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Unformatted text preview: MELLWONS CHAPTER TWO SOLUTIONS Chapter Two Readings
Cohen, LB. “Galileo," Scientiﬁc American, August 1949, p. 40. Drake, S "Galileo's Discovery of the law of Free Fall," Scien tiﬁc American, May
1973, p. 84. Gingerich, "The Galileo Affair," Scientiﬁc American, August 1982, p. 132. Langford, J.J., Galileo, Science and the Church, 3rd ed., The University of
Michigan Press, Ann Arbor, Michigan, 1992. Salow, R., Thornton, J. and Siegel, P., "Is the Yellow Light Long Enough?," The
Physics Teacher, 31, 80, 1993 Zandy, J.F., "Galileo, Einstein, and the Church," American Journal of Physics,
61, 202, 1993 2.1 Distances traveled are
x1 = v1 t1 = (80.0 kin/h) (0.5 h) = 40.0 km
X2 = vz t2 = (100.0 km/h) (0.2 h) = 20.0 km
X3 = V3t3  (40.0 km/h) (0.75 h) = 30.0 km Thus, the total time is 1.7 h, and the total distance traveled is 90.0 km. (a) 9—f=%= 52.9 km/h (b) x — 90.0 km (see above) 2.2 (a) In the ﬁrst half of the trip, the average velocity is v = (X2 — x1)/ 20.0 s = +50.0 m/20.0 s  +2.50 m/s
(b) On the return leg, we have v = (X3  X2)/22.0 s  (0  50.0 m)/22.0 s =  2.27 m/s
(c) For the entire trip, 17 = (X3  x1)/42.0 s = 0/42.0 s  0 2.3 (a) Boat A requires 1 h to cross the lake and 1 h to return, total time 2h.
Boat B requires 2 h to cross the lake at which time the race is over, Boat
B being on the other side of the lake or 60 km from the ﬁnish .
(b) Average velocity is the displacement of the boat divided by the time
required to accomplish the displacement The winning boat is back
where it started, its diSplacement thus being zero yielding a zero
average velocity. x 84 x 10'2 m
2.4 =————‘ = . 4 = . 7
t v 3.5 x 106 5 24x10 8 66 h 9 min 21 s _
60 min/h + 3600 s/h ' 1156 h
. 385 yd .
and 26 miles + 1760 yd/mile = 26.22 miles
Thus, the average speed is 26.22 miles/2.156 h = 12.2 mph. 2.5 211+ 2.6 2.7 2.8 2.9 2.10 2.11 mﬁmm (a) V0,1==(X]_ — x0)/(At) = (4.0 m 0)/1.0 s = +4.0 m/s
(b) V0’4 = (X4  xo)/At= (—2.0 m» 0)/ 4.0 s =  0.5 m/s
(c) v15 = (X5  x1)/At= (0  4.0 m)/4.0 s =  1.0 m/s
(d) v05 = (xs  xo)/At= (0  0)/5.0 s = 0 (a) The time for the faster car is 10 miles/70 mph — 0.14 h, or 8.57 min.
The time for the slower car is 10 miles/55 mph = 0.18 h or 10.9 min.
The difference in time is 2.34 min. (b) When the faster car has a 15.0 min lead, it is ahead by a distance
equal to that traveled by the slower car in a time of 15.0 min. This
distance is given by: x= vt= (55 0 mi/h)(15 0 mix—M) = 13 3 mi. ’ ‘ 60.0 min ' The faster car pulls ahead of the slower car at a rate of: Vrelauve = 70.0 mph  55.0 mph = 15.0 mph. Thus, the time
required for it to get 13.8 mi ahead is: x 13.8 mi ts Vrelative _ 150 Eli/h 0.92011
Finally, the distance the faster car has traveled during the time it is
gainmg' a 13.8 mi lead is given by: X: Vt .. (70.0 mi/h)(0.920 h) = 64.4 mi. The distance traveled by the space shuttle in one orbit is
21c(Earth's radius + 200 miles) = 2::(3963 + 200) = 26,1563 miles . . . 26156.9 miles
and so the required time IS —_—“—19800 miles/h= 1.32 h The total time for the trip is t= t1 + 22.0 min = t1 + 0.367 h, where I1 is the
time spent traveling at 89.5 km/h. Thus, the distance traveled is
x= v t= (89.5 km/h)t1 = (77.8 1cm/h)(t1 + 0.367 h)
or, (89.5 km/h)t1 = (77.8 km/h) t1 + 28.5 km
From which, :1  2.44 h for a total time of t”= t1 + 0.367 h  2.81 h. Therefore, x= l7 t= (77.8 km/h)(2.81 h) = 218 km (a) The speed of the tortoise is Vt —— 0.1 m/s, and the speed of the hare is
Vh .. 20x0.1 m/s= 2.0 m/s.
Xt=Xh + 0.2 m
vtt= vh( t 120 s) + 0.2 In, or
(0.1 m/s)t= 2.0 m/s(t 120 s) + (0.2 m).
From which, t= 126 s.
(b) xt= Vtt= (0.1 m/s)(126 s) = 12.6 m Let It be the maximum time to complete the trip.
