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1
MECHANICS PART MOTION IN A STRAIGHT LINE
course in 2 h 12 min 36 s. What was Thugwame's average speed, in meters per second?
CHAPTER 2
ActivPhysics can help with these problems: Activities 1.21.9 Section 21: Distance, Time, Speed, and Velocity Problem
1. In 1996 Donovan Bailey of Canada set a world record in the 100m dash, with a time of 9.84 s. What was his average speed?
Solution
v= r (26 + 385/1760) mi mi m = = 11.9 5.30 , t (2 + 756/3600) h h s
Solution
Bailey's average speed was (Equation 21) v = x/t = 100 m/9.84 s = 10.16 m/s. (One can assume that the race distance was known to more than four significant figures.)
or a little over half the speed of Bailey's 100 m dash in Problem 1. (Runners usually compute their average pace, 1/, which in this case was 5 min 3.4 s per mile. v See Appendix C for the appropriate conversion factors.)
Problem
4. Human nerve impulses travel at about 102 m/s. Estimate the minimum time that must elapse between the time you perceive a stalled car in front of you and the time you can activate the muscles in your leg to brake your car. (Your actual "reaction time" is much longer than this estimate.) Moving at 90 km/h, how far would your car travel in this time?
Problem
2. When races in a track meet are timed manually, timers start their watches when they see smoke from the starting gun, rather than when they hear the gun. How much error is introduced in timing a 200m dash over a straight track if the watch is started on the sound rather than the smoke? The speed of sound is about 340 m/s.
Solution
Suppose the neural path length from the brain to the leg muscles (the quadriceps) is about 1 m long. The travel time for nerve impulses is about 1 m/102 m/s = 102 s. A car moving at 90 km/h would travel r = v t = (90103 m/3600 s)(102 s) = 25.0 cm during this interval.
Solution
Suppose the timer stands at the finish line, 200 m from where the starting gun is fired. The time required for a signal to travel this distance at speed v is t = r/v = 200 m/v. Because the speed of light is so great, the puff of smoke is seen after a negligible delay, as far as conventional watches are concerned, t = 200 m/3108 m/s = 0.67 s. The travel time for a sound signal, however, t = 200 m/340 m/s = 0.59 s, would introduce a significant error, if times are recorded to the nearest hundredth of a second. (For a manually operated timing device, the error due to human reaction time is about 0.2 s.)
Problem
5. Starting from home, you bicycle 24 km north in 2.5 h, then turn around and pedal straight home in 1.5 h. What are your (a) displacement at the end of the first 2.5 h, (b) average velocity over the first 2.5 h, (c) average velocity for the homeward leg of the trip, (d) displacement for the entire trip, and (e) average velocity for the entire trip?
Problem
3. In 1996, Josia Thugwame of South Africa won the Olympic Marathon, completing the 26mi, 385yd
Solution
(a) rout = 24 km (north). (b) vout = 24 km (north)/2.5 h = 9.6 km/h(north). (c) vback =
CHAPTER 2 24 km(south)/1.5 h = 16 km/h (south). (d) rout and back = 0. (e) vround trip = 0.
11
Solution
At an average speed of 20 mi/h for the first 1 1 15 min = 4 h, you travel only (20 mi/h)( 4 h) = 5 mi. Therefore, you must cover the remaining 5 (25  5) mi = 20 mi in (40  15) min = 25 min = 12 h. 5 This implies an average speed of 20 mi/ 12 h = 48 mi/h. (Note that your overall average speed was predetermined to be 25 mi/(40 h/60) = 37.5 mi/h, and that this equals the timeweighted average of the average speeds for the two parts of the trip: (15 min/40 min)(20 mi/h) + (25 min/40 min) (48 mi/h).)
Problem
6. The Pathfinder spacecraft landed on Mars on July 4, 1997, at which time Mars was approximately 190 million km from Earth. How long did it take Pathfinder's radio signals, traveling at the speed of light, to reach Earth?
Solution
t = r/ = (190106 km)/(3105 km/s) = v 6.33102 s or about 10.6 min.
Problem
10. Taking Earth's orbit to be a circle of radius 1.5108 km, determine the speed of Earth's orbital motion in (a) meters per second and (b) miles per second.
Problem
7. Australian Chris McCormack won the 1997 world triathlon championship, completing the 1500m swim, 40km bicycle ride, and 10km run in 1 h, 48 min, 29 s. What was McCormack's average speed?
Solution
(a) Since it takes 1 y for the earth to travel the full circumference of its nearly circular orbit, v = 2R/t = 2(1.51011 m)/(3.156107 s) = 2.99104 m/s 30 km/s (an easily remembered figure). (b) (2.99104 m/s) (1 mi/1609 m) = 18.6 mi/s (about 104 times the speed of light).
Solution
v = r/t = (1.5 + 40 + 10) km/(1 + 48/60 + 29/3600) h = (28.5 km/h)(1 m/s)(3.6 km/h) = 7.91 m/s.
Problem
8. (a) Find a value, good to one significant figure, for the speed of light in feet per nanosecond (ft/ns)(1 ns = 109 s). (b) Electrical signals in wires travel at about half the speed of light. What is the maximum possible separation between a computer's central processing unit and its memory if the central processor is to be able to get a signal to memory requesting data, and have the data return, all in 8 ns?
Problem
11. What is the conversion factor from meters per second to miles per hour?
Solution
1 mi/h = 1609 m/3600 s = 0.447 m/s = (2.24)1 m/s.
Problem
12. If the average American driver goes 5000 mi each year on interstate highways, how much less time did the average driver spend on interstate highways each year as a result of the 1995 increase in the speed limit from 55 mi/h to 65 mi/h?
Solution
(a) (3108 m/s)(3.281 ft/m)(109 s/ns) 1 ft/ns. (b) The time for an electrical signal to make a round trip over a length of wire, x, is 8 ns = t = 2 x/, v where v is one half the speed of light from part (a). 1 Thus, x = 1 v t = 2 1 ft/ns (8 ns) = 2 ft. 2 2
Solution
With an average speed equal to the highway speed limit, t = r/ = (5000 mi)/(65 mi/h) = 76.9 h v after 1995, and t = (5000 mi)/(55 mi/h) = 90.9 h before, a difference of 14.0 h.
Problem
9. You allow yourself 40 min to drive 25 mi to the airport, but are caught in heavy traffic and average only 20 mi/h for the first 15 min. What must your average speed be on the rest of the trip if you are to get there on time?
Problem
13. A fast baserunner can get from first to second base in 3.4 s. If he leaves first base as the pitcher throws a 90 mi/h fastball the 61ft distance to the catcher, and if the catcher takes 0.45 s to catch
12
CHAPTER 2 and rethrow the ball, how fast does the catcher have to throw the ball to second base to make an out? Home plate to second base is the diagonal of a square 90 ft on a side. 210 km, so it would take you 21 2.5 h = 52.5 h to travel 21 210 km = 4410 km. You could drive the final 190 km in (190 km)/(105 km/h) = 1.81 h, so the complete trip would take 54.3 h. (b) Overall, v = 4600 km/54.3 h = 84.7 km/h.
Solution
At 90 mi/h = 132 ft/s, the ball takes 61 ft/(132 ft/s) = 0.462 s to travel from the pitcher to the catcher. (We are keeping extra significant figures in the intermediate calculations as suggested in Section 17.) After the catcher throws the ball, it has 3.4 s  0.462 s  0.45 s = 2.49 s to reach second base at the same time as the runner. The distance is 2(90 ft), so the minimum speed is v = 2(90 ft)/2.49 s = 51.2 ft/s = 35 mi/h. A prudent catcher would allow extra time for the player covering second base to make the tag.
Problem
16. I can run 9.0 m/s, 20% faster than my kid brother. How much head start should I give him in order to have a tie race over 100 m?
