zero) We obtain for the two-way trip

discussion We briefly describe the graph (with kilometers-per-hour understood for the slopes): two contiguous line segments, the first having a slope of 55 and

connecting the origin to (t1, x1) = (T/2, 55T/2) and the second having a slope of 90 and connecting (t1, x1) to (T, D) where D = (55 + 90)T/2 The average velocity, from the graphical point of view, is the slope of a line drawn from the origin to (T, D) The

graph (not drawn to scale) is depicted below:

14 Using the general property d

dxexp( )bx =bexp( )bx , we write

v dx dt

de dt

we find x = 5.9 m

15 We use Eq 2-4 to solve the problem

(a) The velocity of the particle is

v dx dt

d

= = ( 4 12 − + 3 2 ) = − 12 6 +

Thus, at t = 1 s, the velocity is v = (–12 + (6)(1)) = –6 m/s

(b) Since v < 0, it is moving in the –x direction at t = 1 s

(c) At t = 1 s, the speed is |v| = 6 m/s

(d) For 0 < t < 2 s, |v| decreases until it vanishes For 2 < t < 3 s, |v| increases from

zero to the value it had in part (c) Then, |v| is larger than that value for t > 3 s

(e) Yes, since v smoothly changes from negative values (consider the t = 1 result) to positive (note that as t → + ∞, we have v → + ∞) One can check that v = 0 when

2 s.

t=

(f) No In fact, from v = –12 + 6t, we know that v > 0 for t > 2 s

16 We use the functional notation x(t), v(t), and a(t) in this solution, where the latter

two quantities are obtained by differentiation:

v t dx t

dv t dt

b g b g= = − 12 and b g b g= = − 12 with SI units understood

(a) From v(t) = 0 we find it is (momentarily) at rest at t = 0

(b) We obtain x(0) = 4.0 m

(c) and (d) Requiring x(t) = 0 in the expression x(t) = 4.0 – 6.0t2 leads to t = ±0.82 s

for the times when the particle can be found passing through the origin

(e) We show both the asked-for graph (on the left) as well as the “shifted” graph that

is relevant to part (f) In both cases, the time axis is given by –3 ≤ t ≤ 3 (SI units understood)

(f) We arrived at the graph on the right (shown above) by adding 20t to the x(t)

expression

(g) Examining where the slopes of the graphs become zero, it is clear that the shift

causes the v = 0 point to correspond to a larger value of x (the top of the second curve

shown in part (e) is higher than that of the first)

17 We use Eq 2-2 for average velocity and Eq 2-4 for instantaneous velocity, and work with distances in centimeters and times in seconds

(a) We plug into the given equation for x for t = 2.00 s and t = 3.00 s and obtain x2 =

21.75 cm and x3 = 50.25 cm, respectively The average velocity during the time interval 2.00 ≤ t ≤ 3.00 s is

which yields vavg = 28.5 cm/s

(b) The instantaneous velocity is v dx t

dt

= = 4 5 2, which, at time t = 2.00 s, yields v =

(4.5)(2.00)2 = 18.0 cm/s

(c) At t = 3.00 s, the instantaneous velocity is v = (4.5)(3.00)2 = 40.5 cm/s

(d) At t = 2.50 s, the instantaneous velocity is v = (4.5)(2.50)2 = 28.1 cm/s

(e) Let t m stand for the moment when the particle is midway between x2 and x3 (that is,

when the particle is at x m = (x2 + x3 )/2 = 36 cm) Therefore,

x m = 9 75 + 15 t m3 ⇒ t m = 2 596

in seconds Thus, the instantaneous speed at this time is v = 4.5(2.596)2 = 30.3 cm/s

(f) The answer to part (a) is given by the slope of the straight line between t = 2 and t

= 3 in this x-vs-t plot The answers to parts (b), (c), (d), and (e) correspond to the

slopes of tangent lines (not shown but easily imagined) to the curve at the appropriate points

18 (a) Taking derivatives of x(t) = 12t2 – 2t3 we obtain the velocity and the acceleration functions:

v (t) = 24t – 6t2 and a(t) = 24 – 12t with length in meters and time in seconds Plugging in the value t = 3 yields

