# Solution Manual For Fundamentals Of Physics Extended, 9th Edition

Preview Extract

Chapter 2
1. The speed (assumed constant) is v = (90 km/h)(1000 m/km) โ (3600 s/h) = 25 m/s.
Thus, in 0.50 s, the car travels a distance d = vt = (25 m/s)(0.50 s) โ 13 m.
2. (a) Using the fact that time = distance/velocity while the velocity is constant, we
find
73.2 m + 73.2 m
vavg = 73.2 m 73.2 m = 1.74 m/s.
1.22 m/s + 3.05 m
(b) Using the fact that distance = vt while the velocity v is constant, we find
vavg =
(122
. m / s)(60 s) + (3.05 m / s)(60 s)
= 2.14 m / s.
120 s
(c) The graphs are shown below (with meters and seconds understood). The first
consists of two (solid) line segments, the first having a slope of 1.22 and the second
having a slope of 3.05. The slope of the dashed line represents the average velocity (in
both graphs). The second graph also consists of two (solid) line segments, having the
same slopes as before โ the main difference (compared to the first graph) being that
the stage involving higher-speed motion lasts much longer.
3. Since the trip consists of two parts, let the displacements during first and second
parts of the motion be ฮx1 and ฮx2, and the corresponding time intervals be ฮt1 and ฮt2,
respectively. Now, because the problem is one-dimensional and both displacements
are in the same direction, the total displacement is ฮx = ฮx1 + ฮx2, and the total time
for the trip is ฮt = ฮt1 + ฮt2. Using the definition of average velocity given in Eq. 2-2,
we have
ฮx ฮx1 + ฮx2
=
.
vavg =
ฮt ฮt1 + ฮt2
To find the average speed, we note that during a time ฮt 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 = | ฮx | = vฮt .
21
22
CHAPTER 2
(a) During the first part of the motion, the displacement is ฮx1 = 40 km and the time
interval is
(40 km)
t1 =
= 133
. h.
(30 km / h)
Similarly, during the second part the displacement is ฮx2 = 40 km and the time
interval is
(40 km)
t2 =
= 0.67 h.
(60 km / h)
The total displacement is ฮx = ฮx1 + ฮx2 = 40 km + 40 km = 80 km, and the total time
elapsed is ฮt = ฮt1 + ฮt2 = 2.00 h. Consequently, the average velocity is
vavg =
ฮx (80 km)
=
= 40 km/h.
ฮt
(2.0 h)
(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 (ฮt1, ฮx1) =
(1.33 h, 40 km) and the second having a slope of 60 km/h and connecting (ฮt1, ฮx1)
to (ฮt, ฮx) = (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
Dup + Ddown
t up + tdown
=
2D
D
D
+
vup vdown
which, after canceling D and plugging in vup = 40 km/h and vdown = 60 km/h, yields 48
km/h for the average speed.
5. Using x = 3t โ 4t2 + t3 with SI units understood is efficient (and is the approach we
23
will use), but if we wished to make the units explicit we would write
x = (3 m/s)t โ (4 m/s2)t2 + (1 m/s3)t3.
We will quote our answers to one or two significant figures, and not try to follow the
significant figure rules rigorously.
(a) Plugging in t = 1 s yields x = 3 โ 4 + 1 = 0.
(b) With t = 2 s we get x = 3(2) โ 4(2)2+(2)3 = โ2 m.
(c) With t = 3 s we have x = 0 m.
(d) Plugging in t = 4 s gives x = 12 m.
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 ฮx = 12 m.
(f) The position at t = 2 s is subtracted from the position at t = 4 s to give the
displacement ฮx = 14 m. Eq. 2-2, then, leads to
vavg =
ฮx 14 m
=
= 7 m/s.
ฮt
2s
(g) The position of the object for the interval 0 โค t โค 4 is plotted below. The straight
line drawn from the point at (t, x) = (2 s , โ2 m) to (4 s, 12 m) would represent the
average velocity, answer for part (f).
6. Huberโs 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
24
CHAPTER 2
ฮt =
ฮ x 200 m
=
= 5.554 s.
v1 36 m/s
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 vs is
ฮt = L / vs . With each additional person, the depth increases by one body depth d
(a) The rate of increase of the layer of people is
R=
dv (0.25 m)(3.50 m/s)
d
d
=
= s =
= 0.50 m/s
ฮt L / vs
L
1.75 m
(b) The amount of time required to reach a depth of D = 5.0 m is
t=
5.0 m
D
=
= 10 s
R 0.50 m/s
9. Converting to seconds, the running times are t1 = 147.95 s and t2 = 148.15 s,
respectively. If the runners were equally fast, then
savg1 = savg 2
โ
L1 L2
= .
t1 t2
From this we obtain
โt
โ
โ 148.15 โ
L2 โ L1 = โ 2 โ 1 โ L1 = โ
โ 1 โ L1 = 0.00135 L1 โ 1.4 m
โ 147.95 โ
โ t1
โ
where we set L1 โ 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 vw be the speed of the wind and vc be the speed of the car.
(a) Suppose during time interval t1 , the car moves in the same direction as the wind.
Then the effective speed of the car is given by veff ,1 = vc + vw , and the distance traveled
is d = veff ,1t1 = (vc + vw )t1 . 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
veff ,2 = vc โ vw . The distance traveled is d = veff ,2t2 = (vc โ vw )t2 . The two expressions
can be rewritten as
25
vc + vw =
d
t1
and vc โ vw =
d
t2
1โd d โ
Adding the two equations and dividing by two, we obtain vc = โ + โ . Thus,
2 โ t1 t2 โ
method 1 gives the carโs speed vc a in windless situation.
(b) If method 2 is used, the result would be
d
2d
vcโฒ =
=
=
(t1 + t2 ) / 2 t1 + t2
2d
d
d
+
vc + vw vc โ vw
โก โ v โ2 โค
vc2 โ vw2
=
= vc โข1 โ โ w โ โฅ .
vc
โขโฃ โ vc โ โฅโฆ
The fractional difference is
2
vc โ vcโฒ โ vw โ
= โ โ = (0.0240) 2 = 5.76 ร10โ4 .
vc
โ vc โ
11. The values used in the problem statement make it easy to see that the first part of
the trip (at 100 km/h) takes 1 hour, and the second part (at 40 km/h) also takes 1 hour.
Expressed in decimal form, the time left is 1.25 hour, and the distance that remains is
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 / vs = (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
d = vt โ L = (25 m/s)(2.4 s) โ 12.0 m = 48.0 m.
(b) Let the initial separation at t = 0 be d = 96.0 m. At a later time t, the slow and
the fast cars have traveled x = vs t and the fast car joins the line by moving a distance
d + x . From
t=
x d+x
=
,
vs
v
we get
x=
vs
5.00 m/s
(96.0 m) = 24.0 m,
d=
25.0 m/s โ 5.00 m/s
v โ vs
which in turn gives t = (24.0 m) /(5.00 m/s) = 4.80 s. Since the rear of the slow-car
pack has moved a distance of ฮx = x โ L = 24.0 m โ 12.0 m = 12.0 m downstream, the
speed of the rear of the slow-car pack, or equivalently, the speed of the shock wave, is
vshock =
ฮx 12.0 m
=
= 2.50 m/s.
t
4.80 s
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CHAPTER 2
(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
savg1 =
D (55 km/h)(T/2) + (90 km/h)(T / 2)
=
= 72.5 km/h
T
T
which should be rounded to 73 km/h.
(b) Using the fact that time = distance/speed while the speed is constant, we find
savg2 =
D
D
= D/2
= 68.3 km/h
/2
T 55 km/h + 90Dkm/h
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
savg =
2D
D
D
72.5 km/h + 68.3 km/h
= 70 km/h.
(d) Since the net displacement vanishes, the average velocity for the trip in its entirety
is zero.
(e) In asking for a sketch, the problem is allowing the student to arbitrarily set the
distance D (the intent is not to make the student go to an atlas to look it up); the
student can just as easily arbitrarily set T instead of D, as will be clear in the following
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:
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14. Using the general property
v=
d
dx
exp(bx ) = b exp(bx ) , we write
FG
H
IJ
K
FG IJ .
H K
dx
d (19t )
de โ t
=
โ
e โ t + (19t ) โ
dt
dt
dt
If a concern develops about the appearance of an argument of the exponential (โt)
apparently having units, then an explicit factor of 1/T where T = 1 second can be
inserted and carried through the computation (which does not change our answer).
The result of this differentiation is
v = 16(1 โ t )e โ t
with t and v in SI units (s and m/s, respectively). We see that this function is zero
when t = 1 s. Now that we know when it stops, we find out where it stops by
plugging our result t = 1 into the given function x = 16teโt with x in meters. Therefore,
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 d
=
(4 โ 12t + 3t 2 ) = โ12 + 6t .
dt dt
Thus, at t = 1 s, the velocity is v = (โ12 + (6)(1)) = โ6 m/s.
(b) Since v < 0, it is moving in the โx 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.
(e) Yes, since v smoothly changes from negative values (consider the t = 1 result) to
positive (note that as t โ + โ, we have v โ + โ). One can check that v = 0 when
t = 2 s.
(f) No. In fact, from v = โ12 + 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:
b g dxdtbt g = โ 12t and abt g = dvdtbt g = โ 12
vt =
with SI units understood.
(a) From v(t) = 0 we find it is (momentarily) at rest at t = 0.
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CHAPTER 2
(b) We obtain x(0) = 4.0 m.
(c) and (d) Requiring x(t) = 0 in the expression x(t) = 4.0 โ 6.0t2 leads to t = ยฑ0.82 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 โshiftedโ graph that
is relevant to part (f). In both cases, the time axis is given by โ3 โค t โค 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 โค t โค 3.00 s is
vavg =
ฮx 50.25 cm โ 2175
. cm
=
ฮt
3.00 s โ 2.00 s
which yields vavg = 28.5 cm/s.
