Chapter 2 Motion in One Dimension

P2.1 (a) 10 m 5 m s2 savg

xvt

Δ= = =Δ

(b) 5 m 1.2 m s4 savgv = =

(c) 2 1

2 1

5 m 10 m 2.5 m s4 s 2 savg

x xvt t− −

= = = −− −

(d) 2 1

2 1

5 m 5 m 3.3 m s7 s 4 savg

x xvt t− − −

= = = −− −

(e) 2 1

2 1

0 0 0 m s8 0avg

x xvt t− −

= = =− −

P2.3 (a) Let d represent the distance between A and B. Let be the time for which the

walker has the higher speed in

1t

1

5.00 m s dt

= . Let represent the longer time

for the return trip in

2t

2

3.00 m s dt

− = − . Then the times are ( )1 5.00 m s

dt = and

( )2 3.00 m sdt = . The average speed is:

( ) ( ) ( ) ( )( )

2 2

2 2

Total distance 2Total time / 5.00 m s / 3.00 m s 8.00 m s / 15.0 m s

2 15.0 m s3.75 m s

8.00 m s

avg

avg

d d dvd d d

v

+= = =

+

= =

(b) She starts and finishes at the same point A. With total displacement = 0,

average velocity 0= .

P2.4 210x t= : By substitution, for ( )( )s 2.0 2.1 3.0m 40 44.1

tx

== 90

(a) 50 m 50.0 m s1.0 savg

xvt

Δ= = =Δ

(b) 4.1 m 41.0 m s0.1 savg

xvt

Δ= = =Δ

P2.5 (a) at 1. , (Point A) 5 sit = 8.0 mix = at 4. , (Point B) 0 sft = 2.0 mfx =

( )( )2.0 8.0 m 6.0 m 2.4 m s

4 1.5 s 2.5 sf i

avgf i

x xv

t t− −

= = = − = −− −

(b) The slope of the tangent line can be found from points C

and D. and ( )1.0 s, 9.5 mC Ct x= = ( )3.5 s, 0D Dt x= = ,

3.8 m sv ≈ − .

FIG. P2.5

(c) The velocity is zero when x is a minimum. This is at 4 st ≈ .

P2.8 (a) ( )( )5 0 m

5 m s1 0 s

v−

= =−

(b) ( )( )5 10 m

2.5 m s4 2 s

v−

= = −−

(c) ( )( )5 m 5 m

05 s 4 s

v−

= =−

(d) ( )( )0 5 m

5 m s8 s 7 s

v− −

= = +−

FIG. P2.8

P2.9 Once it resumes the race, the hare will run for a time of

1 000 m 800 m25 s

8 m sf i

x

x xt

v− −

= = = .

In this time, the tortoise can crawl a distance

( )( )0.2 m s 25 s 5.00 mf ix x− = = .

P2.13 22.00 3.00x t t= + − , so 3.00 2.00dxv tdt

= = − , and 2.00dvadt

= = −

At 3. : 00 st = (a) ( )2.00 9.00 9.00 m 2.00 mx = + − =

(b) ( )3.00 6.00 m s 3.00 m sv = − = −

(c) 22.00 m sa= −

*P2.14 The acceleration is zero whenever the marble is on a horizontal section. The

acceleration has a constant positive value when the marble is rolling on the 20-to-40-cm

section and has a constant negative value when it is rolling on the second sloping section. The position graph is a straight sloping line whenever the speed is constant and a section of a parabola when the speed changes.

Position as a function of time

0

20

40

60

80

100

time

Posi

tion

alon

g tra

ck, c

m

Velocity as a function of time

-15

0

15

time

x co

mpo

nent

of v

eloc

ity,

arbi

trar

y un

its

 

Acceleration as a function of time

0

time

acce

lera

tion,

arb

itrar

y un

its3

-3

P2.15 (a) At 2. , . 00 st = ( ) ( )23.00 2.00 2.00 2.00 3.00 m 11.0 mx ⎡ ⎤= − + =⎣ ⎦

At 3. , 00 st = ( ) ( )23.00 9.00 2.00 3.00 3.00 m 24.0 mx ⎡ ⎤= − + =⎣ ⎦

so

24.0 m 11.0 m 13.0 m s3.00 s 2.00 savg

xvt

Δ −= = =Δ −

.