Solution
The older brother can run 100 m in 100 m/(9.0 m/s) = 11.1 s, while the younger brother takes 20% longer or 13.3 s for the same distance (younger = volder /(120%)). v Therefore, the slower brother should be given a head start in time of 2.2 s. (Another way to produce a tie is to give the slower brother a 16.7 m head start in distance.)
Problem
14. Despite the fact that jet airplanes fly at about 1000 km/h, plane schedules and connections are such that the 4800km trip from Burlington, Vermont, to San Franscisco ends up taking about 11 h. (a) What is the average speed of such a trip? (b) How much time is spent on the ground, assuming that the actual distance covered by the several aircraft involved in connecting flights is 6700km and that the planes maintain a steady 960 km/h in flight?
Problem
17. A jetliner leaves San Francisco for New York, 4600 km away. With a strong tailwind, its speed is 1100 km/h. At the same time, a second jet leaves New York for San Francisco. Flying into the wind, it makes only 700 km/h. When and where do the two planes pass each other?
Solution
(a) v = r/t = 4800 km/11 h = 436 km/h. (b) The actual flight time is t = 6700 km/(960 km/h) = 6.98 h, so 11 h  6.98 h = 4.02 h is spent on the ground.
Solution
When the planes pass, the total distance traveled by both is 4600 km. Therefore, 4600 km = (1100 km/h)t + (700 km/h)t, or t = 4600 km (1800 km/h) = 2.56 h. (The planes meet 2.56 h after taking off.) The encounter occurs at a point about (700 km/h)(2.56 h) 1790 km from New York City or (1100 km/h)(2.56 h) 2810 km from San Francisco.
Problem
15. If you drove the 4600 km from coast to coast of the United States at 65 mi/h (105 km/h), stopping an average of 30 min for rest and refueling after every 2 h of driving, (a) What would be your average velocity for the entire trip? (b) How long would it take?
Problem
18. Figure 219 shows the position of an object as a function of time. Determine the average velocity for (a) the first 2 s; (b) the first 4 s; (c) the first 6 s; (d) the interval from 3 s to 4 s.
Solution
If you stopped 30 min for every 2 h of driving at 105 km/h, your average speed would be v = (2 h/2.5 h)(105 km/h) + (0.5 h/2.5 h)(0) = 84.0 km/h, and a coasttocoast trip would take (4600 km) (84.0 km/h) = 54.8 h. However, this is only approximate, because the exact travel time does not include a 30min stop after the final segment. (a) To find the total time, note that every 2 h 30 min you would cover a distance x = v t = (105 km/h)(2 h) =
Solution
From Fig. 219, one can determine the positions of the object at the times indicated, and the average velocities from Equation 21. (a) v = [x(2 s)  x(0)] (2 s  0) = (2.5 m  0)/(2 s  0) = 1.25 m/s. (b) v = [x(4 s)  x(0)]/(4 s  0) = 0/4 s = 0. (c) v = [x(6 s)  x(0)]/(6 s  0) = 2 m/6 s = 0.33 m/s. (d) v = [x(4 s)  x(3 s)]/(4 s  3 s) = (0 m  3 m)/1 s = 3.00 m/s.
CHAPTER 2
x (m)
13
and velocities vAC and vC B were chosen arbitrarily.)
x C C B
0
3 2 1
tA xC xB
tC
tC
tB
t
0 1 2
1 2 3 5 6 7
t (s)
xA A
Problem 19 Solution. figure 219 Problem 18.
Problem
20. For the motion plotted in Fig. 220, estimate (a) the greatest velocity in the positive x direction; (b) the greatest velocity in the negative x direction; (c) any times when the object is instantaneously at rest; and (d) the average velocity over the interval shown.
Section 22: Instantaneous Velocity Problem
19. On a single graph, plot distance versus time for the two trips from Houston to Des Moines described on page 24. For each trip, identify graphically the average velocity and, for each segment of the trip, the instantaneous velocity.
Solution
(a) A little before 2 s the curve has its greatest positive slope: x/t [x(2.3 s)  x(1.6 s)]/0.7 s (4  2) m/0.7 s 2.9 m/s. (b) The greatest negative slope occurs around 4 s: v(4 s) [x(4.3 s)  (3.6 s)] (4.3 s  3.6 s) (3  4) m/0.7 s 1.4 m/s. (c) v (t) = 0 where the tangent is horizontal, near 3 s and 5 s. (d) v = [x(6 s)  x(0)]/6 s 3 m/6 s = 0.5 m/s.
5 4 3 2 1 1 (c) (a) t 2 3 Time (s) 4 Distance (m) t (b) x (c) 5 t 6 x (d)
Solution
Both trips start at the same place (Houston, point A) xA = 1000 km at time tA = 0, and end at the same place (Des Moines, point B) xB = 300 km at tB = 2.6 h. (We are using the coordinate system in Fig. 22.) They have the same overall displacement, x = xB  xA = 1300 km, in the same time period, t = tB  tA = 2.6 h, and thus the same average velocity vAB = 500 km/h, as explained in the text. vAB is the slope of the straight line AB. AB is also the graph of the first trip, a direct flight at constant velocity, x1 (t) = xA + vAB t for 0 t 2.6 h. (Short intervals of acceleration at takeoff and landing are ignored.) The second trip, using a faster plane (steeper slopes when flying), stops for a while in Minneapolis at xC = 650 km (this segment is flat) and then proceeds south to Des Moines (negative velocity and slope). This trip is shown by three straight segments ACC B, and is given analytically by the equations xA + xC  xA t = xA + vAC t, tC  0 for tA = 0 t tC x2 (t) = xC , for tC t tC x + xB  xC (t  t ) = x + v (tt ), C C B C C C tB  tC for tC t tB . (In the graph, we assumed each segment of the second trip was executed with constant velocity and ignored takeoffs and landings as before. The times tC and tC
x
figure 220 Problem 20 Solution.
Problem
21. Figure 221 shows the position of an object as a function of time. From the graph, determine the instantaneous velocity at (a) 1.0 s; (b) 2.0 s; (c) 3.0 s; (d) 4.5 s. (e) What is the average velocity over the interval shown?
Solution
The instantaneous velocity at a particular time is the slope of the graph of x versus t at that point, v(t) = dx/dt. For a straight line segment of graph,
14
CHAPTER 2
x (m)
3 2 1 0 1 2
1 2 4 6 7 t (s)
surrounding 2 s gets smaller, the average and instantaneous velocities agree better; the values in parts (c) and (d) differ by less than 0.02%.
Problem
23. A model rocket is launched straight upward; its altitude y as a function of time is given by 2 y = bt  ct2 , where b = 82 m/s, c = 4.9 m/s , t is the time in seconds, and y is in meters. (a) Use differentiation to find a general expression for the rocket's velocity as a function of time. (b) When is the velocity zero?
figure 221 Problem 21.
Solution
v equals the average velocity over that segment, v = x/t. Each of the times specified in this problem falls on a different straight segment of the graph in Fig. 221, whose slopes we determine from the coordinates of the endpoints of that segment. (a) v(1 s) = (3  0) m/(1.5  0) s = 2 m/s; (b) v(2 s) = (3  3) m/(2.5  1.5) s = 0; (c) v(3 s) = (2  3) m/(3.5  2.5) s = 5 m/s; (d) v(4.5 s) = [1  (2)] m/(6  3.5) s = 1.2 m/s. (e) The overall average velocity is v = [x(6 s)  x(0)] (6 s  0) = 1 m/6 s = 0.167 m/s. (a) Equation 23 can be used to find the derivative of each term in the altitude: v(t) = dy/dt = b  2ct. (b) The velocity is zero when b = 2ct, or t = b/2c = (82 m/s)/2(4.9 m/s2 ) = 8.37 s.
Problem
24. The position of an object as a function of time is given by x = bt4 , where b is a constant. Find an expression for the instantaneous velocity as a function of time, and show that the average velocity over the interval from t = 0 to any time t is onefourth of the instantaneous velocity at t.