2
(b) The instantaneous velocity is v = dx
dt = 4.5t , 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 tm stand for the moment when the particle is midway between x2 and x3 (that is,
when the particle is at xm = (x2 + x3)/2 = 36 cm). Therefore,
xm = 9.75 + 15
. tm3
โ
tm = 2.596
in seconds. Thus, the instantaneous speed at this time is v = 4.5(2.596)2 = 30.3 cm/s.
29
(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 โ 2t3 we obtain the velocity and the
acceleration functions:
v(t) = 24t โ 6t2
and
a(t) = 24 โ 12t
with length in meters and time in seconds. Plugging in the value t = 3 yields
x(3) = 54 m .
(b) Similarly, plugging in the value t = 3 yields v(3) = 18 m/s.
(c) For t = 3, a(3) = โ12 m/s2.
(d) At the maximum x, we must have v = 0; eliminating the t = 0 root, the velocity
equation reveals t = 24/6 = 4 s for the time of maximum x. Plugging t = 4 into the
equation for x leads to x = 64 m for the largest x value reached by the particle.
(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 โ 12(4) = โ24 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)).
Thus,
54 โ 0
vavg =
= 18 m/s.
3โ 0
19. We represent the initial direction of motion as the +x direction. The average
acceleration over a time interval t1 โค t โค t2 is given by Eq. 2-7:
30
CHAPTER 2
ฮv v(t2 ) โ v(t1 )
=
.
t2 โ t1
ฮt
Let v1 = +18 m/s at t1 = 0 and v2 = โ30 m/s at t2 = 2.4 s. Using Eq. 2-7 we find
aavg =
aavg =
v(t2 ) โ v(t1 ) (โ30 m/s) โ (+1 m/s)
=
= โ 20 m/s 2 .
t2 โ t1
2.4 s โ 0
The average acceleration has magnitude 20 m/s2 and is in the opposite direction to the
particleโs initial velocity. This makes sense because the velocity of the particle is
decreasing over the time interval.
20. We use the functional notation x(t), v(t) and a(t) and find the latter two quantities
by differentiating:
b g dxtbt g = โ 15t + 20 and abt g = dvdtbt g = โ 30t
vt =
2
with SI units understood. These expressions are used in the parts that follow.
(a) From 0 = โ 15t 2 + 20 , we see that the only positive value of t for which the
. s.
particle is (momentarily) stopped is t = 20 / 15 = 12
(b) From 0 = โ 30t, we find a(0) = 0 (that is, it vanishes at t = 0).
(c) It is clear that a(t) = โ 30t is negative for t > 0.
(d) The acceleration a(t) = โ 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
31
direction of motion during 5 min โค t โค 10 min is taken to be the positive x direction.
We also use the fact that ฮx = vฮt ' when the velocity is constant during a time
interval ฮt' .
(a) The entire interval considered is ฮt = 8 โ 2 = 6 min, which is equivalent to 360 s,
whereas the sub-interval in which he is moving is only ฮt' = 8 โ 5 = 3min = 180 s.
His position at t = 2 min is x = 0 and his position at t = 8 min is x = vฮ t โฒ =
(2.2)(180) = 396 m . Therefore,
396 m โ 0
vavg =
= 110
. m / s.
360 s
(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,
aavg =
2.2 m / s โ 0
= 0.00611 m / s2 .
360 s
(c) Now, the entire interval considered is ฮt = 9 โ 3 = 6 min (360 s again), whereas the
sub-interval in which he is moving is ฮ t โฒ = 9 โ 5 = 4 min = 240 s ). His position at
t = 3 min is x = 0 and his position at t = 9 min is x = vฮ t โฒ = (2.2)(240) = 528 m .
Therefore,
528 m โ 0
. m / s.
vavg =
= 147
360 s
(d) The man is at rest at t = 3 min and has velocity v = +2.2 m/s at t = 9 min.
Consequently, aavg = 2.2/360 = 0.00611 m/s2 just as in part (b).
(e) The horizontal line near the bottom of this x-vs-t graph represents the man
standing at x = 0 for 0 โค t < 300 s and the linearly rising line for 300 โค t โค 600 s
represents his constant-velocity motion. The lines represent the answers to part (a)
and (c) in the sense that their slopes yield those results.
The graph of v-vs-t is not shown here, but would consist of two horizontal โstepsโ
(one at v = 0 for 0 โค t 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 โ (2.0 m / s 3 )(4.0 s) 3 = โ 80 m .
The total path length it travels is 1.0 m + 1.0 m + 80 m = 82 m.
(e) Its displacement is ฮx = x2 โ x1, where x1 = 0 and x2 = โ80 m. Thus, ฮx = โ80 m .
The velocity is given by v = 2ct โ 3bt2 = (6.0 m/s2)t โ (6.0 m/s3)t2.
(f) Plugging in t = 1 s, we obtain
v(1 s) = (6.0 m/s 2 )(1.0 s) โ (6.0 m/s3 )(1.0 s) 2 = 0.
(g) Similarly, v(2 s) = (6.0 m/s 2 )(2.0 s) โ (6.0 m/s3 )(2.0 s) 2 = โ 12m/s .
(h) v(3 s) = (6.0 m/s 2 )(3.0 s) โ (6.0 m/s3 )(3.0 s)2 = โ 36 m/s .
(i) v(4 s) = (6.0 m/s 2 )(4.0 s) โ (6.0 m/s3 )(4.0 s) 2 = โ72 m/s .
The acceleration is given by a = dv/dt = 2c โ 6b = 6.0 m/s2 โ (12.0 m/s3)t.
(j) Plugging in t = 1 s, we obtain
a (1 s) = 6.0 m/s 2 โ (12.0 m/s3 )(1.0 s) = โ 6.0 m/s 2 .
(k) a (2 s) = 6.0 m/s 2 โ (12.0 m/s3 )(2.0 s) = โ 18 m/s 2 .
33
(l) a (3 s) = 6.0 m/s 2 โ (12.0 m/s3 )(3.0 s) = โ30 m/s 2 .
(m) a (4 s) = 6.0 m/s 2 โ (12.0 m/s3 )(4.0 s) = โ 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:
v = v0 + at
(2 โ 11)
1
x โ x0 = v0t + at 2
2
(2 โ 15)
v 2 = v02 + 2a ( x โ x0 )
(2 โ 16)
The acceleration can be found by solving Eq. (2-16). With v0 = 1.50 ร105 m/s ,
v = 5.70 ร 106 m/s , x0 = 0 and x = 0.010 m, we find the average acceleration to be
a=
v 2 โ v02 (5.7 ร 106 m/s) 2 โ (1.5 ร 105 m/s) 2
=
= 1.62 ร1015 m/s 2 .
2x
2(0.010 m)
24. In this problem we are given the initial and final speeds, and the displacement, and
are asked to find the acceleration. We use the constant-acceleration equation given in
Eq. 2-16, v2 = v20 + 2a(x โ x0).
(a) Given that v0 = 0 , v = 1.6 m/s, and ฮx = 5.0 ฮผ m, the acceleration of the spores
during the launch is
v 2 โ v02
(1.6 m/s) 2
=
= 2.56 ร 105 m/s 2 = 2.6 ร 104 g
a=
โ6
2x
2(5.0 ร 10 m)
(b) During the speed-reduction stage, the acceleration is
v 2 โ v02
0 โ (1.6 m/s) 2
a=
=
= โ1.28 ร 103 m/s 2 = โ1.3 ร 102 g
โ3
2x
2(1.0 ร10 m)
The negative sign means that the spores are decelerating.
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 = โ1.0 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,
20 = 0 + 2.0t1 yields t1 = 10 s. We obtain the duration t2 of part 2 from the same
equation. Thus, 0 = 20 + (โ1.0)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 โ x0) from Table 2-1
34
CHAPTER 2
and find
x=
v 2 โ v02 (20 m/s) 2 โ (0) 2
=
= 100 m .
2a
2(2.0 m/s 2 )
This position is then the initial position for part 2, so that when the same equation is
used in part 2 we obtain
v 2 โ v02 (0) 2 โ (20 m/s) 2
.
x โ 100 m =
=
2a
2(โ1.0 m/s 2 )
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 v 2 = v02 + 2a ( x โ x0 ) , we find
x= โ
1 v02
1 (5.00 ร 106 ) 2
=โ
= 0.100 m .
2 a
2 โ1.25 ร 1014
Since the muon is slowing, the initial velocity and the acceleration must have opposite
signs.
(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 = v0 + at and
x = v0 t + 12 at 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 = โ2.5 s.
Thus, Eq. 2-11 gives
c
h
v = (9.6 m / s) + 3.2 m / s2 ( โ2.5 s) = 16
. m / s.
35
(b) Now, t = +2.5 s and we find
c
h
v = (9.6 m / s) + 3.2 m / s2 (2.5 s) = 18 m / s.
28. We take +x in the direction of motion, so v0 = +24.6 m/s and a = โ 4.92 m/s2. We
also take x0 = 0.
(a) The time to come to a halt is found using Eq. 2-11:
0 = v0 + at โ t =
24.6 m/s
= 5.00 s .
โ 4.92 m/s 2
(b) Although several of the equations in Table 2-1 will yield the result, we choose Eq.
2-16 (since it does not depend on our answer to part (a)).
0 = v02 + 2ax โ x = โ
(24.6 m/s) 2
= 61.5 m .