(b) At all times the instantaneous velocity is

( ) ( )23.00 2.00 3.00 6.00 2.00 m sdv t t tdt

= − + = −

At 2. , 00 st = ( )6.00 2.00 2.00 m s 10.0 m sv = − =⎡ ⎤⎣ ⎦ .

At 3. , 00 st = ( )6.00 3.00 2.00 m s 16.0 m sv = − =⎡ ⎤⎣ ⎦ .

(c) 216.0 m s 10.0 m s6.00 m s

3.00 s 2.00 savgvat

−Δ= = =Δ −

(d) At all times ( ) 26.00 2.00 6.00 m sda tdt

= − = . This includes both

and .

2.00 st =

3.00 st =

P2.23 (a) 100 m siv = , 25.00 m sa= − , f iv v at= + so 0 100 5t= − ,

so

( )2 2 2f i f iv v a x x= + −

( ) ( )( )20 . Thus x100 2 5.00 0fx= − − 1000 mf = and 20.0 st = .

(b) 1 000 m is greater than 800 m. With this acceleration the plane would overshoot the runway: it cannot land .

P2.25 In the simultaneous equations:

( )12

xf xi x

f i xi xf

v v a t

x x v v

= +

t

⎧ ⎫⎪ ⎪⎨ ⎬

− = +⎪ ⎪⎩ ⎭

we have ( )( )

( )( )

25.60 m s 4.20 s162.4 m 4.20 s2

xf xi

xi xf

v v

v v

⎧ ⎫= −⎪ ⎪⎨ ⎬

= +⎪ ⎪⎩ ⎭

.

So substituting for gives xiv ( )( ) ( )2162.4 m 5.60 m s 4.20 s 4.20 s2 xf xfv v⎡ ⎤= + +⎣ ⎦

( )( )2114.9 m s 5.60 m s 4.20 s2xfv= + .

Thus

3.10 m sxfv = .

P2.28 (a) Compare the position equation 22.00 3.00 4.00x t= + − t to the general form

212f i ix x v t at= + +

to recognize that 2.00 mix = , 3.00 m siv = , and 28.00 m sa= − . The velocity

equation, , is then f iv v a= + t

( )23.00 m s 8.00 m sfv t= − .

The particle changes direction when 0fv = , which occurs at 3 s8

t = . The

position at this time is

( ) ( )2

23 32.00 m 3.00 m s s 4.00 m s s 2.56 m8 8

x ⎛ ⎞ ⎛ ⎞= + − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

.

(b) From 212f i ix x v t at= + + , observe that when f ix x= , the time is given by

2 ivta

= − . Thus, when the particle returns to its initial position, the time is

( )

2

2 3.00 m s 3 s8.00 m s 4

t−

= =−

and the velocity is ( )2 33.00 m s 8.00 m s s 3.00 m s4fv ⎛ ⎞= − = −⎜ ⎟

⎝ ⎠.

P2.32 Take the original point to be when Sue notices the van. Choose the origin of the x-axis at Sue’s car. For her we have 0isx = , 30.0 m sisv = , 22.00 m ssa = − so her position is given by

( ) ( ) ( )2 21 130.0 m s 2.00 m s2 2s is is s

2x t x v t a t t t= + + = + − .

For the van, , 155 mivx = 5.00 m sivv = , 0va = and

( ) ( )21 155 5.00 m s 02v iv iv vx t x v t a t t= + + = + + .