Solution
From Equations 22 and 3, v = dx/dt = 4bt3 . The average velocity over an interval from 0 to t, from Equation 21, is v = [x(t)  x(0)]/(t  0) = bt4 /t = bt3 , 1 which is just 4 of v(t) from above. (Note that v is not equal to the average of v(0) and v(t).)
Problem
22. The position of an object as a function of time is given by x = bt + ct3 , where b = 1.50 m/s and 3 c = 0.640 m/s . To study the limiting process leading to the definition of instantaneous velocity, calculate the average velocity of the object over time intervals from (a) 1.00 s to 3.00 s; (b) 1.50 s to 2.50 s; (c) 1.95 s to 2.05 s. (d) Obtain the instantaneous velocity as a function of time by differentiating, and compare its value at 2 s with your average velocities.
Problem
25. The position of an object is given by x = bt3  ct2 + dt, with x in meters and t in seconds. The constants b, c, and d are 3 2 b = 3.0 m/s , c = 8.0 m/s , and d = 1.0 m/s. (a) Find all times when the object is at position x = 0. (b) Determine a general expression for the instantaneous velocity as a function of time, and from it find (c) the initial velocity and (d) all times when the object is instantaneously at rest. (e) Graph the object's position as a function of time, and identify on the graph the quantities you found in (a) to (d).
Solution
From the given function, x(t) = bt + ct3 , and the definition of average velocity (Equation 21), v = x/t, one obtains: (a) va = [x(3 s)  x(1 s)] (3 s  1 s) = [(1.5 m/s)(3 s  1 s) + (0.64 m/s2 )(9 s3  1 s3 )]/2 s = 9.82 m/s; (b) vb = [x(2.5 s)  x(1.5 s)] (2.5 s  1.5 s) = 9.34 m/s; (c) vc = [x(2.05 s)  x(1.95 s)]/(2.05 s  1.95 s) = 9.18 m/s. (d) The instantaneous velocity (Equation 22b) is v(t) = dx/dt = b + 3ct2 . At t = 2.00 s, v(2 s) = 1.5 m/s + 3 3(0.64 m/s )(2 s)2 = 9.18 m/s. As the interval
Solution
(a) With the aid of the quadratic formula and factorization, x = t(bt2  ct + d) = 0 implies t = 0, or t = (c c2  4bd)/2b. Substituting the given constants, t = 0, t = (4 13) s/3 = 0.131 s and
CHAPTER 2 2.54 s. (b) v(t) = dx/dt = 3bt2  2ct + d. (c) When t = 0, v(0) = d = 1 m/s. (d) v = 3bt2  2ct + d = 0 implies t = (c c2  3bd)/3b = (8 55) s/9 = 64.9 ms and 1.71 s. (e) The graph of this cubic has roots from part (a), slope at the origin from part (c), and relative maximum and minimum from part (d), as shown.
x
15
Solution
Equation 24, the definition of average linear acceleration, for this onehour time interval, gives a = v/t = (v2  v1 )/(t2  t1 ) = 2 (450 km/s  0)/3600 s = 125 m/s .
Problem
28. Starting from rest, a subway train first accelerates to 25 m/s, then begins to brake. Fortyeight seconds after starting, it is moving at 17 m/s. What is its average acceleration in this 48s interval?
slope 1 m/s 0.13 s 1.71 s 0 0.065 s 2.54 s t
Solution
During an interval of 48 s = t2  t1 , the velocity of the train (along a linear track) changes from v1 = 0 (starting at rest) to v2 = 17 m/s. The average acceleration is a = (v2  v1 )/(t2  t1 ) = 2 (17  0) m/s/48 s = 0.354 m/s . (Note that the intermediate velocity is irrelevant.)
Problem 25 Solution.
Problem
26. In a drag race, the position of a car as a function of time is given by x = bt2 , with b = 2.000 m/s2 . In an attempt to determine the car's velocity midway down a 400m track, two observers stand 20 m on either side of the 200m mark and note the time when the car passes them. (a) What value do the two observers compute for the car's velocity? Give your answer to four significant figures. (b) By what percentage does this observed value differ from the actual instantaneous value at x = 200 m?
Problem
29. A space shuttle's main engines cut off 8.5 min after launch, at which time the shuttle's speed is 7.6 km/s. What is the shuttle's average acceleration during this interval?
Solution
The average acceleration of the shuttle along its trajectory, from liftoff until its main engines stop, was (Equation 24) a = v/t = 3 (7.610 m/s  0)/(8.560 s) = 14.9 m/s2 1.5 g, where g = 9.8 m/s2 , the acceleration due to gravity at the surface of the Earth, is a frequently used anthropomorphic unit.
Solution
(a) The car passes the first observer at time t1 = x1 /b = (200  20) m/(2 m/s ) = 9.4868 s, and the second observer at 2 t2 = (200 + 20) m/(2 m/s ) = 10.4881 s. They compute the car's (average) velocity to be vobs = (x2  x1 )/(t2  t1 ) = 40 m/1.0013 s = 39.95 m/s. (b) The car reaches x = 200 m at t = 10 s, where v(10 s) = (dx/dt)10 s = (2bt)10 s = 40.00 m/s. The observed velocity differs from this by only 100 (40.00  39.95)/40.00 = 1 % 0.13%. 8
2
Problem
30. An egg drops from a secondstory window, taking 1.12 s to fall and reaching a speed of 11.0 m/s just before hitting the ground. On contact with the ground, the egg stops completely in 0.131 s. Calculate the average magnitudes of its acceleration while falling and of its deceleration while stopping.
Section 23: Acceleration Problem
27. A giant eruption on the Sun propels solar material from rest to a final speed of 450 km/s over a period of 1 h. What is the average acceleration of this material, in m/s2 ?
Solution
The velocity (positive downward) changes from v1 = 0 to v2 = 11.0 m/s in 1.12 s while falling with average acceleration a = v/t = (11.0 m/s) = (1.12 s) = 9.82 m/s2 . When stopping, the egg's velocity changes from v2 = 11.0 m/s to v3 = 0 in 0.131 s, with average acceleration a = (0  11.0 m/s)/0.131 s = 84.0 m/s2 .
16
CHAPTER 2
1 107 s
v (104 m/s)
(A negative sign means that the acceleration is upward; when the speed is decreasing, this is called a deceleration.)
2 1 0
2
2 104 m/s
Problem
31. An airplane's takeoff speed is 320 km/h. If its 2 average acceleration is 2.9 m/s , how long is it on the runway after starting its takeoff roll?
4
8
t (107 s)
1 2
figure 222 Problem 34.
Solution
If we assume that the airplane starts from rest, v = 320 km/h = 88.9 m/s at the end of a time interval t, during which the average acceleration was 2 a = 2.9 m/s = v/t. Solving for t, we find t = (88.9 m/s)/(2.9 m/s2 ) = 30.7 s.
Solution
The graph of v versus t has its steepest slope when crossing the taxis. An estimate of the slope at t = 0 gives amax . = (dv/dt)0 (2 104 m/s)/(1 107 s) = 2 1011 m/s2 .
Problem
32. ThrustSSC, the world's first supersonic car, accelerates from rest to 1000 km/h in 16 s. What is its acceleration, in m/s2 ?
Problem
35. Determine the instantaneous acceleration as a function of time for the motion in Problem 25.
Solution
a = v/t = (1000 km/h  0)/16 s = 2 (62.5 km/h/s) (1 m/s)/(3.6 km/h) = 17.4 m/s .
Solution
From the answer to Problem 25(b), we find: a(t) = dv/dt = (d/dt)(3bt2  2ct + d) = 6bt  2c.
Problem
33. Your plane reaches its takeoff runway and then holds for 4.0 min because of airtraffic congestion. The plane then heads down the runway with an average acceleration of 3.6 m/s2 . It is airborne 35 s later. What are (a) its takeoff speed and (b) its average acceleration from the time it reaches the takeoff runway until it's airborne?
Problem
36. The position of an object is given by x = bt3 , where x is in meters, t is in seconds, and where the constant b is 1.5 m/s3 . Determine (a) the instantaneous velocity and (b) the instantaneous acceleration at the end of 2.5 s. Find (c) the average velocity and (d) the average acceleration during the first 2.5 s.