2 ( โ 4.92 m/s 2 )
(c) Using these results, we plot v0t + 12 at 2 (the x graph, shown next, on the left) and
v0 + at (the v graph, on the right) over 0 โค t โค 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/s2 and a2 = โ1.22 m/s2. We use SI units so the velocity at t
= t1 is v = 305/60 = 5.08 m/s.
(a) We denote ฮx as the distance moved during t1, and use Eq. 2-16:
(5.08 m/s) 2
= 10.59 m โ 10.6 m.
v = v + 2a1ฮx โ ฮx =
2(1.22 m/s 2 )
2
2
0
(b) Using Eq. 2-11, we have
t1 =
v โ v0 5.08 m/s
=
= 4.17 s.
a1
1.22 m/s 2
The deceleration time t2 turns out to be the same so that t1 + t2 = 8.33 s. The distances
36
CHAPTER 2
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 โ 21.18 m = 168.82 m, the elevator is traveling at
constant velocity. This time of constant velocity motion is
t3 =
168.82 m
= 33.21 s.
5.08 m / s
Therefore, the total time is 8.33 s + 33.21 s โ 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 = โ5.2 m/s2
into v = v0 + at, we obtain
t=
25 m / s โ 38 m / s
= 2.5 s .
โ5.2 m / s2
(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
x = ( 38 m/s ) t +
1
( โ5.2 m/s2 ) t 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 โ v0 101 (3.0 ร 10 8 m / s)
= 31
=
. ร 10 6 s
9.8 m / s 2
a
which is equivalent to 1.2 months.
(b) We evaluate x = x0 + v0 t + 12 at 2 , with x0 = 0. The result is
x=
1
9.8 m/s 2 ) (3.1ร106 s) 2 = 4.6 ร1013 m .
(
2
Note that in solving parts (a) and (b), we did not use the equation v 2 = v02 + 2a( x โ x0 ) .
This equation can be employed for consistency check. The final velocity based on this
37
equation is
v = v02 + 2a ( x โ x0 ) = 0 + 2(9.8 m/s 2 )(4.6 ร 1013 m โ 0) = 3.0 ร 107 m/s ,
which is what was given in the problem statement. So we know the problems have
been solved correctly.
32. The acceleration is found from Eq. 2-11 (or, suitably interpreted, Eq. 2-7).
a=
F 1000 m / kmIJ
1020 km / hg G
b
H 3600 s / h K
ฮv
ฮt
=
. s
14
= 202.4 m / s 2 .
In terms of the gravitational acceleration g, this is expressed as a multiple of 9.8 m/s2
as follows:
โ 202.4 m/s 2 โ
a=โ
g = 21g .
2 โ
โ 9.8 m/s โ
33. The problem statement (see part (a)) indicates that a = constant, which allows us
to use Table 2-1.
(a) We take x0 = 0, and solve x = v0t + 21 at2 (Eq. 2-15) for the acceleration: a = 2(x โ
v0t)/t2. Substituting x = 24.0 m, v0 = 56.0 km/h = 15.55 m/s and t = 2.00 s, we find
a=
2( x โ v0t ) 2 ( 24.0m โ ( 15.55m/s ) ( 2.00s ) )
=
= โ 3.56m/s 2 ,
2
t2
( 2.00s )
or | a | = 3.56 m/s 2 . The negative sign indicates that the acceleration is opposite to
the direction of motion of the car. The car is slowing down.
(b) We evaluate v = v0 + at as follows:
c
v = 1555
. m / s โ 356
. m / s2
h b2.00 sg = 8.43 m / s
which can also be converted to 30.3 km/h.
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):
1
where t1 = x1 โ v1
1
where t2 = x2 โ v2
d โ x1 = vo t1 + 2 a t12
d โ x2 = vo t2 + 2 a t22
38
CHAPTER 2
We simultaneously solve these equations and obtain the following results:
(a) The initial velocity of the green car is vo = โ 13.9 m/s. or roughly โ 50 km/h (the
negative sign means that itโs along the โx direction).
(b) The corresponding acceleration of the car is a = โ 2.0 m/s2 (the negative sign
means that itโs along the โx direction).
35. The positions of the cars as a function of time are given by
1
1
xr (t ) = xr 0 + ar t 2 = (โ35.0 m) + ar t 2
2
2
xg (t ) = xg 0 + vg t = (270 m) โ (20 m/s)t
where we have substituted the velocity and not the speed for the green car. The two
cars pass each other at t = 12.0 s when the graphed lines cross. This implies that
1
(270 m) โ (20 m/s)(12.0 s) = 30 m = (โ35.0 m) + ar (12.0 s) 2
2
which can be solved to give ar = 0.90 m/s 2 .
36. (a) Equation 2-15 is used for part 1 of the trip and Eq. 2-18 is used for part 2:
1
where a1 = 2.25 m/s2 and ฮx1 =
1
where a2 = โ0.75 m/s2 and ฮx2 =
ฮx1 = vo1 t1 + 2 a1 t12
ฮx2 = v2 t2 โ 2 a2 t22
900
4 m
3(900)
m
4
In addition, vo1 = v2 = 0. Solving these equations for the times and adding the results
gives t = t1 + t2 = 56.6 s.
(b) Equation 2-16 is used for part 1 of the trip:
โ 900 โ
2 2
v2 = (vo1)2 + 2a1ฮx1 = 0 + 2(2.25) โ
โ = 1013 m /s
โ 4 โ
which leads to v = 31.8 m/s for the maximum speed.
37. (a) From the figure, we see that x0 = โ2.0 m. From Table 2-1, we can apply
x โ x0 = v0t +
1
2
at2
with t = 1.0 s, and then again with t = 2.0 s. This yields two equations for the two
unknowns, v0 and a:
39
1
2
0.0 โ ( โ2.0 m ) = v0 (1.0 s ) + a (1.0 s )
2
1
2
6.0 m โ ( โ2.0 m ) = v0 ( 2.0 s ) + a ( 2.0 s ) .
2
Solving these simultaneous equations yields the results v0 = 0 and a = 4.0 m/s2.
(b) The fact that the answer is positive tells us that the acceleration vector points in
the +x direction.
38. We assume the train accelerates from rest ( v0 = 0 and x0 = 0 ) at
a1 = +134
. m / s2 until the midway point and then decelerates at a2 = โ134
. m / s2
until it comes to a stop v2 = 0 at the next station. The velocity at the midpoint is v1,
which occurs at x1 = 806/2 = 403m.
b
g
(a) Equation 2-16 leads to
v12 = v02 + 2a1 x1 โ v1 = 2 (1.34 m/s 2 ) ( 403 m ) = 32.9 m/s.
(b) The time t1 for the accelerating stage is (using Eq. 2-15)
x1 = v0t1 +
2 ( 403 m )
1 2
a1t1 โ t1 =
= 24.53 s .
2
1.34 m/s 2
Since the time interval for the decelerating stage turns out to be the same, we double
this result and obtain t = 49.1 s for the travel time between stations.
(c) With a โdead timeโ of 20 s, we have T = t + 20 = 69.1 s for the total time between
start-ups. Thus, Eq. 2-2 gives
806 m
vavg =
= 117
. m/s .
69.1 s
(d) The graphs for x, v and a as a function of t are shown below. The third graph, a(t),
consists of three horizontal โstepsโ โ one at 1.34 m/s2 during 0 < t < 24.53 s, and
the next at โ1.34 m/s2 during 24.53 s < t < 49.1 s and the last at zero during the โdead
timeโ 49.1 s < t 5/2
then there are no (real) solutions to the equation; the cars are never side by side.
(e) Here we have 102 โ 2(โ20)(aB) > 0 โ
at two different times.
two real roots.
The cars are side by side
41
40. We take the direction of motion as +x, so a = โ5.18 m/s2, and we use SI units, so
v0 = 55(1000/3600) = 15.28 m/s.
(a) The velocity is constant during the reaction time T, so the distance traveled during
it is
dr = v0T โ (15.28 m/s) (0.75 s) = 11.46 m.
We use Eq. 2-16 (with v = 0) to find the distance db traveled during braking:
v 2 = v02 + 2adb โ db = โ
(15.28 m/s) 2
2 ( โ5.18 m/s 2 )
which yields db = 22.53 m. Thus, the total distance is dr + db = 34.0 m, which means
that the driver is able to stop in time. And if the driver were to continue at v0, the car
would enter the intersection in t = (40 m)/(15.28 m/s) = 2.6 s, which is (barely)
enough time to enter the intersection before the light turns, which many people would
consider an acceptable situation.
(b) In this case, the total distance to stop (found in part (a) to be 34 m) is greater than
the distance to the intersection, so the driver cannot stop without the front end of the
car being a couple of meters into the intersection. And the time to reach it at constant
speed is 32/15.28 = 2.1 s, which is too long (the light turns in 1.8 s). The driver is
caught between a rock and a hard place.
41. The displacement (ฮx) for each train is the โareaโ in the graph (since the
displacement is the integral of the velocity). Each area is triangular, and the area of
a triangle is 1/2( base) ร (height). Thus, the (absolute value of the) displacement for
one train (1/2)(40 m/s)(5 s) = 100 m, and that of the other train is (1/2)(30 m/s)(4 s) =
60 m. The initial โgapโ between the trains was 200 m, and according to our
displacement computations, the gap has narrowed by 160 m. Thus, the answer is
200 โ 160 = 40 m.
42. (a) Note that 110 km/h is equivalent to 30.56 m/s. During a two-second interval,
you travel 61.11 m. The decelerating police car travels (using Eq. 2-15) 51.11 m. In
light of the fact that the initial โgapโ between cars was 25 m, this means the gap has
narrowed by 10.0 m โ that is, to a distance of 15.0 m between cars.