To test for a collision, we look for an instant when both are at the same place: ct

2

2

30.0 155 5.000 25.0 155.

c c c

c c

t t tt t

− = +

= − + From the quadratic formula

( ) ( )225.0 25.0 4 155

13.6 s2ct

± −= = or 11.4 s .

The roots are real, not imaginary, so there is a collision. The smaller value is the

collision time. (The larger value tells when the van would pull ahead again if the vehicles could move through each other). The wreck happens at position

( ) ( )155 m 5.00 m s 11.4 s 212 m+ = .

*P2.33 (a) Starting from rest and accelerating at 13.0 mi h sba = ⋅ , the bicycle reaches its

maximum speed of ,max 20.0 mi hbv = in a time

,max,1

0 20.0 mi h1.54 s

13.0 mi h sb

bb

vt

a−

= = =⋅

Since the acceleration of the car is less than that of the bicycle, the car cannot catch the bicycle until some time (that is, until the bicycle is at its maximum speed and coasting). The total displacement of the bicycle at time t is

ca

,1bt t>

( )

( ) ( )( )

( )

2,1 ,max ,1

2

121.47 ft s mi h1 13.0 1.54 s 20.0 mi h 1.54 s1 mi h 2 s

29.4 ft s 22.6 ft

b b b b bx a t v t t

t

t

Δ = + −

⎛ ⎞ ⎡ ⎤⎛ ⎞= +⎜ ⎟ −⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦⎝ ⎠= −

The total displacement of the car at this time is

( )2 21.47 ft s mi h1 1 9.00 6.62 ft s2 1 mi h 2 sc c

2x a t t t⎛ ⎞ ⎡ ⎤⎛ ⎞

Δ = = =⎜ ⎟ ⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦⎝ ⎠

At the time the car catches the bicycle c bx xΔ = Δ . This gives

( ) ( )2 26.62 ft s 29.4 ft s 22.6 ftt t= − or

( )2 24.44 s 3.42 s 0t t− + = that has only one physically meaningful solution . This solution gives

the total time the bicycle leads the car and is ,1bt t>

3.45 st = .

(b) The lead the bicycle has over the car continues to increase as long as the bicycle

is moving faster than the car. This means until the car attains a speed of ,max 20.0 mi hc bv v= = . Thus, the elapsed time when the bicycle’s lead ceases

to increase is

,max 20.0 mi h2.22 s

9.00 mi h sb

c

vt

a= = =

⋅

At this time, the lead is

( ) ( ) ( )( ) ( )( )22max 2.22 s

29.4 ft s 2.22 s 22.6 ft 6.62 ft s 2.22 sb c b c tx x x x

=⎡ ⎤⎡ ⎤Δ − Δ = Δ − Δ = − −⎣ ⎦ ⎣ ⎦

or ( )max 10.0 ftb cx xΔ −Δ = .

P2.41 (a) : when f iv v gt= − 0fv = 3.00 st = , and 29.80 m sg = . Therefore,

( )( )29.80 m s 3.00 s 29.4 m siv gt= = = .

(b) ( )12f i f iy y v v t− = +

( )( )1 29.4 m s 3.00 s 44.1 m2f iy y− = =

P2.43 Time to fall 3.00 m is found from the equation describing position as a function of time,

with , thus: 0iv = ( 2 213.00 m 9.80 m s2

t= ) , giving 0.782 st = .

(a) With the horse galloping at 10.0 m s , the horizontal distance is 7.82 mvt = .

