Tài liệu Physics exercises_solution:Chapter 02 doc

2.3: Although the distance could be found, the intermediate calculation can be avoided by considering that the time will be inversely proportional to the speed, and the extra time will b

2.1: a) During the later 4.75-s interval, the rocket moves a distance

m63m

754

m63m1000

)m105150





The direction has been defined to be the –x-direction ( iˆ)

b) Because the bird ends up at the starting point, the average velocity for the round

trip is 0.

2.3: Although the distance could be found, the intermediate calculation can be avoided

by considering that the time will be inversely proportional to the speed, and the extra time will be

min

701hrkm70

hrkm105min)140

,s)m(6.20m

200s)

Fast runner has run (6.20m s)t1770m

Slow runner has run (5.50m s)t 1570m

2.6: The s-waves travel slower, so they arrive 33 s after the p-waves.

km250

s335

.65.3

s33

s33

s

km s

km

p s

p s

v

d v d

v

d t vt d

t t

2.7: a) The van will travel 480 m for the first 60 s and 1200 m for the next 60 s, for a total distance of 1680 m in 120 s and an average speed of 14.0m s b) The first stage of the journey takes 8.0240mms 30s and the second stage of the journey takes

,s12s)m

0.10(

s)0.10)(

sm120.0(s)0.10)(

sm4.2

b) From Eq (2.3), the instantaneous velocity as a function of time is

,)sm360.0()sm80.4(3

t t

ct bt



t

2.10: a) IV: The curve is horizontal; this corresponds to the time when she stops b) I:

This is the time when the curve is most nearly straight and tilted upward (indicating postive velocity) c) V: Here the curve is plainly straight, tilted downward (negative velocity) d) II: The curve has a postive slope that is increasing e) III: The curve is still tilted upward (positive slope and positive velocity), but becoming less so

16 1.7m s (to two significant figures) b) 2

s 10 s m 7 16

0  1.7m s c) No change in speed, so the acceleration

is zero d) The final speed is the same as the initial speed, so the average acceleration is zero

2.13: a) The plot of the velocity seems to be the most curved upward near t = 5 s.

b) The only negative acceleration (downward-sloping part of the plot) is between t = 30 s and t = 40 s c) At t = 20 s, the plot is level, and in Exercise 2.12 the car is said to be

cruising at constant speed, and so the acceleration is zero d) The plot is very nearly a straight line, and the acceleration is that found in part (b) of Exercise 2.12, 1.7m s2.e)

2.14: (a) The displacement vector is:

j i

r(t)(5.0m s)tˆ(10.0m s)tˆ (7.0m s)t(3.0m s2)t2 ˆ

The velocity vector is the time derivative of the displacement vector:

k j

i

))sm0.3(2sm0.7(ˆsm0.10(ˆsm0.5()

t dt

r(t)(5.0m s)(5.0s)ˆ(10.0m s)(5.0s ˆ (7.0m s)(5.0s)(3.0m s2)(25.0s2) ˆ

(25.0m iˆ(50.0m ˆj(40.0m)kˆ

k j

i

k j

i r

ˆ)sm0.23(ˆsm0.10(ˆsm0.5

(

ˆ))s0.5)(

sm0.6(sm0.7((

ˆsm0.10(ˆsm0.5

(b) The velocity in both the x- and the y-directions is constant and nonzero; thus

the overall velocity can never be zero

(c) The object's acceleration is constant, since t does not appear in the acceleration

vector

b) Set v x 0andsolvefor t:t16.0s.

c) Set x = 50.0 cm and solve for t This gives t0and t32.0s The turtle returns

to the starting point after 32.0 s

d) Turtle is 10.0 cm from starting point when x = 60.0 cm or x = 40.0 cm.

s

8.25ands20.6:for solveandcm

In all cases, the negative acceleration indicates an acceleration to the left

2.17: a) Assuming the car comes to rest from 65 mph (29 m/s) in 4 seconds,

.sm25.7)s4()0s

)sm00.3()sm50.5

b) The instantaneous acceleration is obtained by using Eq (2.5),

.)sm2.0(

2.19: a)

b)

2.20: a) The bumper’s velocity and acceleration are given as functions of time by

5 6 2

)sm600.0()sm60.9

t dt

dv

There are two times at which v = 0 (three if negative times are considered), given by t =