Solution
(a) During the 35 s the plane is actually taking off, Equation 24 gives v = v  0 = at = 2 (3.6 m/s )(35 s) = 126 m/s = 454 km/h. (b) If we include the fourminute wait before taking off, the average acceleration for the entire interval on the runway is only a = v/t = (126 m/s  0) (4 min + 35 s) = (126 m/s)/(275 s) = 0.458 m/s2 .
Solution
(a) v(t) = dx/dt = 3bt2 , and v(2.5 s) = 3 3(1.5 m/s )(2.5 s)2 = 28.1 m/s. (b) a(t) = dv/dt = 3 2 6bt, and a(2.5 s) = 6(1.5 m/s ) (2.5 s) = 22.5 m/s . 3 (c) v = [x(2.5 s)  x(0)]/2.5 s = (1.5 m/s )(2.5 s)2 = 9.38 m/s. (d) a = [v(2.5 s)  v(0)]/2.5 s = 3 2 3(1.5 m/s )(2.5 s) = 11.3 m/s .
Problem
34. Under the influence of a radio wave, an electron in an antenna undergoes backandforth motion whose velocity as a function of time is described by Fig. 222. From the graph, estimate the electron's maximum acceleration.
Section 24: Constant Acceleration Problem
37. A car accelerates from rest to 25 m/s in 8.0 s. Determine the distance it travels in two ways: (a) by multiplying the average velocity given in Equation 28 by the time and (b) by calculating the acceleration from Equation 27 and using the result in Equation 210.
CHAPTER 2
17
Solution
(a) For constant acceleration, Equation 28 can be combined with Equation 21 to yield x = v t = 1 (v0 + v)t = 1 (0 + 25 m/s)(8.0 s) = 100 m. 2 2 (b) Alternatively, for constant acceleration, Equation 27 gives a = a = v/t = (25 m/s  0) 2 8.0 s = 3.13 m/s , so Equation 210 yields x = 2 x x0 = v0 t+ 1 at2 = 0 + 1 (3.13 m/s )(8.0 s)2 = 100 m. 2 2
2.8 km/s. (a) What is its acceleration? (b) How long does the ascent take?
Solution
(a) In Equation 211 (with x positive upward) we are given that x  x0 = 85 km, v0 = 0 (the rocket starts from rest), and v = 2.8 km/s. Therefore, we can solve 2 for the acceleration, a = (v 2  v0 )/2(x  x0 ) = 2 2 (2.8 km/s) /2(85 km) = 46.1 m/s (note the change of units). (b) From Equation 29, we can solve for the time of flight, t = 2(x  x0 )/(v0 + v) = 2(85 km) (2.8 km/s) = 60.7 s. (We chose to relate t directly to the given data, but once the acceleration is known, Equation 27 or 210 could have been used to find t = v/a or t = 2(x  x0 )/a, respectively.)
Problem
38. Differentiate both sides of Equation 210, and show that you get Equation 27.
Solution
v = dx d 1 1 = (x0 + v0 t + 2 at2 ) = 0 + v0 + 2 a2t dt dt = v0 + at.
Problem
42. On packed snow, use of computerized antilock brakes can reduce the stopping distance for a car by 55%. By what percentage is the stopping time reduced?
Problem
39. If you square Equation 27, you'll have an expression for v 2 . Equation 211 also gives an expression for v 2 . Equate the two expressions for v 2 , and show that the resulting equation reduces to Equation 210.
Solution
The stopping distance and the stopping time are related by Equation 29, for motion with constant deceleration. When stopping, v0 is the initial velocity and v = 0. Therefore x  x0 = 1 v0 t, or the stopping 2 distance, x  x0 , is proportional to the stopping time, t, and both are reduced by the same amount. (Antilock brakes optimize the deceleration by controlling the wheels to roll just at the point of skidding.)
Solution
2 Squaring Equation 27, v 2 = (v0 + at)2 = v0 + 2v0 at + 2 2 2 2 a t , and equating to Equation 211, v = v0 + 2a 2 2 (x  x0 ), one finds 2v0 at + a t = 2a(x  x0 ), or 1 x  x0 = v0 t + 2 at2 , which is Equation 210.
Section 25: Using the Equations of Motion Problem
40. Electrons that "paint" the picture in a TV tube undergo constant acceleration over a distance of 3.8 cm. If they reach a final speed of 1.2107 m/s, what are (a) the electrons' acceleration and (b) the time spent accelerating?
Problem
43. Starting from rest, a car accelerates at a constant rate, reaching 88 km/h in 12 s. (a) What is its acceleration? (b) How far does it go in this time?
Solution
(a) From Equation 27, a = (v  v0 )/t = (88 km/h  0)/12 s = 7.33 km/h/s = 2.04 m/s2 . 1 (b) From Equation 29, x  x0 = 2 (v0 + v)t = 1 2 (0 + 88 km/h)(12 s) = 147 m. (Note the change in units. Again, we chose equations that relate the answers directly to the given data; see solution to Problem 41.)
Solution
(a) From Equation 211, with v0 = 0 assumed, a = v 2 /2(x  x0 ) = (1.2107 m/s)2 /2(0.038 m) = 2 1.891015 m/s . (b) The time can be found from Equation 29 and the given data: t = 2(x  x0 )/v = 2(3.8 cm)/(1.2107 m/s) = 6.33 ns. (Alternatively, the value of acceleration from part (a) can be substituted into Equation 27 or 210: t = v/a = 2(x  x0 )/a.)
Problem
44. A car moving initially at 50 mi/h begins decelerating at a constant rate 100 ft short of a stoplight. If the car comes to a full stop just at the light, what is the magnitude of its deceleration?
Problem
41. A rocket rises with constant acceleration to an altitude of 85 km, at which point its speed is
18
CHAPTER 2
Solution
Since the car stops (v = 0), after traveling 100 ft = x  x0 , from an initial speed of (55 mi/h)(22 ft/s 15 mi/h) = 73.3 ft/s, Equation 211 gives a = 2 (73.3 ft/s)2 /2(100 ft) = 26.9 ft/s . The magnitude of the deceleration is the absolute value of a.
Solution
For a particular fragment (which followed a straightline path to the bottom, perpendicular to the desert surface), we can use Equation 211 to find the 2 2 initial speed: v0 = 2(4105 m/s )(180 m) or v0 = (144106 m2 /s2 ) = 12 km/s.
Problem
45. In an Xray tube, electrons are accelerated to a velocity of 108 m/s, then slammed into a tungsten target. The electrons undergo rapid deceleration, producing X rays. If the stopping time for an electron is on the order of 1019 s, approximately how far does an electron move while decelerating? Assume constant deceleration.
Problem
48. A gazelle accelerates from rest at 4.1 m/s2 over a distance of 60 m to outrun a predator. What is its final speed?
Solution
From Equation 211, v 2 = 2a(x  x0 ) = 2(4.1 m/s ) (60 m), or v = 22.2 m/s (almost 50 mi/h).
2
Solution
Assuming the electrons travel in a straight line while coming to rest (v = 0), Equation 29 gives 1 x  x0 = 2 (v0 + v)t = 1 (108 m/s)(1019 s) = 2 12 510 m for the stopping distance. (The X rays emitted are called bremsstrahlung.)
Problem
49. A hockey puck moving at 32 m/s slams through a wall of snow 35 cm thick. It emerges moving at 18 m/s. (a) How much time does it spend in the snow? (b) How thick a wall of snow would be needed to stop the puck entirely?
Problem
46. A particle leaves its initial position x0 at time t = 0, moving in the positive x direction with speed v0 , but undergoing acceleration of magnitude a in the negative xdirection. Find expressions for (a) the time when it returns to the position x0 and (b) its speed when it passes that point.