(b) from above 0.782 st =

P2.51 Let point 0 be at ground level and point 1 be at the end of the engine burn. Let

point 2 be the highest point the rocket reaches and point 3 be just before impact. The data in the table are found for each phase of the rocket’s motion. (0 to 1) ( ) ( )( )22 80.0 2 4.00 1 000fv − = so 120 m sfv =

giving 10( )120 80.0 4.00 t= + .0 st = (1 to 2) ( ) ( )( )20 120 2 9.80 f ix x− = − − giving 735 mf ix x− = giving 120 120 9.80t− = − .2 st = This is the time of maximum height of the rocket. (2 to 3) ( )( )2 0 2 9.80 1735fv − = − −

( )184 9.80fv = − = − t giving 18.8 st =

FIG. P2.51

(a) total 10 12.2 18.8 41.0 st = + + =

(b) ( )

total1.73 kmf ix x− =

(c) final 184 m sv = −

t x v a 0 Launch 0.0 0 80 +4.00 #1 End Thrust 10.0 1 000 120 +4.00 #2 Rise Upwards 22.2 1 735 0 –9.80 #3 Fall to Earth 41.0 0 –184 –9.80 P2.53 (a) Let x be the distance traveled at acceleration a until maximum speed v is

reached. If this is achieved in time we can use the following three equations: 1t

( )= + 112 ix v v t , ( )− = − 1100 10.2x v t , and 1iv v at= + .

The first two give

( )

⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=−

1 1

1 1

1 1100 10.2 10.22 2

200 .20.4

t v t at

at t

1

( ) ( )

( ) ( )

2

2

200For Maggie: 5.43 m s18.4 2.00

200For Judy: 3.83 m s17.4 3.00

a

a

= =

= =

(b) 1v at=

( ) ( )

( )( )

Maggie: 5.43 2.00 10.9 m s

Judy: 3.83 3.00 11.5 m s

v

v

= =

= =

(c) At the six-second mark

( )21 1

1 6.002

x at v t= + −

( )( ) ( )( )

( ) ( ) ( )( )

2

2

1Maggie: 5.43 2.00 10.9 4.00 54.3 m21Judy: 3.83 3.00 11.5 3.00 51.7 m2

x

x

= + =

= + =

Maggie is ahead by 54.3 m − 51.7 m = 2.62 m . Note that your students may

need a reminder that to get the answer in the back of the book they must use

calculator memory or a piece of paper to save intermediate results without ''rounding off'' until the very end.

*P2.54 (a) We first find the distance sstop over which you can stop. The car travels this distance during your reaction time: Δx1 = v0(0.6 s). As you brake to a stop, the average speed of the car is v0/2 , the interval of time is

(vf − vi)/a = −v0/(−2.40 m/s2) = v0 s2/2.40 m, and the braking distance is

Δx2 = vavgΔt = (v0 s2/2.40 m)( v0/2) = v02 s2/4.80 m. The total stopping distance is

then sstop = Δx1 + Δx2 = v0(0.6 s) + v0

2 s2/4.80 m. If the car is at this distance from the intersection, it can barely brake to a stop, so it

should also be able to get through the intersection at constant speed while the light is yellow, moving a total distance sstop + 22 m = v0(0.6 s) + v0

2

s2/4.80 m + 22 m. This constant-speed motion requires time Δty = (sstop + 22 m)/v0 = (v0(0.6 s) + v0

2 s2/4.80 m + 22 m)/v0 = 0.6 s + v0 s2/4.80 m

+ 22 m/v0. (b) Substituting, Δty = 0.6 s + (8 m/s) s2/4.80 m + 22 m/(8 m/s) = 0.6 s + 1.67 s + 2.75 s

= 5.02 s. (c) We are asked about higher and higher speeds. For 11 m/s instead of 8 m/s, the

time is 0.6 s + (11 m/s) s2/4.80 m + 22 m/(11 m/s) = 4.89 s less than we had at the lower

speed. (d) Now the time 0.6 s + (18 m/s) s2/4.80 m + 22 m/(18 m/s) = 5.57 s begins to

increase (e) 0.6 s + (25 m/s) s2/4.80 m + 22 m/(25 m/s) = 6.69 s (f) As v0 goes to zero, the 22 m/v0 term in the expression for Δty becomes large,

approaching infinity. (g) As v0 grows without limit, the v0

s2/4.80 m term in the expression for Δty becomes large, approaching infinity.