0 and t4 = 16 s4 At t = 0, x = 2.17 m and a x = 9.60 m s 2 When t4 = 16 s4,

x = (2.17 m) + (4.80 m s 2) (16s4)– (0.100) m s6)(16 s4)3/2 = 14.97 m,

a x = (9.60 m s 2) – (3.000 m s6)(16 s4) = –38.4 m s 2 b)

2.21: a) Equating Equations (2.9) and (2.10) and solving for v0,

.sm00.5sm0.15s007

m)70(2)

x x

b) The above result for v 0x may be used to find

,sm43.1s

00.7

sm00.5sm0

x

or the intermediate calculation can be avoided by combining Eqs (2.8) and (2.12) to

eliminate v 0x and solving for a x,

.sm43.1s)007(

m0.70s

00.7

sm0.152

x x t

v

x

2.22: a) The acceleration is found from Eq (2.13), which v0x= 0;

2 ft

3.2811m

2 hr mi 1 s m 4470 0 0

v

x

where the conversions are from Appendix E

b) The time can be found from the above acceleration,

.s42.2s

m0.32

)hrmi173(

2 hr mi 1 s m 4470 0

The intermediate calculation may be avoided by using Eq (2.14), again with v 0x = 0,

  2.42s.)

hrmi173(

ft307(2)(2

hr mi 1 s m 4470 0 ft 3.2811m

2.23: From Eq (2.13), with 0, 2  max.Taking 0 0,

m.70.1)

sm250(2

))hrkms)(3.6mhr)(1km105((

2 max

2.24: In Eq (2.14), with x – x0 being the length of the runway, and v 0x = 0 (the plane

starts from rest), 2  0 22808sm 70.0m s

t x x x

v

2.25: a) From Eq (2.13), with v0x 0,

.sm67.1m)120(2

)sm20()(2

2 2

2.26: a) x0 < 0, v 0x < 0, a x< 0

b) x0 > 0, v 0x < 0, a x > 0

c) x0 > 0, v 0x > 0, a x < 0

2.27: a) speeding up:

?s,9.19,0ft,

2

1 0

)(

)(

2.28: a) Interpolating from the graph:

left) the(toscm3.1,

s0.7

At

right) the(toscm72s,

0.4

s 0 6 s / cm 0

8 1.3cm sgraph

of

1s

cm2s5

1s

cm8s62

2.29: a)

b)

2.30: a)

2.31: a) At t = 3 s the graph is horizontal and the acceleration is 0 From t = 5 s to t =

9 s, the acceleration is constant (from the graph) and equal to 45 m s4s20 m s 6.3m s2 From

t = 9 s to t = 13 s the acceleration is constant and equal to

.sm2

b) In the first five seconds, the area under the graph is the area of the rectangle, (20

m)(5 s) = 100 m Between t = 5 s and t = 9 s, the area under the trapezoid is (1/2)(45 m/s

+ 20 m/s)(4 s) = 130 m (compare to Eq (2.14)), and so the total distance in the first 9 s is

230 m Between t = 9 s and t = 13 s, the area under the triangle is

m90s)4)(

2.33: a) The maximum speed will be that after the first 10 min (or 600 s), at which time

the speed will be

.skm18sm101.8s)900)(

sm0.20

b) During the first 15 minutes (and also during the last 15 minutes), the ship will travel

km8100s)

900)(

(  , so the distance traveled at non-constant speed is 16,200

km and the fraction of the distance traveled at constant speed is

,958.0km384,000

km200,16

keeping an extra significant figure

c) The time spent at constant speed is 2.04 104s

s km 18

km 200 , 16 km 000 ,

during both the period of acceleration and deceleration is 900 s, so the total time required for the trip is 2.22104s, about 6.2 hr

2.34: After the initial acceleration, the train has traveled

m8.156)s0.14)(

sm60.1(2

(from Eq (2.12), with x0 = 0, v 0x = 0), and has attained a speed of

.sm4.22)s0.14)(

sm60.1

During the 70-second period when the train moves with constant speed, the train travels

22.4m s 70s 1568m The distance traveled during deceleration is given by Eq (2.13), with v x0,v0x 22.4m s and a x3.50m s2, so the train moves a distance

m 68 71

) m/s

3.50

2(

) s

In terms of the initial acceleration a1, the initial acceleration time t1, the time t2 during

which the train moves at constant speed and the magnitude a2 of the final acceleration,

the total distance xT is given by

which yields the same result

,

|

|

22

|

|

)(2

1)(2

1

2

1 1 2 1 1 2

2 1 1 2

1 1

2 1 1

T  a t  a t t  a a t  a t t  t  a a t 

x

2.36: a) The truck’s position as a function of time is given by xT = vTt, with vT being the

truck’s constant speed, and the car’s position is given by xC = (1/2) aCt 2 Equating the two expressions and dividing by a factor of t (this reflects the fact that the car and the truck are at the same place at t = 0) and solving for t yields

s5.12s

m20.3

)sm0.20(22

2 C



a

v t

and at this time

2.37: a)