Solution
(a) If we assume a constant linear deceleration for the puck, Equation 29 can be used to find the time it spends traversing 35 cm of snow: t = 2(x  x0 )/(v0 + v) = 2(0.35 m)/(32 + 18)(m/s) = 14 ms. (b) If we assume the same deceleration for penetrating any wall of snow, Equation 211, with v = 0, gives the thickness necessary to stop a puck moving with the 2 same initial speed: x  x0 = v0 /2a. The acceleration (which is negative when the puck is decelerating) can be found from Equation 27 with the time from part (a) (or from a second application of Equation 211 with data from part (a), etc.): a = (18  32)(m/s) 2 (0.014 s) = 103 m/s . Then any wall of snow thicker 2 than (32 m/s)2 /2(103 m/s ) = 51.2 cm would stop this puck.
Solution
In this problem we must use a for the acceleration in Table 21. (a) A return to the initial position means x(t) = x0 for t > 0. From Equation 210, x = x0 = 1 x0 + v0 t + 2 (a)t2 , or 2v0 t = at2 . Since t = 0, we can divide to get t = 2v0 /a. (b) The speed, or magnitude of the velocity, can be found from Equation 27: v = v0 + (a)t = v0  a(2v0 /a) =   v0  = v0 . The speed is the same, but the direction of motion is reversed.
Problem
50. Amtrak's 20thCentury Limited is en route from Chicago to New York at 110 km/h, when the engineer spots a cow on the track. The train brakes to a halt in 1.2 min with constant deceleration, stopping just in front of the cow. (a) What is the magnitude of the train's acceleration? (b) What is the direction of the acceleration? (c) How far was the train from the cow when the engineer first applied the brakes?
Problem
47. The Barringer meteor crater in northern Arizona is 180 m deep and 1.2 km in diameter. The fragments of the meteor lie just below the bottom of the crater. If these fragments decelerated at a 2 constant rate of 4105 m/s as they ploughed through the Earth in forming the crater, what was the speed of the meteor's impact at Earth's surface?
CHAPTER 2
19
Solution
(a) and (b) The train goes from velocity v0 = 110 km/h = 30.6 m/s (positive eastward) at t0 = 0, to a stop, v = 0, at t = 1.2 min = 72 s. The constant acceleration was a = (v  v0 )/(t  t0 ) = (30.6 m/s) (72 s) = 0.424 m/s2 . The magnitude of the acceleration is the absolute value of this, while its direction, indicated by the negative sign, was westward. (c) Equation 29 gives the stopping 1 distance: x  x0 = 2 (v0 + v)t = 1 (30.6 m/s)(72 s) = 2 1.10 km. (Equations 210 or 211 and the acceleration from part (a) could also have been used to obtain the same result.)
Problem
53. The maximum acceleration that a human being can survive even for a short time is about 200g. In a highway accident, a car moving at 88 km/h slams into a stalled truck. The front end of the car is squashed by 80 cm on impact. If the deceleration during the collision is constant, will a passenger wearing a seatbelt survive?
Solution
The passenger, originally moving with velocity v0 = 88 km/h = 24.4 m/s, comes to rest, v = 0, in a distance x  x0 = 0.8 m, so the acceleration (from 2 Equation 211) was a = (v 2  v0 )/2(x  x0 ) = 2 2 (24.4 m/s) /1.6 m = 373 m/s = 38.1g. Such a person could survive. Without a seatbelt, however, the stopping distance would not have been 0.8 m (think about it!) and the passenger would surely not survive the secondary collision with the interior of the car (see Problem 86).
Problem
51. A jetliner touches down at 220 km/h, reverses its engines to provide braking, and comes to a halt 29 s later. What is the shortest runway on which this aircraft can land, assuming constant deceleration starting at touchdown?
Solution
From Equation 29 with v = 0, we find x  x0 = 1 1 2 v0 t = 2 (220 km/h)(29 h/3600) = 886 m (over half a mile).
Problem
54. A racing car undergoing constant acceleration covers 140 m in 3.6 s. (a) If it is moving at 53 m/s at the end of this interval, what was its speed at the beginning of the interval? (b) How far did it travel from rest to the end of the 140m distance?
Problem
52. A motorist suddenly notices a stalled car and slams on the brakes, decelerating at the rate of 6.3 m/s2 . Unfortunately this isn't good enough, and a collision ensues. From the damage sustained, police estimate that the car was moving at 18 km/h at the time of the collision. They also measure skid marks 34 m long. (a) How fast was the motorist going when the brakes were first applied? (b) How much time elapsed from the initial braking to the collision?
Solution
1 1 (a) x  x0 = 140 m = 2 (v0 + v)t = 2 (v0 + 53 m/s) (3.6 s); therefore v0 = 24.8 m/s. (b) The acceleration 2 is a = (v  v0 )/t = (53  24.8) m/s/3.6 s = 7.84 m/s . Starting from rest, the distance traveled while reaching a velocity v = 53 m/s is v 2 /2a = (53 m/s)2 2(7.84 m/s2 ) = 179 m.
Problem
55. The maximum deceleration of a car on a dry road is about 8 m/s2 . If two cars are moving headon toward each other at 88 km/h (55 mi/h), and their drivers apply their brakes when they are 85 m apart, will they collide? If so, at what relative speed? If not, how far apart will they be when they stop? On the same graph, plot distance versus time for both cars.
Solution
(a) From the given acceleration, 6.3 m/s , the distance traveled, 34 m, and the final velocity, 18 km/h = 5 m/s (just before the collision), the initial velocity (when the braking began) can be calculated: 2 v0 = v 2  2a(x  x0 ), or v0 =
2 2
(5 m/s)2  2(6.3 m/s )(34 m) = 21.3 m/s = 76.7 km/h. (b) The deceleration time interval was t = (v  v0 )/a = (5 m/s  21.3 m/s)/(6.3 m/s2 ) = 2.59 s. (The positive x direction is the direction in which the car was moving.)
Solution
The minimum distance a car needs to stop (v = 0) from an initial speed v0 = 88 km/h = 24.4 m/s, with a constant acceleration a = 8 m/s2 , is (Equation 211) 2 2 x x0 = v0 /2a = (24.4 m/s)2 /2(8 m/s ) = 37.3 m (positive in the direction of v0 ). Since 85 m is greater
20
CHAPTER 2 and t = (2.8 m/s)2  4(.0375 m/s )(95 m) (2.8 m/s) (.075 m/s ) = 25.3 s
2 2
than twice this distance, the cars can avoid a collision, and they will 85 be m  2(37.3 m) = 10.3 m apart when stopped. To plot x versus t, using Equation 210 for each car, we need to choose an origin, say x = 0 at the midpoint of the separation between the cars, with positive x in the direction of the initial velocity of the first car, and t = 0 when the brakes are applied. Then x10 = 42.5 m = x20 , v10 = 24.4 m/s = v20 , and a1 = 8 m/s = a2 . A graph of x1 (t) and x2 (t) is as shown.
50
(where only the solution with t > 0 is meaningful in this problem). When they meet, Amy and George are a distance xBldg  xAmy = v0,Amy t = (1.6 m/s)(25.3 s) = 40.5 m from the physics building.
X xBldg xAmy
25 x 5.154 m x(m) 0 x 5.154 m 25
t 3.056 s
xDorm
xGeo t
50 0 1 t(s) 2 3
Problem 56 Solution.
Problem 55 Solution.
Problem Problem
56. George, a physics student, leaves his dormitory at a speed of 1.2 m/s, heading for the physics building 95 m away. Just as he leaves his dorm, Amy, another physics student, leaves the physics building and heads toward George at a steady 1.6 m/s. George immediately spots her and begins accelerating at 0.075 m/s2 . Where and when do the two meet? Plot positionversustime curves for both students on a single graph. 57. After 35 minutes of running, at the 9km point in a 10km race, you find yourself 100 m behind the leader and moving at the same speed. What should your acceleration be if you are to catch up by the finish line? Assume that the leader maintains constant speed throughout the entire race.