(b) First, we add 0.4 s to the considerations of part (a). During a 2.4 s interval, you
travel 73.33 m. The decelerating police car travels (using Eq. 2-15) 58.93 m during
that time. The initial distance between cars of 25 m has therefore narrowed by 14.4
m. Thus, at the start of your braking (call it t0) the gap between the cars is 10.6 m.
The speed of the police car at t0 is 30.56 โ 5(2.4) = 18.56 m/s. Collision occurs at time
t when xyou = xpolice (we choose coordinates such that your position is x = 0 and the
police carโs position is x = 10.6 m at t0). Eq. 2-15 becomes, for each car:
1
xpolice โ 10.6 = 18.56(t โ t0) โ 2 (5)(t โ t0)2
1
xyou = 30.56(t โ t0) โ 2 (5)(t โ t0)2 .
42
CHAPTER 2
Subtracting equations, we find
10.6 = (30.56 โ 18.56)(t โ t0) โ
0.883 s = t โ t0.
At that time your speed is 30.56 + a(t โ t0) = 30.56 โ 5(0.883) โ 26 m/s (or 94 km/h).
43. In this solution we elect to wait until the last step to convert to SI units. Constant
acceleration is indicated, so use of Table 2-1 is permitted. We start with Eq. 2-17 and
denote the trainโs initial velocity as vt and the locomotiveโs velocity as vA (which is
also the final velocity of the train, if the rear-end collision is barely avoided). We note
that the distance ฮx consists of the original gap between them, D, as well as the
forward distance traveled during this time by the locomotive vA t . Therefore,
vt + v A ฮ x D + vA t D
=
=
= + vA .
2
t
t
t
We now use Eq. 2-11 to eliminate time from the equation. Thus,
vt + v A
D
=
+ vA
v A โ vt / a
2
b
g
which leads to
F v + v โ v IJ FG v โ v IJ = โ 1 bv โ v g .
a=G
H 2 K H D K 2D
Hence,
1
FG 29 km โ 161 kmIJ = โ12888 km / h
a=โ
h
h K
2(0.676 km) H
t
A
A
A
2
t
A
t
2
2
which we convert as follows:
F 1000 mIJ FG 1 h IJ = โ0.994 m / s
a = c โ12888 km / h h G
H 1 km K H 3600 sK
2
2
2
so that its magnitude is |a| = 0.994 m/s2. A graph is
shown here for the case where a collision is just
avoided (x along the vertical axis is in meters and t
along the horizontal axis is in seconds). The top
(straight) line shows the motion of the locomotive
and the bottom curve shows the motion of the
passenger train.
The other case (where the collision is not quite
avoided) would be similar except that the slope of
the bottom curve would be greater than that of the
top line at the point where they meet.
44. We neglect air resistance, which justifies setting a = โg = โ9.8 m/s2 (taking down
as the โy direction) for the duration of the motion. We are allowed to use Table 2-1
(with ฮy replacing ฮx) because this is constant acceleration motion. The ground level
43
is taken to correspond to the origin of the y axis.
(a) Using y = v0t โ 21 gt 2 , with y = 0.544 m and t = 0.200 s, we find
v0 =
y + gt 2 / 2 0.544 m + (9.8 m/s 2 ) (0.200 s) 2 / 2
=
= 3.70 m/s .
t
0.200 s
(b) The velocity at y = 0.544 m is
v = v0 โ gt = 3.70 m/s โ (9.8 m/s 2 ) (0.200 s) = 1.74 m/s .
(c) Using v 2 = v02 โ 2 gy (with different values for y and v than before), we solve for
the value of y corresponding to maximum height (where v = 0).
y=
v02
(3.7 m/s) 2
=
= 0.698 m.
2 g 2(9.8 m/s 2 )
Thus, the armadillo goes 0.698 โ 0.544 = 0.154 m higher.
45. In this problem a ball is being thrown vertically upward. Its subsequent motion is
under the influence of gravity. We neglect air resistance for the duration of the motion
(between โlaunchingโ and โlandingโ), so a = โg = โ9.8 m/s2 (we take downward to be
the โy direction). We use the equations in Table 2-1 (with ฮy replacing ฮx) because
this is a = constant motion:
v = v0 โ gt
(2 โ 11)
1
y โ y0 = v0t โ gt 2
2
v 2 = v02 โ 2 g ( y โ y0 )
(2 โ 15)
(2 โ 16)
We set y0 = 0. Upon reaching the maximum height y, the speed of the ball is
momentarily zero (v = 0). Therefore, we can relate its initial speed v0 to y via the
equation 0 = v 2 = v02 โ 2 gy .
The time it takes for the ball to reach maximum height is given by v = v0 โ gt = 0 , or
t = v0 / g . Therefore, for the entire trip (from the time it leaves the ground until the
time it returns to the ground), the total flight time is T = 2t = 2v0 / g .
(a) At the highest point v = 0 and v0 = 2 gy . Since y = 50 m we find
v0 = 2 gy = 2(9.8 m/s 2 )(50 m) = 31.3 m/s.
(b) Using the result from (a) for v0, we find the total flight time to be
44
CHAPTER 2
T=
2v0 2(31.3 m/s)
=
= 6.39 s โ 6.4 s .
g
9.8 m/s 2
(c) SI units are understood in the x and v graphs shown. The acceleration graph is a
horizontal line at โ9.8 m/s2.
In calculating the total flight time of the ball, we could have used Eq. 2-15. At
t = T > 0 , the ball returns to its original position ( y = 0 ). Therefore,
2v
1
y = v0T โ gT 2 = 0 โ T = 0 .
2
g
46. Neglect of air resistance justifies setting a = โg = โ9.8 m/s2 (where down is our โy
direction) for the duration of the fall. This is constant acceleration motion, and we
may use Table 2-1 (with ฮy replacing ฮx).
(a) Using Eq. 2-16 and taking the negative root (since the final velocity is downward),
we have
v = โ v02 โ 2 g ฮy = โ 0 โ 2(9.8 m/s 2 )(โ1700 m) = โ183 m/s .
Its magnitude is therefore 183 m/s.
(b) No, but it is hard to make a convincing case without more analysis. We estimate
the mass of a raindrop to be about a gram or less, so that its mass and speed (from part
(a)) would be less than that of a typical bullet, which is good news. But the fact that
one is dealing with many raindrops leads us to suspect that this scenario poses an
unhealthy situation. If we factor in air resistance, the final speed is smaller, of course,
and we return to the relatively healthy situation with which we are familiar.
47. We neglect air resistance, which justifies setting a = โg = โ9.8 m/s2 (taking down
as the โy direction) for the duration of the fall. This is constant acceleration motion,
which justifies the use of Table 2-1 (with ฮy replacing ฮx).
(a) Starting the clock at the moment the wrench is dropped (v0 = 0), then
v 2 = v02 โ 2 g ฮy leads to
ฮy = โ
(โ24 m/s) 2
= โ29.4 m
2(9.8 m/s 2 )
so that it fell through a height of 29.4 m.
45
(b) Solving v = v0 โ gt for time, we find:
t=
v0 โ v 0 โ (โ24 m/s)
=
= 2.45 s.
g
9.8 m/s 2
(c) SI units are used in the graphs, and the initial position is taken as the coordinate
origin. The acceleration graph is a horizontal line at โ9.8 m/s2.
As the wrench falls, with a = โ g 0), to the result:
t=
โ12 m/s + (โ12 m/s) 2 โ 2(9.8 m/s 2 )(โ30 m)
= 1.54 s.
9.8 m/s 2
(b) Enough information is now known that any of the equations in Table 2-1 can be
used to obtain v; however, the one equation that does not use our result from part (a)
is Eq. 2-16:
v = v02 โ 2 gฮy = 27.1 m / s
where the positive root has been chosen in order to give speed (which is the
magnitude of the velocity vector).
49. We neglect air resistance, which justifies setting a = โg = โ9.8 m/s2 (taking down
as the โy direction) for the duration of the motion. We are allowed to use Table 2-1
(with ฮy replacing ฮx) because this is constant acceleration motion. We are placing
the coordinate origin on the ground. We note that the initial velocity of the package is
46
CHAPTER 2
the same as the velocity of the balloon, v0 = +12 m/s, and that its initial coordinate is
y0 = +80 m.
(a) We solve y = y0 + v0t โ 21 gt 2 for time, with y = 0, using the quadratic formula
(choosing the positive root to yield a positive value for t).
t=
v0 + v02 + 2 gy0
g
=
12 m/s + (12 m/s) 2 + 2 ( 9.8 m/s 2 ) ( 80 m )
9.8 m/s 2
= 5.4 s
(b) If we wish to avoid using the result from part (a), we could use Eq. 2-16, but if
that is not a concern, then a variety of formulas from Table 2-1 can be used. For
instance, Eq. 2-11 leads to
v = v0 โ gt = 12 m/s โ (9.8 m/s 2 )(5.447 s) = โ41.38 m/s
Its final speed is about 41 m/s.
1
50. The y coordinate of Apple 1 obeys y โ yo1 = โ 2 g t2 where y = 0 when t = 2.0 s.
This allows us to solve for yo1, and we find yo1 = 19.6 m.
The graph for the coordinate of Apple 2 (which is thrown apparently at t = 1.0 s with
velocity v2) is
1
y โ yo2 = v2(t โ 1.0) โ 2 g (t โ 1.0)2
where yo2 = yo1 = 19.6 m and where y = 0 when t = 2.25 s. Thus, we obtain |v2| = 9.6
m/s, approximately.
51. (a) With upward chosen as the +y direction, we use Eq. 2-11 to find the initial
velocity of the package:
v = vo + at โ
0 = vo โ (9.8 m/s2)(2.0 s)
which leads to vo = 19.6 m/s. Now we use Eq. 2-15:
1
ฮy = (19.6 m/s)(2.0 s) + 2 (โ9.8 m/s2)(2.0 s)2 โ 20 m .