I total distance 1 1600 m 1 km/h
t' needed average speed ' 250 km/h 0.278 m/s
The time spent to complete the ﬁrst half, q, is
t half distance 800 m 1 km/h _1251&
1 = vlave 230 km/h 0273 m/s  = 23.02 5. Thus, the maximum time that can be spent on second half of the trip is I: .. tt  t1 = 23.02 s  12.51 s = 10.51 s,
and the required average speed on the second half is 2.16 2.17 2.18 2.19 2.20 2.21 2.22 2.23 2.24 CHAPTER TWO SOLUTIQNS = (—2.0 In — 4.0 m)/1.5 s
= 6.0 W15 3 = 4.0 m/s
(c) V(t=3 s) =2 [X(t= 4 s)  x(t= 2.5 s)]/(4.0s  2.5 s)
= (—2.0m (2.0m))/ 1.5 s = 0
(d) V(t=4.5 s) = {x(t= 5 s) x(t=4s)]/(5 s4s)
= (0 — (—2.0 m))/1.0 s = + 2.0 m/s From the definition of acceleration, Av == amt) = (0.80 m/sz)(2.0 s) = 1.6
m/s. From this, the final velocity is Vf = 7.0 m/s + 1.6 m/s = 8.6 m/s The average acceleration is found as
a =AV/A t (+8.0 m/s  5.0 m/s)/(4.0 s) = 0.75 m/sz. —  I  I
a= AV/At= WC ms 100 m/s =1.5x103 m/sZ 12 x 10'3 s
_ __n_11_'(0.447 m/s) .
v1=55 h (1 [Di/h) =24.58m/s, and by the same method, Vf = 26.82 m/s. Thus; Av = 2.24 m/s, and from
the deﬁnition of acceleration At= AV/a = (2.24 m/s)/ (0.6 m/sz) = 3.73 s. (a) am to 5 s) = AV/At= (0 0)/5.0 s = 0
3(5 s to 15 s) = (+8.0 m/s  (8.0 m/s))/(10.0 s) = +1.6 m/sZ. 3(0 to 20 s) = (+8.0 m/s (—8.0 m/s))/20.0 s = +0.80 111/32.
(‘0) At t== 2 s, the slope of the tangent line to the curve is 0. At t= 10.0 s, the slope of the tangent line is +1.6 m/sz.
At t = 18.0 s, the slope of the tangent line is 0 (a) The average acceleration can be found from the curve, and its value will be a=Av/At= (16 m/s)/ 2.0s=+ 8.00m/sz.
(b) The instantaneous acceleration at t = 1.5 3 equals the slope of the
tangent line to the curve at that time. The line will have a slope of about +11.0 m/sZ. From v2  v02 + 23x, we have (10.97 x 103 m/s)2 =0+ 2a(220 m),
so that a = 2.74 x 105 m/s2 which is 2.79 x 104 times g! (a) From the deﬁnition of acceleration, we have V_‘KQ_0__M/_§_ 2
a“ t ‘ 5.0s "3'0W5' (b) From x= v0t+%at2 , we have x = (40 m/s)(5.0 s) + % (8.0 m/32)(5.0 s)2 = 100111. v VQ_0100 tn/s
(a) t 3 _ 5 W52
(b) x=9r+(z—‘”2—VQ)1=10%2’5"—020s=1000m=1.0m Therefore, the minimum distance to stop exceeds the length of the
runway, so it cannot land safely. =205 11 EME—Two §6__‘Lunous 2.35 The initial velocity of the train is Vi = 82.4 km/h = 22.9 m/s and the ﬁnal velocity is Vf = 16.4 km/h — 4.56 m/s. We also know that Ax= if It, so tA—f‘ and ix =V—i+2—Vf— = 13.73 m/s. v
_400_m _
Therefore, t= 13.73 m/s — 29.1 s. 2.36 (a) Take t= 0 at the time when the ﬁrst player starts to chase the second
player. At this time, the second player is 36 m in front of the ﬁrst player. Let us write down x= vot+ % at2 for both players. For the
ﬁrst player, we have X1 =Vot+ % at2 = 0+% (4 m/sz)t2, (1) and for the second player, X2= Veti % at2 = (12 m/s)t+ 0 (2) When the players are sideby—side, x1 =xz + 36 m (3) From Equations (1), (2), and (3), we ﬁnd r2  6t18 = 0.
The roots of this equation are t=  2.2 s and t= + 8.2 s. We must
choose the 8.2 5 answer since the time must be greater than zero. (b) AX1= V01t+%a1 :2 =0+ % (4 m/s2)(8.2 s)2 = 134m 2.37 (3) Use vzV02+2aywith vO. We have 0 = (25.0 m/s)2 + 2(9.80 m/s2)ym This gives the maximum height,
ym, as ym = 31.9 m.
(b) The time to reach the highest point is found from the deﬁnition of
0  25.0 m/s_255$
(9.8 m/sz) ' (c) From the symmetry of the motion, the ball takes the same amount of
time to reach the ground from its highest point as it does to move
from the ground to its highest point. Thus, t 2.55 s. (d) We can use v= vo+ at, with the position of the ball at its highest
point as the Origin of our coordinate system. Thus, v0 = O, and t is
the time for the ball to move from its maximum height to ground
level. This was found in part (c) to be 2.55 s. Thus, v=0 + (9.80 m/s2)(2.55 s) = 25 m/s. acceleration as t= 2.38 Use y=voyr+%a:2 , or 76.0m=0+%(9.80 m/s2)t2. This givest=3.94s 2.39 (a) After 2.00 s, we ﬁnd that v= V0+ gt= 1.50 + (—9.80)2.00 = 21.1 m/s,
so the speed is 21.1 m/s. 2
(b) Using X= vot+ gg we have the distance the ball falls in 2.00 s as
IQ, = (1.50)(2.00) + (9.80)(2.00)2/2 =  22.6 In, and the distance
the helicopter moves is Xc = (~l.50 m/s)(2.00 s) =  3.0 m. Therefore, the mailbag is {22.6 m)  (— 3.0 m) = — 19.6 m below the
helicopter after two seconds.
(c) Here we have after 2.00 s, v v0 + gt: +1.50 + (9.80)(2.00)2/2 = 18.1 m/s, or a speed of 18.1 W8 14 ...
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 Fall '01
 Shapiro
 Physics

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