Solution
Taking x0 = 0 and t = 0 at the 9km point (and assuming a straight path to the finish), we can express your position (runner A) and that of the leader 1 (runner B) as xA = v0 t + 2 at2 , and xB = 100 m + v0 t. Since B's speed was constant, v0 = x/t = (9 km + 100 m)/35 min = 0.26 km/min. If both runners finish simultaneously, xA = xB = 1 km, so a = 2(1 km  2 v0 t)v0 /(v0 t)2 . We multiplied and divided by v0 because the quantity v0 t (at the finish) equals xB  100 m = 1 km  100 m = 0.9 km. Therefore a = 2(1 km  0.9 km)(0.26 km/ min)2 /(0.9 km)2 = 1.67102 km/min = 4.64103 m/s .
2 2
Solution
Let the (assumed) straightline path between the dormitory and the building be along the xaxis, with positive direction the way George walks. If both students leave at t = 0, then xGeo = xDorm + v0,Geo t + 1 2 2 aGeo t , and xAmy = xBldg + v0,Amy t, where v0,Geo = 2 1.2 m/s, aGeo = .075m/s , v0,Amy = 1.6 m/s, and xBldg  xDorm = 95 m. The students meet when xGeo = xAmy , or xBldg  xDorm = (v0,Geo  v0,Amy )t + 1 2 2 aGeo t . If we substitute the given values and use the quadratic formula to solve for t, then
1 95 m = (2.8 m/s)t + 2 (.075 m/s )t2 , 2
Problem
58. You're speeding at 85 km/h when you notice that you're only 10 m behind the car in front of you,
CHAPTER 2 which is moving at the legal speed limit of 60 km/h. You slam on your brakes, and your car decelerates at 4.2 m/s2 . Assuming the car in front of you continues at constant speed, will you collide? If so, at what relative speed? If not, what will be the distance between the cars at their closest approach?
21
Solution
See the solution to the next problem.
Problem
59. Repeat the preceding problem, now assuming your initial speed is 95 km/h.
Solution
The position as a function of time for either car, moving with constant acceleration, is given by Equation 210. Let us choose our origin t = 0 and x = 0 at the time and place the speeding driver in car number one notices car number two in front and applies the brakes, with the direction of initial motion positive. Then x10 = 0, x20 = 10 m, v10 > v20 = 2 60 km/h = 16.7 m/s, a1 = 4.2 m/s , and a2 = 0. The position of the cars is x1 (t) = v10 t + 1 a1 t2 and x2 (t) = 2 x20 + v20 t, valid for 0 t t , where t is the time for which the accelerations remain constant. (Thus, t is either the time the cars collide, if this happens, or the time when car number one stops decelerating.) The distance between the cars is x21 (t) = x2 (t)  x1 (t). The condition for a collision is that the quadratic equation x21 (t) = 0 have a real root (in which case the smaller root is t ), and the condition for no collision is that this equation have no real roots. The solution of the equation x21 (t) = 0 =  1 a1 t2  2 (v10  v20 )t + x20 follows from the quadratic formula, t = [(v10  v20 ) (v10  v20 )2  2 a1  x20 ]/ a1 . (Since a1 is negative, we wrote it explicitly as a1 =  a1 .) Thus, if (v10  v20 )2 2 a1  x20 , there is a collision at time t = [(v10  v20 )  (v10  v20 )2  2 a1  x20 ]/ a1 , from which the relative velocity at collision, v1 (t )  v20 , can be calculated. On the other hand, if (v10  v20 )2 < 2 a1  x20 , there is no collision, and the minimum distance x21 can be found by setting the derivative of x21 (t) equal to zero, or by physical reasoning. When v10 = 95 km/h = 26.4 m/s, (v10  v20 )2 = (26.4 m/s  16.7 m/s)2 = 94.5 m2 /s2 > 2 2(4.2 m/s )(10 m) = 84 m2 /s2 , so there is a collision at t = (9.72 m/s  10.5 m2 /s2 )/(4.2 m/s2 ) = 1.54 s. The relative speed at collision is v1 (t )  v20 = 2 v10  v20  a1  t = 9.72 m/s  (4.2 m/s )(1.54 s) = 3.24 m/s = 11.7 km/h, where we used Equation 27 for
the velocities. When v10 = 85 km/h, (v10  v20 )2 = 25 km/h)2 = 48.2 m2 /s2 < 2 a1  x20 = 84 m2 /s2 , and there is no collision. The relative distance is the quadratic 1 x21 (t) = 2 a1  t2  (v10  v20 )t + x20 . One way to obtain the distance of closest approach is to minimize this function of time. Setting the derivative equal to zero gives us dx21 /dt = a1  t  (v10  v20 ) = 0, or 1 tmin = (v10  v20 )/ a1  . Then x21 (tmin ) = 2 a1  t2  min 2 (v10  v20 )tmin + x20 = x20  (v10  v20 ) /2 a1  = 2 10 m  (48.2 m2 /s2 )/2(4.2 m/s ) = 4.26 m. This is in fact a minimum because d2 x21 /dt2 = a1  > 0. Another way to obtain the minimum x21 , without using calculus, relies on purely physical reasoning. As long as the velocity of car number one, v1 (t), is greater than 60 km/h (the velocity of car number two), it is gaining ground on car number two, so the relative distance x21 is decreasing. When v1 (t) falls below 60 km/h, car number one loses ground to car number two and x21 starts increasing. Therefore, the closest approach occurs when v1 (t) = v10  a1  t = v20 = 60 km/h, which gives the same tmin as above.
Section 26: The Constant Acceleration of Gravity Problem
60. You drop a rock into a deep well and 4.4 s later hear the splash. How far down is the water? Neglect the travel time of the sound.
Solution
If we neglect the travel time of the sound, the rock fell for a time t = 4.4 s, from rest, v0 = 0, through a 2 1 1 height y0  y = 2 gt2 = 2 (9.8 m/s )(4.4 s)2 = 94.9 m.
Problem
61. Your friend is sitting 6.5 m above you in a tree branch. How fast should you throw an apple so that it just reaches her?
Solution
Equation 211 describes the vertical motion of the apple, whose acceleration is g (positive upward), if one ignores air resistance, intervening leaves, etc. The difference in height between your friend and you is y  y0 = 6.5 m, v0 is the initial velocity we desire, and v is the velocity of the apple when it reaches your friend. If the apple just reaches her, v = 0. Then
2 v0 = 0 + 2g(y  y0 ), or v0 = 2(9.8 m/s )(6.5 m) = 11.3 m/s. (We chose the positive square root because v0 is upward.) 2
22
CHAPTER 2
Problem
62. A model rocket leaves the ground, heading straight up at 49 m/s. (a) What is its maximum altitude? What are its speed and altitude at (b) 1 s; (c) 4 s; (d) 7 s?
Problem
65. Space pirates kidnap an earthling and hold him imprisoned on one of the planets of the solar system. With nothing else to do, the prisoner amuses himself by dropping his watch from eye level (170 cm) to the floor. He observes that the watch takes 0.95 s to fall. On what planet is he being held? Hint: Consult Appendix E.
Solution
(a) At its maximum altitude, the rocket's vertical 2 speed is instantaneously zero, so 0 = v0  2g 2 2 (ymax  y0 ), or ymax  y0 = (49 m/s) /2(9.8 m/s ) = 123 m. (b) At any time t, the rocket's velocity and 1 altitude are v = v0  gt, and y  y0 = v0 t  2 gt2 . 2 When t = 1 s, v = 49 m/s  (9.8 m/s )(1 s) = 2 39.2 m/s, and y  y0 = (49 m/s)(1 s)  1 (9.8 m/s ) 2 2 (1 s) = 44.1 m. (Note: The altitude is the height above the ground, y0 , and v is positive upward.) 2 (c) v = 49 m/s  (9.8 m/s )(4 s) = 9.8 m/s, y  y0 = 2 1 (49 m/s)(4 s)  2 (9.8 m/s )(4 s)2 = 118 m. (d) v = 19.6 m/s, y  y0 = 103 m.