(e) From (d), we see that the x reaches its maximum at t = 4.0 s

(f) A maximum v requires a = 0, which occurs when t = 24/12 = 2.0 s This, inserted into the velocity equation, gives vmax = 24 m/s

(g) From (f), we see that the maximum of v occurs at t = 24/12 = 2.0 s

(h) In part (e), the particle was (momentarily) motionless at t = 4 s The acceleration at

that time is readily found to be 24 – 12(4) = –24 m/s2

(i) The average velocity is defined by Eq 2-2, so we see that the values of x at t = 0 and t = 3 s are needed; these are, respectively, x = 0 and x = 54 m (found in part (a))

19 We represent the initial direction of motion as the +x direction The average

acceleration over a time interval t1≤ ≤ is given by Eq 2-7: t t2

2 1 avg

v t v t v

a

− Δ

20 We use the functional notation x(t), v(t) and a(t) and find the latter two quantities

(a) From 0 = − 15t2 + 20, we see that the only positive value of t for which the particle is (momentarily) stopped is t= 20 15 12 / = s

(b) From 0 = – 30t, we find a(0) = 0 (that is, it vanishes at t = 0)

(c) It is clear that a(t) = – 30t is negative for t > 0

(d) The acceleration a(t) = – 30t is positive for t < 0

(e) The graphs are shown below SI units are understood

21 We use Eq 2-2 (average velocity) and Eq 2-7 (average acceleration) Regarding our coordinate choices, the initial position of the man is taken as the origin and his

direction of motion during 5 min ≤ t ≤ 10 min is taken to be the positive x direction

We also use the fact that Δx v t= Δ ' when the velocity is constant during a time interval Δt'

(a) The entire interval considered is Δt = 8 – 2 = 6 min, which is equivalent to 360 s,

whereas the sub-interval in which he is moving is only Δ = − =t' 8 5 3min 180 s.=

His position at t = 2 min is x = 0 and his position at t = 8 min is x v t′= Δ =

(b) The man is at rest at t = 2 min and has velocity v = +2.2 m/s at t = 8 min Thus,

keeping the answer to 3 significant figures,

indications of the average accelerations found in parts (b) and (d) would be dotted

lines connecting the “steps” at the appropriate t values (the slopes of the dotted lines representing the values of aavg)

22 In this solution, we make use of the notation x(t) for the value of x at a particular t The notations v(t) and a(t) have similar meanings

(a) Since the unit of ct2 is that of length, the unit of c must be that of length/time2, or m/s2 in the SI system

(b) Since bt3 has a unit of length, b must have a unit of length/time3, or m/s3

(c) When the particle reaches its maximum (or its minimum) coordinate its velocity is

zero Since the velocity is given by v = dx/dt = 2ct – 3bt2, v = 0 occurs for t = 0 and

For t = 0, x = x0 = 0 and for t = 1.0 s, x = 1.0 m > x0 Since we seek the maximum, we

reject the first root (t = 0) and accept the second (t = 1s)

(d) In the first 4 s the particle moves from the origin to x = 1.0 m, turns around, and

goes back to

x(4 s) = ( 3 0 m / s 2 )( 4 0 s) 2 − ( 2 0 m / s 3 )( 4 0 s) 3 = − 80 m

The total path length it travels is 1.0 m + 1.0 m + 80 m = 82 m

(e) Its displacement is Δx = x 2 – x1, where x1 = 0 and x2 = –80 m Thus, Δ = −x 80 m

The velocity is given by v = 2ct – 3bt2 = (6.0 m/s2)t – (6.0 m/s3)t2

(l) a(3 s) 6.0 m/s = 2 − (12.0 m/s )(3.0 s) 3 = − 30 m/s 2

(m) a(4 s) 6.0 m/s = 2 − (12.0 m/s )(4.0 s) 3 = − 42 m/s 2

23 Since the problem involves constant acceleration, the motion of the electron can

be readily analyzed using the equations in Table 2-1:

25 We separate the motion into two parts, and take the direction of motion to be positive In part 1, the vehicle accelerates from rest to its highest speed; we are

given v0 = 0; v = 20 m/s and a = 2.0 m/s2 In part 2, the vehicle decelerates from its

highest speed to a halt; we are given v0 = 20 m/s; v = 0 and a = –1.0 m/s2 (negative because the acceleration vector points opposite to the direction of motion)

(a) From Table 2-1, we find t1 (the duration of part 1) from v = v0 + at In this way,

1

20 0 2.0t= + yields t1 = 10 s We obtain the duration t2 of part 2 from the same

equation Thus, 0 = 20 + (–1.0)t2 leads to t2 = 20 s, and the total is t = t1 + t2 = 30 s

(b) For part 1, taking x0 = 0, we use the equation v2 = v20 + 2a(x – x0 ) from Table 2-1

This position is then the initial position for part 2, so that when the same equation is

used in part 2 we obtain

Thus, the final position is x = 300 m That this is also the total distance traveled

should be evident (the vehicle did not "backtrack" or reverse its direction of motion)

26 The constant-acceleration condition permits the use of Table 2-1

(a) Setting v = 0 and x0 = 0 in 2 2

(b) Below are the time plots of the position x and velocity v of the muon from the

moment it enters the field to the time it stops The computation in part (a) made no

reference to t, so that other equations from Table 2-1 (such as v v= 0 + at and

x v t= 0 + 1at

2

2 ) are used in making these plots

27 We use v = v0 + at, with t = 0 as the instant when the velocity equals +9.6 m/s

(a) Since we wish to calculate the velocity for a time before t = 0, we set t = –2.5 s

Thus, Eq 2-11 gives

v= ( 9 6 m / s) +c3 2 m / s 2h( − 2 5 s) = 16 m / s.

(b) Now, t = +2.5 s and we find

v t+ at (the x graph, shown next, on the left) and

v0 + at (the v graph, on the right) over 0 ≤ t ≤ 5 s, with SI units understood

29 We assume the periods of acceleration (duration t1) and deceleration (duration t2 )

are periods of constant a so that Table 2-1 can be used Taking the direction of motion

to be +x then a1 = +1.22 m/s 2 and a2 = –1.22 m/s 2 We use SI units so the velocity at t

traveled during t1 and t2 are the same so that they total to 2(10.59 m) = 21.18 m This implies that for a distance of 190 m – 21.18 m = 168.82 m, the elevator is traveling at constant velocity This time of constant velocity motion is

Therefore, the total time is 8.33 s + 33.21 s ≈ 41.5 s

30 We choose the positive direction to be that of the initial velocity of the car

(implying that a < 0 since it is slowing down) We assume the acceleration is constant

and use Table 2-1

(a) Substituting v0 = 137 km/h = 38.1 m/s, v = 90 km/h = 25 m/s, and a = –5.2 m/s2

into v = v0 + at, we obtain

(b) We take the car to be at x = 0 when the brakes

are applied (at time t = 0) Thus, the coordinate of

the car as a function of time is given by

(38 m/s) 1( 5.2 m/s2) 2

2

in SI units This function is plotted from t = 0 to t

= 2.5 s on the graph to the right We have not

shown the v-vs-t graph here; it is a descending

straight line from v0 to v

31 The constant acceleration stated in the problem permits the use of the equations in Table 2-1

(a) We solve v = v0 + at for the time:

t v v a

(a) We take x0 = 0, and solve x = v0t + 1

2at2 (Eq 2-15) for the acceleration: a = 2(x –

v0t)/t2 Substituting x = 24.0 m, v0 = 56.0 km/h = 15.55 m/s and t = 2.00 s, we find

3.56m/s , 2.00s

x v t a

34 Let d be the 220 m distance between the cars at t = 0, and v1 be the 20 km/h = 50/9

m/s speed (corresponding to a passing point of x1 = 44.5 m) and v2 be the 40 km/h

=100/9 m/s speed (corresponding to a passing point of x2 = 76.6 m) of the red car

We have two equations (based on Eq 2-17):

d – x1 = vo t1 + 12 a t1 2 where t1 = x1 ⁄ v1

d – x2 = vo t2 + 12 a t2 2 where t2 = x2 ⁄ v2