We note that the โ2.0 sโ in this second computation refers to the time interval 2 < t < 4
in the graph (whereas the โ2.0 sโ in the first computation referred to the 0 < t < 2 time
interval shown in the graph).
(b) In our computation for part (b), the time interval (โ6.0 sโ) refers to the 2 < t < 8
portion of the graph:
1
ฮy = (19.6 m/s)(6.0 s) + 2 (โ9.8 m/s2)(6.0 s)2 โ โ59 m ,
or | ฮy |= 59 m .
47
52. The full extent of the boltโs fall is given by
1
y โ y0 = โ2 g t2
where y โ y0 = โ90 m (if upward is chosen as the positive y direction). Thus the time
for the full fall is found to be t = 4.29 s. The first 80% of its free-fall distance is given
by โ72 = โg ฯ2/2, which requires time ฯ = 3.83 s.
(a) Thus, the final 20% of its fall takes t โ ฯ = 0.45 s.
(b) We can find that speed using v = โgฯ. Therefore, |v| = 38 m/s, approximately.
(c) Similarly, vfinal = โ g t โ |vfinal| = 42 m/s.
53. The speed of the boat is constant, given by vb = d/t. Here, d is the distance of the
boat from the bridge when the key is dropped (12 m) and t is the time the key takes in
falling. To calculate t, we put the origin of the coordinate system at the point where
the key is dropped and take the y axis to be positive in the downward direction.
Taking the time to be zero at the instant the key is dropped, we compute the time t
when y = 45 m. Since the initial velocity of the key is zero, the coordinate of the key
is given by y = 12 gt 2 . Thus,
t=
2y
2(45 m)
=
= 3.03 s .
g
9.8 m / s2
Therefore, the speed of the boat is
vb =
12 m
= 4.0 m / s .
3.03 s
54. (a) We neglect air resistance, which justifies setting a = โg = โ9.8 m/s2 (taking
down as the โy direction) for the duration of the motion. We are allowed to use Eq.
2-15 (with ฮy replacing ฮx) because this is constant acceleration motion. We use
primed variables (except t) with the first stone, which has zero initial velocity, and
unprimed variables with the second stone (with initial downward velocity โv0, so that
v0 is being used for the initial speed). SI units are used throughout.
ฮyโฒ = 0 ( t ) โ
1 2
gt
2
ฮy = ( โv0 )( t โ 1) โ
1
2
g ( t โ 1)
2
Since the problem indicates ฮyโ = ฮy = โ43.9 m, we solve the first equation for t
(finding t = 2.99 s) and use this result to solve the second equation for the initial speed
of the second stone:
48
CHAPTER 2
โ43.9 m = ( โv0 ) (1.99 s ) โ
1
2
9.8 m/s 2 ) (1.99 s )
(
2
which leads to v0 = 12.3 m/s.
(b) The velocity of the stones are given by
vโฒy =
d (ฮyโฒ)
= โ gt ,
dt
vy =
d ( ฮy )
= โv0 โ g (t โ 1)
dt
The plot is shown below:
55. During contact with the ground its average acceleration is given by
ฮv
aavg =
ฮt
where ฮv is the change in its velocity during contact with the ground and
ฮt = 20.0 ร10โ3 s is the duration of contact. Thus, we must first find the velocity of the
ball just before it hits the ground (y = 0).
(a) Now, to find the velocity just before contact, we take t = 0 to be when it is dropped.
Using Eq. (2-16) with y0 = 15.0 m , we obtain
v = โ v02 โ 2 g ( y โ y0 ) = โ 0 โ 2(9.8 m/s 2 )(0 โ 15 m) = โ17.15 m/s
where the negative sign is chosen since the ball is traveling downward at the moment
of contact. Consequently, the average acceleration during contact with the ground is
aavg =
ฮv 0 โ (โ17.1 m/s)
=
= 857 m/s 2 .
โ3
ฮt
20.0 ร 10 s
49
(b) The fact that the result is positive indicates that this acceleration vector points
upward. In a later chapter, this will be directly related to the magnitude and direction
of the force exerted by the ground on the ball during the collision.
56. We use Eq. 2-16,
vB2 = vA2 + 2a(yB โ yA),
1
with a = โ9.8 m/s2, yB โ yA = 0.40 m, and vB = 3 vA. It is then straightforward to solve:
vA = 3.0 m/s, approximately.
57. The average acceleration during contact with the floor is aavg = (v2 โ v1) / ฮt,
where v1 is its velocity just before striking the floor, v2 is its velocity just as it leaves
the floor, and ฮt is the duration of contact with the floor (12 ร 10โ3 s).
(a) Taking the y axis to be positively upward and placing the origin at the point where
the ball is dropped, we first find the velocity just before striking the floor, using
v12 = v02 โ 2 gy . With v0 = 0 and y = โ 4.00 m, the result is
v1 = โ โ2 gy = โ โ2(9.8 m/s 2 ) (โ4.00 m) = โ8.85 m/s
where the negative root is chosen because the ball is traveling downward. To find the
velocity just after hitting the floor (as it ascends without air friction to a height of 2.00
m), we use v 2 = v22 โ 2 g ( y โ y0 ) with v = 0, y = โ2.00 m (it ends up two meters
below its initial drop height), and y0 = โ 4.00 m. Therefore,
v2 = 2 g ( y โ y0 ) =
2(9.8 m/s 2 ) (โ2.00 m + 4.00 m) = 6.26 m/s .
Consequently, the average acceleration is
aavg =
v2 โ v1 6.26 m/s โ (โ 8.85 m/s)
=
= 1.26 ร 103 m/s 2 .
ฮt
12.0 ร 10โ3 s
(b) The positive nature of the result indicates that the acceleration vector points
upward. In a later chapter, this will be directly related to the magnitude and direction
of the force exerted by the ground on the ball during the collision.
58. We choose down as the +y direction and set the coordinate origin at the point
where it was dropped (which is when we start the clock). We denote the 1.00 s
duration mentioned in the problem as t โ t' where t is the value of time when it lands
and t' is one second prior to that. The corresponding distance is y โ y' = 0.50h, where y
denotes the location of the ground. In these terms, y is the same as h, so we have h โy'
= 0.50h or 0.50h = y' .
(a) We find t' and t from Eq. 2-15 (with v0 = 0):
50
CHAPTER 2
1
2 yโฒ
yโฒ = gt โฒ2 โ t โฒ =
g
2
y=
1 2
2y
.
gt โ t =
g
2
Plugging in y = h and y' = 0.50h, and dividing these two equations, we obtain
b
g
2 0.50h / g
tโฒ
=
= 0.50 .
2h / g
t
Letting t' = t โ 1.00 (SI units understood) and cross-multiplying, we find
t โ 100
. = t 0.50 โ t =
100
.
1 โ 0.50
which yields t = 3.41 s.
(b) Plugging this result into y = 12 gt 2 we find h = 57 m.
(c) In our approach, we did not use the quadratic formula, but we did โchoose a rootโ
when we assumed (in the last calculation in part (a)) that 0.50 = +0.707 instead
of โ0.707. If we had instead let 0.50 = โ0.707 then our answer for t would have
been roughly 0.6 s, which would imply that t' = t โ 1 would equal a negative number
(indicating a time before it was dropped), which certainly does not fit with the
physical situation described in the problem.
59. We neglect air resistance, which justifies setting a = โg = โ9.8 m/s2 (taking down
as the โy direction) for the duration of the motion. We are allowed to use Table 2-1
(with ฮy replacing ฮx) because this is constant acceleration motion. The ground level
is taken to correspond to the origin of the y-axis.
(a) The time drop 1 leaves the nozzle is taken as t = 0 and its time of landing on the
floor t1 can be computed from Eq. 2-15, with v0 = 0 and y1 = โ2.00 m.
1
โ2 y
โ2(โ2.00 m)
y1 = โ gt12 โ t1 =
=
= 0.639 s .
2
g
9.8 m/s 2
At that moment, the fourth drop begins to fall, and from the regularity of the dripping
we conclude that drop 2 leaves the nozzle at t = 0.639/3 = 0.213 s and drop 3 leaves
the nozzle at t = 2(0.213 s) = 0.426 s. Therefore, the time in free fall (up to the
moment drop 1 lands) for drop 2 is t2 = t1 โ 0.213 s = 0.426 s. Its position at the
moment drop 1 strikes the floor is
1
1
y2 = โ gt22 = โ (9.8 m/s 2 )(0.426 s) 2 = โ0.889 m,
2
2
or about 89 cm below the nozzle.
51
(b) The time in free fall (up to the moment drop 1 lands) for drop 3 is t3 = t1 โ0.426 s
= 0.213 s. Its position at the moment drop 1 strikes the floor is
1
1
y3 = โ gt32 = โ (9.8 m/s 2 )(0.213 s) 2 = โ0.222 m,
2
2
or about 22 cm below the nozzle.
60. To find the โlaunchโ velocity of the rock, we apply Eq. 2-11 to the maximum
height (where the speed is momentarily zero)
v = v0 โ gt โ 0 = v0 โ ( 9.8 m/s 2 ) ( 2.5 s )
so that v0 = 24.5 m/s (with +y up). Now we use Eq. 2-15 to find the height of the
tower (taking y0 = 0 at the ground level)
y โ y0 = v0t +
1 2
1
2
at โ y โ 0 = ( 24.5 m/s )(1.5 s ) โ ( 9.8 m/s 2 ) (1.5 s ) .
2
2
Thus, we obtain y = 26 m.