The car and the motorcycle have gone the same distance during the same time, so their

average speeds are the same The car's average speed is its constant speed vC, and for constant acceleration from rest, the motorcycle's speed is always twice its

average, or 2vC b) From the above, the motorcyle's speed will be vC after half the time needed to catch the car For motion from rest with constant acceleration, the distance traveled is proportional to the square of the time, so for half the time

one-fourth of the total distance has been covered, or d 4

2.38: a) An initial height of 200 m gives a speed of 60 m when rounded to one ssignificant figure This is approximately 200 km/hr or approximately 150 mi hr

(Different values of the approximate height will give different answers; the above may be interpreted as slightly better than order of magnitude answers.) b) Personal experience will vary, but speeds on the order of one or two meters per second are reasonable c) Air resistance may certainly not be neglected

2.39: a) From Eq (2.13), with v y 0and a y g,

,sm94.2m)440.0)(

sm80.9(2)(

)m440.0(22)(

22)(

y y g t

2.41: a) If the meter stick is in free fall, the distance d is related to the reaction time t by

cm980

2

d g

2.43: a) Using the method of Example 2.8, the time the ring is in the air is

)sm80.9(

)m0.12(sm80.9(2s)m00.5(s)m00.5(

)(2

2

2 2

0

2 0 0

2.156s,keeping an extra significant figure The average velocity is then 2.15612 0 ms 5.57m s, down

As an alternative to using the quadratic formula, the speed of the ring when it hits the ground may be obtained from 2 2 ( 0)

2

1

gt t v y

y  y 

2

2)sm8.9(2

1)sm(5.00m

0.12

2.44: a) Using a y = –g, v 0y = 5.00 m and ys 0 = 40.0 m in Eqs (2.8) and (2.12) gives

)m0.400)(

sm80.9(2)sm00.5()sm00.5(

sm80.9(2)sm00.5()(

0

2 0

)sm00.5(

2 0

e)

,sm232m)0.10)(

sm80.9(2s)m00.6()(

0 t gt v

y  y 

solving for v 0y,

.sm5.14)s00.5)(

sm80.9(2

1)s00.5(

)m0.50(2

2.47: a) (224m s) (0.9s)249m s2 b) 9.80 m s 2 25.4.

s m

find the distance is vavet (224m s) 2)(0.9s)101m

d) (283m s) (1.40s)202 m s2but40g 392 m s2, so the figures are not consistent

2.48: a) From Eq (2.8), solving for t gives (40.0 m s – 20.0 m )/9.80 s m s2 = 2.04 s b) Again from Eq (2.8),

.s12.6s

m80.9

)sm0.20(sm0.40

m80.9

)sm40(sm40





(This ignores the t = 0 solution.)

d) Again from Eq (2.8), (40 m )/(9.80 s m s2) = 4.08 s This is, of course, half the time found in part (c)

e) 9.80 m s2, down, in all cases

f)

2.49: a) For a given initial upward speed, the height would be inversely proportional to

the magnitude of g, and with g one-tenth as large, the height would be ten times higher,

or 7.5 m b) Similarly, if the ball is thrown with the same upward speed, it would go ten times as high, or 180 m c) The maximum height is determined by the speed when hitting the ground; if this speed is to be the same, the maximum height would be ten times as large, or 20 m

2.50: a) From Eq (2.15), the velocity v2 at time t

  t

t αt dt v

v

1 1 2

)(

2

2 1

1 1

to two significant figures

b) From Eq (2.16), the position x2 as a function of time is

x x t v x dt

t1 1

2   

dt t

t

t ((4.40m/s) (0.6m/s ) ))

m0.6

)sm6.0())(

sm40.4()m0.6

1 3 3

2.51: a) From Eqs (2.17) and (2.18), with v0=0 and x0=0,

3 4 3

3 2

2 0

2 (0.75m s ) (0.040 m s )

32)