Solution
The planet's surface gravity can be found, since 1.7 m = 1 g(0.95 s)2 , or g = 3.77 m/s2 . This is closest 2 to the value listed for Mars, in Appendix E.
Problem
66. The Mars Pathfinder spacecraft landed in 1997 and deployed a robot Rover that explored the Martian surface. Pathfinder's landing was cushioned by airbags, and the spacecraft bounced 12 m vertically after its first impact. Assuming no loss of speed at contact with the Martian surface, what was Pathfinder's impact speed?
Problem
63. A foul ball leaves the bat going straight upward at 23 m/s. (a) How high does it rise? (b) How long is it in the air? Neglect the distance between the bat and the ground.
Solution
Equation 211 (with positive up and a = gMars = 2 3.74 m/s ) can be used to describe the vertical motion of the Pathfinder spacecraft. After rebounding with vertical speed v0 from the surface, the spacecraft would attain a maximum height of y  y0 = 12 m when its vertical speed was instantaneously zero, v = 0. Then v0 = 2(y  y0 )gMars = 2(12 m)(3.74 m/s2 ) = 9.47 m/s2 .
Solution
2 (a) At the maximum height, v 2 = 0 = v0  2g 2 (ymax  y0 ), so ymax  y0 = v0 /2g = (23 m/s)2 2 2(9.8 m/s ) = 27.0 m. (b) If we neglect the distance between the bat and the ground (and assume that the foul ball is not caught), the flight of the ball lasts until it falls back to its initial height. Then y  y0 = 0 = 1 v0 t  2 gt2 , or t = 2(23 m/s)/(9.8 m/s2 ) = 4.69 s.
Problem
67. A falling object travels onefourth of its total distance in the last second of its fall. From what height was it dropped?
Problem
64. A Frisbee is lodged in a tree branch, 6.5 m above the ground. A rock thrown from below must be going at least 3 m/s to dislodge the Frisbee. How fast much such a rock be thrown upward, if it leaves the thrower's hand 1.3 m above the ground?
Solution
The total distance traveled by a falling object in a time t is given by Equation 210, with a = g and v0 = 0 1 (the meaning of dropped). Thus y0  y(t) = 2 gt2 . The distance fallen during the last second (an interval from 1 t  1 s to t) is y(t  1 s)  y(t) = 2 gt2  1 g(t  1 s)2 . 2 The latter is onefourth of the former when (cancel off 1 1 the common factors of 2 g) 4 t2 = t2  (t  1 s)2 . Then t  1 s = 3 t, or t = 1 s/(1  3 ) = 7.46 s. (We 4 4 discarded the negative square root because t is obviously greater than 1 s.) Substituting this value of t into the equation for the total distance fallen, we find 1 y0  y(t) = 2 (9.8 m/s2 )(7.46 s)2 = 273 m. (In a real
Solution
When it hits the Frisbee, the rock's velocity and height are v = 3 m/s and y = 6.5 m, while its initial velocity and height are v0 and y0 = 1.3 m. Since 2 v 2 = v0  2g(y  y0 ), we can solve for v0 : v0 =
2
(3 m/s)2 + 2(9.8 m/s )(6.5 m  1.3 m) = 10.5 m/s. (Note: v0 > 0 since the rock must be thrown upwards.)
CHAPTER 2 fall from this height, air resistance should be considered.)
23
Problem
68. The defenders of a castle throw rocks down on their attackers from a 15mhigh wall. If the rocks are thrown with an initial speed of 10 m/s, how much faster are they moving when they hit the ground than if they were simply dropped?
off the platform as the first passes it on the way down. (a) What are their speeds as they hit the water? (b) Which hits the water first and by how much?
Solution
(a) Take t = 0 when the second diver steps from the platform at y0 = 3 m (positive up). The first diver has an "initial" velocity of v0 = 1.80 m/s at t = 0. Then 2 2 2 v1 = v0  2g(y  y0 ), and v2 = 2g(y  y0 ). At the water's surface, y = 0, so v1 = (1.8 m/s)2  2(9.8 m/s )(3 m) = 7.88 m/s, and v2 =  2(9.8 m/s )(3 m) = 7.67 m/s. (b) The divers hit at times we can calculate from Equation 29 for each one: t1 = 2y0 /(v0 + v1 ) = 2(3 m)/(1.8 m/s  7.88 m/s) = 0.620 s; t2 = 2y0 /v2 = 2(3 m)/(7.67 m/s) = 0.782 s. The first diver hits about 162 ms before the second.
2 2
Solution
A stone dropped from a height h = y0  y achieves a speed v = 2g(y  y0 ) = 2gh, while one thrown downward with initial speed v0 , attains a speed v = 2 v0 + 2gh. For h = 15 m and v0 = 10 m/s, the difference in speed is (10 m/s)2 + 2(9.8 m/s )(15 m)  2(9.8 m/s )(15 m) = (19.8  17.1) m/s = 2.70 m/s. (This is the same if the rocks are thrown upward with v0 = 10 m/s, but then the attackers would have more time to get out of the way.)
2 2
Problem
71. A balloon is rising at 10 m/s when its passenger throws a ball straight up at 12 m/s. How much later does the passenger catch the ball?
Problem
69. A kingfisher is 30 m above a lake when it accidentally drops the fish it is carrying. A second kingfisher 5 m above the first dives toward the falling fish. What initial speed should it have if it is to reach the fish before the fish hits the water?
Solution
The initial (positive upward) velocity of the ball is 12 m/s relative to the passenger who throws it. Because the passenger is moving upward with constant velocity of 10 m/s, the initial velocity of the ball relative to the ground is 22 m/s. Assuming the ball is acted upon only by gravity (after being thrown at t = 0), we can write its vertical position as yB (t) = 1 y0 + (22 m/s)t 2 gt2 . The balloon carrying the passenger is acted upon by the buoyant force of the air, in addition to gravity, so that it ascends with constant velocity (see Section 183). Thus, the vertical position of the passenger (in the same coordinate system used for the ball) is yP (t) = y0 + (10 m/s)t. The passenger catches the ball when yB (t) = yP (t) for 1 t > 0. This implies y0 + (22 m/s)t  2 gt2 = 2 y0 + (10 m/s)t, or t = 2(12 m/s)/(9.8 m/s ) = 2.45 s. (Because the balloon is moving with constant velocity, a coordinate system attached to the passenger, yP = 0, is an inertial frame (see Section 35) in which the 1 ball's position is yB = (12 m/s)t  2 gt2 . Setting yB = yP gives one the same time of flight.)
Solution
We are concerned with just the vertical motion of the fish and bird, which we describe with the yaxis positive upward and origin at the water's surface. If the fish is dropped at t = 0, its position is yFish = 30 m  (4.9 m/s2 )t2 , where "drop" means v0,Fish = 0, 1 and we substituted the standard value for 2 g. Suppose the second bird starts its dive at t1 , with an initial velocity v0 . Its position is yBird = 35 m + v0 (t  t1 )  (4.9 m/s2 )(t  t1 )2 , where we assume the dive is a freefall. The bird catches the fish before either hits the water if yFish 0 when yBird = yFish . If the dive started without delay, t1 = 0, and the equation yBird = 2 2 yFish = 30 m  (4.9 m/s )t2 = 35 m + v0 t  (4.9 m/s )t2 gives v0 t = 5 m. When this is substituted into the 2 inequality yFish 0, one obtains 30 m  (4.9 m/s ) (5 m/v0 )2 0, or v0  2.02 m/s. This is the minimum downward speed necessary for the bird to catch the fish.
Problem
72. A conveyer belt moves horizontally at 80 cm/s, carrying empty shoe boxes. Every 3 s, a pair of shoes is dropped from a chute 1.7 m above the belt. (a) How far apart should the boxes be
Problem
70. Two divers jump from a 3.00m platform. One jumps upward at 1.80 m/s, and the second steps
24
CHAPTER 2 spaced? (b) At the instant a pair of shoes drops, where should a box be in relation to a point directly below the chute? 5 min + 2.8 s) = 0.694 m/min = 1.16 cm/s. (b) With origin at the bottom of the spout and the positive direction downward, the total displacement of the spider is x = +0.41 m, so its average velocity is 0.41 m/11.2 min = 3.65 cm/min = 0.609 mm/s.