61. We choose down as the +y direction and place the coordinate origin at the top of
the building (which has height H). During its fall, the ball passes (with velocity v1) the
top of the window (which is at y1) at time t1, and passes the bottom (which is at y2) at
time t2. We are told y2 โ y1 = 1.20 m and t2 โ t1 = 0.125 s. Using Eq. 2-15 we have
b
g 21 g bt โ t g
y2 โ y1 = v1 t2 โ t1 +
2
2
1
which immediately yields
v1 =
1.20 m โ 12 ( 9.8 m/s 2 ) ( 0.125 s )
2
0.125 s
= 8.99 m/s.
From this, Eq. 2-16 (with v0 = 0) reveals the value of y1:
v12 = 2 gy1
โ y1 =
(8.99 m/s) 2
= 4.12 m.
2(9.8 m/s 2 )
It reaches the ground (y3 = H) at t3. Because of the symmetry expressed in the
problem (โupward flight is a reverse of the fallโโ) we know that t3 โ t2 = 2.00/2 = 1.00
s. And this means t3 โ t1 = 1.00 s + 0.125 s = 1.125 s. Now Eq. 2-15 produces
1
y3 โ y1 = v1 (t3 โ t1 ) + g (t3 โ t1 ) 2
2
1
y3 โ 4.12 m = (8.99 m/s) (1.125 s) + (9.8 m/s 2 ) (1.125 s) 2
2
52
CHAPTER 2
which yields y3 = H = 20.4 m.
62. The height reached by the player is y = 0.76 m (where we have taken the origin of
the y axis at the floor and +y to be upward).
(a) The initial velocity v0 of the player is
v0 = 2 gy = 2(9.8 m/s 2 ) (0.76 m) = 3.86 m/s .
This is a consequence of Eq. 2-16 where velocity v vanishes. As the player reaches y1
= 0.76 m โ 0.15 m = 0.61 m, his speed v1 satisfies v02 โ v12 = 2 gy1 , which yields
v1 = v02 โ 2 gy1 = (3.86 m/s) 2 โ 2(9.80 m/s 2 ) (0.61 m) = 1.71 m/s .
The time t1 that the player spends ascending in the top ฮy1 = 0.15 m of the jump can
now be found from Eq. 2-17:
ฮy1 =
2 ( 0.15 m )
1
= 0.175 s
( v1 + v ) t1 โ t1 =
2
1.71 m/s + 0
which means that the total time spent in that top 15 cm (both ascending and
descending) is 2(0.175 s) = 0.35 s = 350 ms.
(b) The time t2 when the player reaches a height of 0.15 m is found from Eq. 2-15:
0.15 m = v0t2 โ
1 2
1
gt2 = (3.86 m/s)t2 โ (9.8 m/s 2 )t22 ,
2
2
which yields (using the quadratic formula, taking the smaller of the two positive roots)
t2 = 0.041 s = 41 ms, which implies that the total time spent in that bottom 15 cm
(both ascending and descending) is 2(41 ms) = 82 ms.
63. The time t the pot spends passing in front of the window of length L = 2.0 m is
0.25 s each way. We use v for its velocity as it passes the top of the window (going
up). Then, with a = โg = โ9.8 m/s2 (taking down to be the โy direction), Eq. 2-18
yields
L 1
1
L = vt โ gt 2 โ v = โ gt .
t 2
2
The distance H the pot goes above the top of the window is therefore (using Eq. 2-16
with the final velocity being zero to indicate the highest point)
( 2.00 m / 0.25 s โ (9.80 m/s )(0.25 s) / 2 ) = 2.34 m.
v 2 ( L / t โ gt / 2 )
H=
=
=
2g
2g
2(9.80 m/s 2 )
2
2
2
53
64. The graph shows y = 25 m to be the highest point (where the speed momentarily
vanishes). The neglect of โair frictionโ (or whatever passes for that on the distant
planet) is certainly reasonable due to the symmetry of the graph.
(a) To find the acceleration due to gravity gp on that planet, we use Eq. 2-15 (with +y
up)
1
1
2
y โ y0 = vt + g p t 2 โ 25 m โ 0 = ( 0 )( 2.5 s ) + g p ( 2.5 s )
2
2
so that gp = 8.0 m/s2.
(b) That same (max) point on the graph can be used to find the initial velocity.
y โ y0 =
1
( v0 + v ) t
2
โ
25 m โ 0 =
1
( v0 + 0 ) ( 2.5 s )
2
Therefore, v0 = 20 m/s.
65. The key idea here is that the speed of the head (and the torso as well) at any given
time can be calculated by finding the area on the graph of the headโs acceleration
versus time, as shown in Eq. 2-26:
โ area between the acceleration curve โ
v1 โ v0 = โ
โ
โ and the time axis, from t0 to t1
โ
(a) From Fig. 2.14a, we see that the head begins to accelerate from rest (v0 = 0) at t0 =
110 ms and reaches a maximum value of 90 m/s2 at t1 = 160 ms. The area of this
region is
1
area = (160 โ 110) ร 10โ3s โ
( 90 m/s 2 ) = 2.25 m/s
2
which is equal to v1, the speed at t1.
(b) To compute the speed of the torso at t1=160 ms, we divide the area into 4 regions:
From 0 to 40 ms, region A has zero area. From 40 ms to 100 ms, region B has the
shape of a triangle with area
1
area B = (0.0600 s)(50.0 m/s 2 ) = 1.50 m/s .
2
From 100 to 120 ms, region C has the shape of a rectangle with area
area C = (0.0200 s) (50.0 m/s 2 ) = 1.00 m/s.
From 110 to 160 ms, region D has the shape of a trapezoid with area
1
(0.0400 s) (50.0 + 20.0) m/s 2 = 1.40 m/s.
2
Substituting these values into Eq. 2-26, with v0 = 0 then gives
area D =
54
CHAPTER 2
v1 โ 0 = 0 + 1.50 m/s + 1.00 m/s + 1.40 m/s = 3.90 m/s,
or v1 = 3.90 m/s.
66. The key idea here is that the position of an object at any given time can be
calculated by finding the area on the graph of the objectโs velocity versus time, as
shown in Eq. 2-25:
โ area between the velocity curve โ
x1 โ x0 = โ
โ.
โ and the time axis, from t0 to t1 โ
(a) To compute the position of the fist at t = 50 ms, we divide the area in Fig. 2-34
into two regions. From 0 to 10 ms, region A has the shape of a triangle with area
area A =
1
(0.010 s) (2 m/s) = 0.01 m.
2
From 10 to 50 ms, region B has the shape of a trapezoid with area
area B =
1
(0.040 s) (2 + 4) m/s = 0.12 m.
2
Substituting these values into Eq. 2-25 with x0 = 0 then gives
x1 โ 0 = 0 + 0.01 m + 0.12 m = 0.13 m,
or x1 = 0.13 m.
(b) The speed of the fist reaches a maximum at t1 = 120 ms. From 50 to 90 ms, region
C has the shape of a trapezoid with area
area C =
1
(0.040 s) (4 + 5) m/s = 0.18 m.
2
From 90 to 120 ms, region D has the shape of a trapezoid with area
area D =
1
(0.030 s) (5 + 7.5) m/s = 0.19 m.
2
Substituting these values into Eq. 2-25, with x0 = 0 then gives
x1 โ 0 = 0 + 0.01 m + 0.12 m + 0.18 m + 0.19 m = 0.50 m,
or x1 = 0.50 m.
67. The problem is solved using Eq. 2-26:
โ area between the acceleration curve โ
v1 โ v0 = โ
โ
โ and the time axis, from t0 to t1
โ
55
To compute the speed of the unhelmeted, bare head at t1 = 7.0 ms, we divide the area
under the a vs. t graph into 4 regions: From 0 to 2 ms, region A has the shape of a
triangle with area
1
area A = (0.0020 s) (120 m/s 2 ) = 0.12 m/s.
2
From 2 ms to 4 ms, region B has the shape of a trapezoid with area
area B =
1
(0.0020 s) (120 + 140) m/s2 = 0.26 m/s.
2
From 4 to 6 ms, region C has the shape of a trapezoid with area
area C =
1
(0.0020 s) (140 + 200) m/s2 = 0.34 m/s.
2
From 6 to 7 ms, region D has the shape of a triangle with area
1
area D = (0.0010 s) (200 m/s 2 ) = 0.10 m/s.
2
Substituting these values into Eq. 2-26, with v0=0 then gives
vunhelmeted = 0.12 m/s + 0.26 m/s + 0.34 m/s + 0.10 m/s = 0.82 m/s.
Carrying out similar calculations for the helmeted head, we have the following
results: From 0 to 3 ms, region A has the shape of a triangle with area
1
(0.0030 s) (40 m/s 2 ) = 0.060 m/s.
2
From 3 ms to 4 ms, region B has the shape of a rectangle with area
area A =
area B = (0.0010 s) (40 m/s 2 ) = 0.040 m/s.
From 4 to 6 ms, region C has the shape of a trapezoid with area
area C =
1
(0.0020 s) (40 + 80) m/s2 = 0.12 m/s.
2
From 6 to 7 ms, region D has the shape of a triangle with area
1
area D = (0.0010 s) (80 m/s 2 ) = 0.040 m/s.
2
Substituting these values into Eq. 2-26, with v0 = 0 then gives
vhelmeted = 0.060 m/s + 0.040 m/s + 0.12 m/s + 0.040 m/s = 0.26 m/s.
56
CHAPTER 2
Thus, the difference in the speed is
ฮv = vunhelmeted โ vhelmeted = 0.82 m/s โ 0.26 m/s = 0.56 m/s.