.)sm010.0()sm25.0(126

32

4 4 3

3 4

3 0

b) For the velocity to be a maximum, the acceleration must be zero; this occurs at t=0

and tB A 12.5s At t=0 the velocity is a minimum, and at t=12.5 s the velocity is

s

m1.39s)5.12)(

sm040.0(s)5.12)(

sm75.0



x

v

a

zero

isslope thebecause0

)ms5.1

a d) h = area under v– t graph

cm100.5)scm100)(

ms0.1(2

1)

cm33)ms5.0(2

1)

3 Triangle

A h

cm0.11

)33.1)(ms2.0(s

cm133)ms3.1(2

1)

ms5.1( Triangle Rectangle

2.53: a) The change in speed is the area under the a x versus t curve between

vertical lines at t 2.5s and t7.5s This area is

scm0.30)s52s5.7)(

scm00.8s

This acceleration is positive so the change in velocity is positive

b) Slope of v x versus t is positive and increasing with t.

2.54: a) To average 4 mi hr, the total time for the twenty-mile ride must be five hours, so the second ten miles must be covered in 3.75 hours, for an average of 2.7

hr

mi b) To average 12 mi hr, the second ten miles must be covered in 25 minutes and the average speed must be 24 mi hr c) After the first hour, only ten of the twenty miles have been covered, and 16 mi hr is not possible as the average speed

2.55: a)

The velocity and acceleration of the particle as functions of time are

)

sm00.20()sm0.18()(

)sm00.9()sm00.20()sm00.9()(

2 3

2 2

a

t t

t v

x x

b) The particle is at rest when the velocity is zero; setting v = 0 in the above expression and using the quadratic formula to solve for the time t,

)sm0.9(2

s)m0.9)(

sm0.9(4)sm0.20()sm0.20(

3

3 2

s

11.1sm00.18

sm00.20

3

2

Note that this time is the numerical average of the times found in part (c) e) The greatest distance is the position of the particle when the velocity is zero and the acceleration is negative; this occurs at 0.63 s, and at that time the particle is at

m

45.2)s630s)(

m00.9(s)63.0)(

sm0.10(s)63.0)(

sm

2.56: a) 2520.0 0 ms 1.25m s.

b)2515ms 1.67m s

c) Her net displacement is zero, so the average velocity has zero magnitude.d) 5035.0 0 ms 1.43m s. Note that the answer to part (d) is the harmonic mean, not the

arithmetic mean, of the answers to parts (a) and (b) (See Exercise 2.5)

2.57: Denote the times, speeds and lengths of the two parts of the trip as t1 and t2, v1 and

v2, and l1 and l2.

a) The average speed for the whole trip is

 km) (34km) 82km h,

76()()

h km 88 km 76 2

2 1 1

2 1 2

l l t

t

l l

or 82.3 km/h, keeping an extra significant figure

b) Assuming nearly straight-line motion (a common feature of Nebraska

highways), the total distance traveled is l1–l2 and

 km) (34km) 31km h

76(

h km

7234kmh

km

8876km2

1

2 1

l l v

(31.4km hrto three significant figures.)

2.58: a) The space per vehicle is the speed divided by the frequency with which the cars

pass a given point;

.vehiclem

40hvehicles2400

hkm96

vehiclem

8.13

hm

2.59: (a) Denote the time for the acceleration (4.0 s) as t1 and the time spent running at

constant speed (5.1 s) as t2 The constant speed is then at1, where a is the unknown acceleration The total l is then given in terms of a, t1 and t2 by

,2

1

2 1

2

1 at t at

and solving for a gives

.sm5.3s)s)(5.10.4()s0.4)(

21(

m)100()

21(

2 2

2 1

2 1

l a

(b) During the 5.1 s interval, the runner is not accelerating, so a = 0.

(c) v t [(3.5m s2)(4s)] (9.1s)1.54 m s2

(d) The runner was moving at constant velocity for the last 5.1 s

2.60: a) Simple subtraction and division gives average speeds during the 2-second

d) With both the acceleration and the speed at the 14.4-m known, either Eq (2.8)

or Eq (2.12) gives the time as 6.0 s

e) From Eq (2.12), x – x0 = (4.8 m )(1.0 s) + s 12(0.8 m s2)(1.0 s)2 = 5.2 m This

is also the average velocity (1/2)(5.6 m + 4.8 s m ) times the time interval of s

1.0 s

2.61: If the driver steps on the gas, the car will travel

m

4.70s)0.3)(

sm)(2.321(s)03s)(

m

If the brake is applied, the car will travel

m,9.42s)0.3)(

sm3.8)(

21(s)03s)(

m

so the driver should apply the brake