(h) Δty decreases steeply from an infinite value at v0 = 0, goes through a rather flat

minimum, and then diverges to infinity as v0 increases without bound. For a very slowly moving car entering the intersection and not allowed to speed up, a very long time is required to get

across the intersection. A very fast-moving car requires a very long time to slow down at the constant acceleration we have assumed.

(i) To find the minimum, we set the derivative of Δty with respect to v0 equal to zero:

2 2

1 20 0 0

0

s s0.6 s + 22 m 0 22 m 04.8 m 4.8 mvd v v

dv− −⎛ ⎞

+ = + −⎜ ⎟⎝ ⎠

=

22 m/ v0

2 = s2/4.8 m v0 = (22m [4.8 m/s2])1/2 = 10.3 m/s (j) Evaluating again, Δty = 0.6 s + (10.3 m/s) s2/4.80 m + 22 m/(10.3 m/s) = 4.88 s ,

just a little less than the answer to part (c). For some students an interesting project might be to measure the yellow-times of traffic

lights on local roadways with various speed limits and compare with the minimum Δtreaction + (width/2 |abraking|)1/2 implied by the analysis here. But do not let the students string a tape measure across

the intersection. *P2.56 (a) From the information in the problem, we model the Ferrari as a particle under constant

acceleration. The important "particle" for this part of the problem is the nose of the car. We use the position equation from the particle under constant acceleration model to find the velocity v0 of the particle as it enters the intersection:

( ) ( )( )

20 0

220 0

12

1228.0 m 0 3.10 s –2.10 m/ s 3.10 s 12.3 m/ s

x x v t at

v v

= + +

→ = + + → =

Now we use the velocity-position equation in the particle under constant acceleration model to find the displacement of the particle from the first edge of the intersection when the Ferrari stops:

( ) ( )( )−−

= + − → − = = = =−

22 22 2 0

0 0 0 2

0 12.3 m/ s2 35.9 m

2 2 2.10 m/ s

v vv v a x x x x x

a

(b) The time interval during which any part of the Ferrari is in the intersection is that time interval between the instant at which the nose enters the intersection and the instant when the tail leaves the intersection. Thus, the change in position of the nose of the Ferrari is 4.52 m + 28.0 m = 32.52 m. We find the time at which the car is at position x = 32.52 m if it is at x = 0 and moving at 12.3 m/s at t = 0:

x = x0 + v0t +12 at2 → 32.52 m = 0 + 12.3 m / s( )t + 1

2 –2.10 m / s2( )t2

→ −1.05t2 + 12.3t − 32.52 = 0 The solutions to this quadratic equation are t = 4.04 s and 7.66 s. Our desired solution is

the lower of these, so t = 4.04 s. (The later time corresponds to the Ferrari stopping and

reversing, which it must do if the acceleration truly remains constant, and arriving again at the position x = 32.52 m.)

(c) We again define t = 0 as the time at which the nose of the Ferrari enters the intersection.

Then at time t = 4.04 s, the tail of the Ferrari leaves the intersection. Therefore, to find the minimum distance from the intersection for the Corvette, its nose must enter the intersection at t = 4.04 s. We calculate this distance from the position equation:

( )( )− + + = + + =22 20 0

1 10 0 5.60 m/ s 4.04 s 45.8 m2 2

x x v t at

(d) We use the velocity equation: P2.59 (a) We require s kx x= when 1.00s kt t= +

( )( ) ( )( )2 22 21 13.50 m s 1.00 4.90 m s2 2

1.00 1.183

5.46 s .

s k

k k

k

k kx t t

t t

t

x= + = =

+ =

=

(b) ( )( )221 4.90 m s 5.46 s 73.0 m2kx = =

(c) ( )( )24.90 m s 5.46 s 26.7 m skv = =

( )( )23.50 m s 6.46 s 22.6 m ssv = =