68. This problem can be solved by noting that velocity can be determined by the
graphical integration of acceleration versus time. The speed of the tongue of the
salamander is simply equal to the area under the acceleration curve:
1 โ2
1
1
(10 s)(100 m/s 2 ) + (10โ2 s)(100 m/s 2 + 400 m/s 2 ) + (10โ2 s)(400 m/s 2 )
2
2
2
= 5.0 m/s.
v = area =
z
69. Since v = dx / dt (Eq. 2-4), then ฮx = v dt , which corresponds to the area
under the v vs t graph. Dividing the total area A into rectangular (base ร height) and
triangular 21 base ร height areas, we have
b
g
A
= A0 <t <2 + A2 < t <10 + A10 < t <12 + A12 < t 0 for times less than t = 2 s, then the spot had been moving rightward.
(e) As implied by our answer to part (c), it moves leftward for times immediately after
t = 2 s. In fact, the expression found in part (a) guarantees that for all t > 2, v 2 s cannot
be the right edge; it is the left edge (x = 0). Solving the expression given in the
problem statement (with x = 0) for positive t yields the answer: the spot reaches the
left edge at t = 12 s โ 3.46 s.
72. We adopt the convention frequently used in the text: that “up” is the positive y
direction.
(a) At the highest point in the trajectory v = 0. Thus, with t = 1.60 s, the equation
v = v0 โ gt yields v0 = 15.7 m/s.
1
(b) One equation that is not dependent on our result from part (a) is y โ y0 = vt + 2gt2;
this readily gives ymax โ y0 = 12.5 m for the highest (“max”) point measured relative to
where it started (the top of the building).
1
(c) Now we use our result from part (a) and plug into y โ y0 = v0t + 2gt2 with t = 6.00
s and y = 0 (the ground level). Thus, we have
1
0 โ y0 = (15.68 m/s)(6.00 s) โ 2 (9.8 m/s2)(6.00 s)2.
Therefore, y0 (the height of the building) is equal to 82.3 m.
73. We denote the required time as t, assuming the light turns green when the clock
reads zero. By this time, the distances traveled by the two vehicles must be the same.
(a) Denoting the acceleration of the automobile as a and the (constant) speed of the
truck as v then
1 2
ฮx =
at
= vt truck
2
car
which leads to
FG
H
IJ b g
K
58
CHAPTER 2
t=
Therefore,
2v 2 ( 9.5 m/s )
=
= 8.6 s .
a
2.2 m/s 2
ฮx = vt = ( 9.5 m/s )( 8.6 s ) = 82 m .
(b) The speed of the car at that moment is
vcar = at = ( 2.2 m/s 2 ) ( 8.6 s ) = 19 m/s .
74. If the plane (with velocity v) maintains its present course, and if the terrain
continues its upward slope of 4.3ยฐ, then the plane will strike the ground after traveling
ฮx =
h
35 m
=
= 4655
. m โ 0.465 km.
tan ฮธ tan 4.3ยฐ
This corresponds to a time of flight found from Eq. 2-2 (with v = vavg since it is
constant)
ฮx
0.465 km
t=
=
= 0.000358 h โ 1.3 s.
v 1300 km / h
This, then, estimates the time available to the pilot to make his correction.
75. We denote tr as the reaction time and tb as the braking time. The motion during tr
is of the constant-velocity (call it v0) type. Then the position of the car is given by
x = v0 t r + v0 t b +
1 2
atb
2
where v0 is the initial velocity and a is the acceleration (which we expect to be
negative-valued since we are taking the velocity in the positive direction and we know
the car is decelerating). After the brakes are applied the velocity of the car is given by
v = v0 + atb. Using this equation, with v = 0, we eliminate tb from the first equation
and obtain
1 v02
v2
1 v02
x = v0 t r โ 0 +
= v0 t r โ
.
2 a
2 a
a
We write this equation for each of the initial velocities:
x1 = v01tr โ
2
1 v01
2 a
x2 = v02 tr โ
2
1 v02
.
2 a
and
Solving these equations simultaneously for tr and a we get
59
tr =
2
2
v02
x1 โ v01
x2
v01v02 v02 โ v01
b
g
and
a=โ
2
2
1 v02 v01
โ v01v02
.
2 v02 x1 โ v01 x2
(a) Substituting x1 = 56.7 m, v01 = 80.5 km/h = 22.4 m/s, x2 = 24.4 m and v02 = 48.3
km/h = 13.4 m/s, we find
tr =
2
2
v02
x1 โ v01
x2
(13.4 m/s) 2 (56.7 m) โ (22.4 m/s) 2 (24.4 m)
=
v01v02 (v02 โ v01 ) (22.4 m/s)(13.4 m/s)(13.4 m/s โ 22.4 m/s)
= 0.74 s.
(b) Similarly, substituting x1 = 56.7 m, v01 = 80.5 km/h = 22.4 m/s, x2 = 24.4 m, and
v02 = 48.3 km/h = 13.4 m/s gives
a=โ
2
2
โ v01v02
1 v02 v01
1 (13.4 m/s)(22.4 m/s) 2 โ (22.4 m/s)(13.4 m/s)2
=โ
2 v02 x1 โ v01 x2
2 (13.4 m/s)(56.7 m) โ (22.4 m/s)(24.4 m)
= โ6.2 m/s 2 .
The magnitude of the deceleration is therefore 6.2 m/s2. Although rounded-off values
are displayed in the above substitutions, what we have input into our calculators are
the โexactโ values (such as v02 = 161
12 m/s).
76. (a) A constant velocity is equal to the ratio of displacement to elapsed time. Thus,
for the vehicle to be traveling at a constant speed v p over a distance D23 , the time
delay should be t = D23 / v p .
(b) The time required for the car to accelerate from rest to a cruising speed v p is
t0 = v p / a . During this time interval, the distance traveled is ฮx0 = at02 / 2 = v 2p / 2a.
The car then moves at a constant speed v p over a distance D12 โ ฮx0 โ d to reach
intersection 2, and the time elapsed is t1 = ( D12 โ ฮx0 โ d ) / v p . Thus, the time delay at
intersection 2 should be set to
v p D12 โ (v 2p / 2a ) โ d
D12 โ ฮx0 โ d
ttotal = tr + t0 + t1 = tr + +
= tr + +
a
vp
a
vp
vp
= tr +
1 v p D12 โ d
+
2 a
vp
77. Since the problem involves constant acceleration, the motion of the rod can be
readily analyzed using the equations in Table 2-1. We take +x in the direction of
motion, so
60
CHAPTER 2
b
v = 60 km / h
m / km I
g FGH 1000
J = + 16.7 m / s
3600 s / h K
and a > 0. The location where it starts from rest (v0 = 0) is taken to be x0 = 0.
(a) Using Eq. 2-7, we find the average acceleration to be
aavg =
ฮv v โ v0 16.7 m/s โ 0
=
=
= 3.09 m/s 2 โ 3.1 m/s 2 .
ฮt t โ t0
5.4 s โ 0
(b) Assuming constant acceleration a = aavg = 3.09 m/s 2 , the total distance traveled
during the 5.4-s time interval is
1
1
x = x0 + v0t + at 2 = 0 + 0 + (3.09 m/s 2 )(5.4 s) 2 = 45 m .
2
2
(c) Using Eq. 2-15, the time required to travel a distance of x = 250 m is:
x=
2 ( 250 m )
1 2
2x
at โ t =
=
= 13 s .
2
a
3.1 m/s 2
Note that the displacement of the rod as a function of time can be written as
1
x(t ) = (3.09 m/s 2 )t 2 . Also we could have chosen Eq. 2-17 to solve for (b):
2
1
1
x = ( v0 + v ) t = (16.7 m/s )( 5.4 s ) = 45 m.
2
2
78. We take the moment of applying brakes to be t = 0. The deceleration is constant so
that Table 2-1 can be used. Our primed variables (such as v0โฒ = 72 km/h = 20 m/s ) refer
to one train (moving in the +x direction and located at the origin when t = 0) and
unprimed variables refer to the other (moving in the โx direction and located at x0 =
+950 m when t = 0). We note that the acceleration vector of the unprimed train points
in the positive direction, even though the train is slowing down; its initial velocity is
v0 = โ144 km/h = โ40 m/s. Since the primed train has the lower initial speed, it should
stop sooner than the other train would (were it not for the collision). Using Eq 2-16, it
should stop (meaning vโฒ = 0 ) at
( vโฒ ) โ ( v0โฒ ) = 0 โ (20 m/s)2 = 200 m .
xโฒ =
2
2
2aโฒ
โ2 m/s 2
The speed of the other train, when it reaches that location, is
v = v02 + 2aฮx =
= 10 m/s
( โ40 m/s ) + 2 (1.0 m/s 2 ) ( 200 m โ 950 m )
2
61
using Eq 2-16 again. Specifically, its velocity at that moment would be โ10 m/s since
it is still traveling in the โx direction when it crashes. If the computation of v had
failed (meaning that a negative number would have been inside the square root) then
we would have looked at the possibility that there was no collision and examined how
far apart they finally were. A concern that can be brought up is whether the primed
train collides before it comes to rest; this can be studied by computing the time it
stops (Eq. 2-11 yields t = 20 s) and seeing where the unprimed train is at that moment
(Eq. 2-18 yields x = 350 m, still a good distance away from contact).
1
79. The y coordinate of Piton 1 obeys y โ y01 = โ 2 g t2 where y = 0 when t = 3.0 s.
This allows us to solve for yo1, and we find y01 = 44.1 m. The graph for the coordinate
of Piton 2 (which is thrown apparently at t = 1.0 s with velocity v1) is
1
y โ y02 = v1(tโ1.0) โ 2 g (t โ 1.0)2
where y02 = y01 + 10 = 54.1 m and where (again) y = 0 when t = 3.0 s.
obtain |v1| = 17 m/s, approximately.
Thus we
80. We take +x in the direction of motion. We use subscripts 1 and 2 for the data. Thus,
v1 = +30 m/s, v2 = +50 m/s, and x2 โ x1 = +160 m.
(a) Using these subscripts, Eq. 2-16 leads to
a=
v22 โ v12
(50 m/s) 2 โ (30 m/s) 2
=
= 5.0 m/s 2 .
2 ( x2 โ x1 )
2 (160 m )
(b) We find the time interval corresponding to the displacement x2 โ x1 using Eq. 2-17:
t2 โ t1 =
2 ( x2 โ x1 )
2 (160 m )
=
= 4.0 s .
v1 + v2
30 m/s + 50 m/s
(c) Since the train is at rest (v0 = 0) when the clock starts, we find the value of t1 from
Eq. 2-11:
30 m/s
v1 = v0 + at1 โ t1 =
= 6.0 s .
5.0 m/s 2
(d) The coordinate origin is taken to be the location at which the train was initially at
rest (so x0 = 0). Thus, we are asked to find the value of x1. Although any of several
equations could be used, we choose Eq. 2-17:
x1 =
1
1
( v0 + v1 ) t1 = ( 30 m/s )( 6.0 s ) = 90 m .
2
2
(e) The graphs are shown below, with SI units understood.
62
CHAPTER 2
81. Integrating (from t = 2 s to variable t = 4 s) the acceleration to get the velocity and
using the values given in the problem leads to
t
t
1
1
v = v0 + โซ adt = v0 + โซ (5.0t )dt = v0 + (5.0)(t 2 โ t02 ) = 17 + 2 (5.0)(42 โ 22) = 47 m/s.
t0
t0
2
82. The velocity v at t = 6 (SI units and two significant figures understood) is
6
1
vgiven + โซ adt . A quick way to implement this is to recall the area of a triangle (2
โ2
base ร height). The result is v = 7 m/s + 32 m/s = 39 m/s.
83. The object, once it is dropped (v0 = 0) is in free fall (a = โg = โ9.8 m/s2 if we take
down as the โy direction), and we use Eq. 2-15 repeatedly.
(a) The (positive) distance D from the lower dot to the mark corresponding to a
certain reaction time t is given by ฮy = โ D = โ 21 gt 2 , or D = gt2/2. Thus,
for t1 = 50.0 ms ,
c9.8 m / s hc50.0 ร 10 sh = 0.0123 m = 1.23 cm.
D =
โ3
2
1
2
2
( 9.8 m/s ) ( 100 ร 10 s ) = 0.049 m = 4D .
(b) For t = 100 ms, D =
โ3
2
2
2
2
1
2
( 9.8 m/s ) ( 150 ร 10 s ) = 0.11m = 9D .
(c) For t = 150 ms, D =
โ3
2
3
3
2
1
2
( 9.8 m/s ) ( 200 ร 10 s ) = 0.196 m =16D .
(d) For t = 200 ms, D =
โ3
2
4
4
2
1
2
c9.8 m / s hc250 ร 10 sh = 0.306 m = 25D .
(e) For t = 250 ms, D =
โ3
2
4
5
2
2
1
84. We take the direction of motion as +x, take x0 = 0 and use SI units, so v =
1600(1000/3600) = 444 m/s.
63
(a) Equation 2-11 gives 444 = a(1.8) or a = 247 m/s2. We express this as a multiple of
g by setting up a ratio:
โ 247 m/s 2 โ
a=โ
g = 25 g .
2 โ
โ 9.8 m/s โ
(b) Equation 2-17 readily yields
x=
1
1
( v0 + v ) t = ( 444 m/s )(1.8 s ) = 400 m.
2
2
85. Let D be the distance up the hill. Then
average speed =
total distance traveled
total time of travel
=
2D
D
D
+
20 km/h
35 km/h
โ 25 km/h .
86. We obtain the velocity by integration of the acceleration:
t
v โ v0 = โซ (6.1 โ 1.2t โฒ)dt โฒ .
0
Lengths are in meters and times are in seconds. The student is encouraged to look at
the discussion in the textbook in ยง2-7 to better understand the manipulations here.
(a) The result of the above calculation is
v = v0 + 6.1 t โ 0.6 t 2 ,
where the problem states that v0 = 2.7 m/s. The maximum of this function is found by
knowing when its derivative (the acceleration) is zero (a = 0 when t = 6.1/1.2 = 5.1 s)
and plugging that value of t into the velocity equation above. Thus, we find
v = 18 m/s .
(b) We integrate again to find x as a function of t:
t
t
0
0
x โ x0 = โซ v dt โฒ = โซ (v0 + 6.1t โฒ โ 0.6 t โฒ2 ) dt โฒ = v0t + 3.05 t 2 โ 0.2 t 3 .
With x0 = 7.3 m, we obtain x = 83 m for t = 6. This is the correct answer, but one has
the right to worry that it might not be; after all, the problem asks for the total distance
traveled (and x โ x0 is just the displacement). If the cyclist backtracked, then his total
distance would be greater than his displacement. Thus, we might ask, “did he
backtrack?” To do so would require that his velocity be (momentarily) zero at some
point (as he reversed his direction of motion). We could solve the above quadratic
equation for velocity, for a positive value of t where v = 0; if we did, we would find
that at t = 10.6 s, a reversal does indeed happen. However, in the time interval we
are concerned with in our problem (0 โค t โค 6 s), there is no reversal and the
displacement is the same as the total distance traveled.
64
CHAPTER 2
87. The time it takes to travel a distance d with a speed v1 is t1 = d / v1 . Similarly, with
a speed v2 the time would be t2 = d / v2 . The two speeds in this problem are
v1 = 55 mi/h = (55 mi/h)
1609 m/mi
= 24.58 m/s
3600 s/h
v2 = 65 mi/h = (65 mi/h)
1609 m/mi
= 29.05 m/s
3600 s/h
With d = 700 km = 7.0 ร105 m , the time difference between the two is
โ1
โ
โ
1โ
1
1
ฮt = t1 โ t2 = d โ โ โ = (7.0 ร 105 m) โ
โ
โ = 4383 s = 73 min
โ 24.58 m/s 29.05 m/s โ
โ v1 v2 โ
or 1 h and 13 min.
88. The acceleration is constant and we may use the equations in Table 2-1.
(a) Taking the first point as coordinate origin and time to be zero when the car is there,
we apply Eq. 2-17:
x=
1
1
( v + v0 ) t = (15.0 m/s + v0 ) ( 6.00 s ) .
2
2
With x = 60.0 m (which takes the direction of motion as the +x direction) we solve for
the initial velocity: v0 = 5.00 m/s.
(b) Substituting v = 15.0 m/s, v0 = 5.00 m/s, and t = 6.00 s into a = (v โ v0)/t (Eq. 2-11),
we find a = 1.67 m/s2.
(c) Substituting v = 0 in v 2 = v02 + 2ax and solving for x, we obtain
x= โ
v02
(5.00 m/s) 2
= โ
= โ 7.50m ,
2a
2 (1.67 m/s 2 )
or | x | = 7.50 m .
(d) The graphs require computing the time when v = 0, in which case, we use v = v0 +
at’ = 0. Thus,
tโฒ =
โv0 โ5.00 m/s
=
= โ 3.0s
a 1.67 m/s 2
indicates the moment the car was at rest. SI units are understood.
65
89. We neglect air resistance, which justifies setting a = โg = โ9.8 m/s2 (taking down
as the โy direction) for the duration of the motion. We are allowed to use Table 2-1
(with ฮy replacing ฮx) because this is constant acceleration motion. When something
is thrown straight up and is caught at the level it was thrown from, the time of flight t
is half of its time of ascent ta, which is given by Eq. 2-18 with ฮy = H and v = 0
(indicating the maximum point).
H = vt a +
1 2
gta
2
โ
ta =
2H
g
Writing these in terms of the total time in the air t = 2ta we have
H=
1 2
gt
8
โ
2H
.
g
t=2
We consider two throws, one to height H1 for total time t1 and another to height H2 for
total time t2, and we set up a ratio:
FG IJ
H K
H2 81 gt22
t
= 1 2 = 2
H1 8 gt1
t1
2
from which we conclude that if t2 = 2t1 (as is required by the problem) then H2 = 22H1
= 4H1.
90. (a) Using the fact that the area of a triangle is 12 (base) (height) (and the fact that
the integral corresponds to the area under the curve) we find, from t = 0 through t = 5
s, the integral of v with respect to t is 15 m. Since we are told that x0 = 0 then we
conclude that x = 15 m when t = 5.0 s.
(b) We see directly from the graph that v = 2.0 m/s when t = 5.0 s.
(c) Since a = dv/dt = slope of the graph, we find that the acceleration during the
interval 4 < t 0. With y = 0 and y0 = h, this becomes
t=
v02 + 2 gh โ v0
.
g
(c) If it were thrown upward with that speed from height h then (in the absence of air
friction) it would return to height h with that same downward speed and would
therefore yield the same final speed (before hitting the ground) as in part (a). An
important perspective related to this is treated later in the book (in the context of
energy conservation).
(d) Having to travel up before it starts its descent certainly requires more time than in
part (b). The calculation is quite similar, however, except for now having +v0 in the
equation where we had put in โv0 in part (b). The details follow:
72
CHAPTER 2
ฮ y = v0 t โ
1 2
gt
2
โ
t=
v0 +
v02 โ 2 gฮy
g
with the positive root again chosen to yield t > 0. With y = 0 and y0 = h, we obtain
t=
v02 + 2 gh + v0
.
g
102. We assume constant velocity motion and use Eq. 2-2 (with vavg = v > 0).
Therefore,
F
GH
ฮx = vฮt = 303
km
h
FG 1000 m / kmIJ I c100 ร 10 sh = 8.4 m.
H 3600 s / h K JK
โ3

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