Solution
(a) Since each pair of shoes takes the same amount of time to fall, a pair lands in a box every 3 s. The distance between boxes must be (80 cm/s)(3 s) = 2.4 m in order that each pair be boxed. (b) A pair of shoes falls in a time t, given by 1.7 m = 1 (9.8 m/s2 )t2 , 2 or t = 0.589 s. During this time, the distance a box moves is (80 cm/s)(.589 s) = 47.1 cm, which is how far the center of a box must be placed (in the opposite direction to the motion of the conveyer belt) from the point directly below the chute, at the instant a pair of shoes drops.
Problem
75. A skier starts from rest, and heads downslope with a constant acceleration of 1.9 m/s2 . How long does it take her to go 20 m, and what is her speed at that point?
Solution
The equations for linear motion with constant acceleration are summarized in Table 21. Since the initial velocity is zero, x(t)  x0 = 1 at2 , and the time 2 to travel 20 m is t = 2(20 m)/(1.9 m/s ) = 4.59 s.
2 2
Paired Problems Problem
73. You drive 14 km to the next town, maintaining a speed of 50 km/h except for a stop lasting 4.1 min at a red light. You shop for 20 min, then head back toward your starting point at a steady 70 km/h. You stop at a gas station 4.4 km from the town. What are (a) your average speed and (b) the magnitude of your average velocity between your starting point and the gas station?
The velocity at this time is v = at = (1.9 m/s ) (4.59 s) = 8.72 m/s.
Problem
76. Landing on the moon, a spacecraft fires its retrorockets and comes to a complete stop just 12 m above the lunar surface. It then drops freely to the surface. How long does it take to fall, and what is its impact speed? (Consult Appendix E.)
Solution
(a) The average speed is the total distance traveled, 14 km + 4.4 km, divided by the total time spent, 14 km/(50 km/h) + 4.1 min + 20 min + 4.4 km (70 km/h). Thus, the average speed is 18.4 km 0.745 h = 24.7 km/h. (b) The average velocity is the total displacement divided by the total time. If we take the origin at the starting point and the positive direction toward the next town, x = 14 km  4.4 km, while t is the same as in part (a). Thus, the average velocity is 9.6 km/0.745 h = 12.9 km/h.
Solution
To drop 12 m from rest on the moon (lunar surface 2 gravity = 1.62 m/s ) takes time t = 2(y0  y)/g = 2(12 m)/(1.62 m/s ) = 3.85 s. The velocity at impact is v = gt = (1.62 m/s )(3.85 s) = 6.24 m/s, and the speed is the magnitude of this.
2 2
Problem
77. A frustrated student drops a book out of his dormitory window, releasing it from rest. After falling 2.3 m, it passes the top of a 1.5m high window on a lower floor. How long does it take to cross the window?
Problem
74. The itsybitsy spider climbed 3.7 m up the water spout, starting at the bottom, in 6.2 minutes. She paused at the top for a 5.0min rest. Then down came the rain, and washed the spider outall the way to the ground, 0.41 m below the bottom of the spout, in 2.8 s. What were the spider's (a) average speed and (b) magnitude of her average velocity for the entire adventure?
Solution
Equation 210 gives the distance fallen by the book 1 when dropped (v0 = 0) at time t = 0: y0  y(t) = 2 gt2 . The book passes the top of the lowerfloor window at time t1 , given by y0  y(t1 ) = 2.3 m = 1 gt2 , or t1 = 2 1
Solution
(a) The average speed = (total distance) (total time) = (3.7 m + 3.7 m + 0.41 m)/(6.2 min +
2(2.3 m)/(9.8 m/s2 ) = 0.685 s. It passes the bottom of the window at time t2 , given by y0  y(t2 ) = 2.3 m + 1.5 m = 1 gt2 , or t2 = 0.881 s. 2 2 The time to cross the window is t2  t1 = 0.196 s.
CHAPTER 2
25
Problem
78. Launched from the ground, a rocket accelerates vertically upward at 4.6 m/s2 . It passes through a band of clouds 5.3 km thick, extending upward from an altitude of 1.9 km. How long is it in the clouds?
Solution
Suppose the orange is released at t = 0 when the parachutist is at an altitude y0 (positive upward). Then yorange (t) = y0 + 8.4 m  (2.2 m/s)t  1 gt2 , and 2 (since the parachutist's velocity is constant) ypara (t) = y0  (11 m/s)t. The orange and parachutist are at the same altitude (ignoring the possible influence of the intervening parachute) when yorange (t) = ypara (t) and t > 0, or 1 gt2  (8.8 m/s)t 2 8.4 m = 0. Using the quadratic formula to solve for the positive root, we find t = [8.8 m/s+ (8.8 m/s)2 + 2(9.8 m/s )(8.4 m)]/(9.8 m/s ) = 2.49 s. The relative speed of the orange and parachutist, when they meet, is 2 vorange (t)  vpara (t) = (2.2 m/s)  (9.8 m/s ) (2.49 s)  (11 m/s) = 15.6 m/s. (In Chapter 3, we will see that in order for the orange to start straight downward, the pilot must throw it downward and backward, to cancel the plane's velocity.)
2 2
Solution
The altitude of the rocket, launched from rest (v0 = 0) at ground zero (y0 = 0 and t = 0), is y(t) = 1 at2 . The 2 rocket enters the clouds at time t1 , given by y(t1 ) = 2 1 1.9 km = 2 (4.6 m/s )t2 , or t1 = 28.7 s. It leaves the 1 clouds at time t2 , given by y(t2 ) = 1.9 km + 5.3 km = 2 2 1 2 (4.6 m/s )t2 , or t2 = 56.0 s. The time spent in the clouds is t2  t1 = 27.2 s.
Problem
79. A subway train is traveling at 80 km/h when it approaches a slower train 50 m ahead traveling in the same direction at 25 km/h. If the faster train begins decelerating at 2.1 m/s2 , while the slower train continues at constant speed, how soon and at what relative speed will they collide?
Problem
81. You toss a hammer over the 3.7mhigh wall of a construction site, starting your throw at a height of 1.2 m above the sidewalk. On the other side of the wall, the hammer falls to the bottom of an excavation 7.9 m below the sidewalk (see Fig. 223). (a) What is the minimum speed at which you must throw the hammer for it to clear the wall? (b) Assuming it's thrown with the speed given in part (a), when will it hit the bottom of the excavation?
Solution
Take the origin x = 0 and t = 0 at the point where the first train begins decelerating, with positive x in the direction of motion. Equation 210 gives the instantaneous position of each train, with x10 = 0, 2 v10 = 80 km/h, a1 = 2.1 m/s , x20 = 50 m, v20 = 1 25 km/h, and a2 = 0 given. Thus x1 (t) = v10 t + 2 a1 t2 , and x2 (t) = x20 + v20 t. The trains collide at the first time that x1 = x2 , or when x20  (v10  v20 )t  1 a1 t2 = 2 0. Using the quadratic formula to solve for the smaller root, we find t = [(v10  v20 )  (v10  v20 )2 + 2a1 x20 ]/(a1 ) = [(55 m/3.6 s)  (55 m/3.6 s)2 + 2(2.1 m/s )(50 m)]/(2.1 m/s ) = 4.97 s. The velocity of the first train at the time of the collision is v1 = v10 + a1 t = (80 km/h)  (2.1 m/s2 ) (4.97 s)(3.6 km/h/m/s) = 42.4 km/h. Therefore, the relative speed at impact is v1  v2 = 42.4 km/h  25 km/h = 17.4 km/h.
2 2
Solution
If we consider just the vertical motion of the hammer, and ignore air resistance, etc., the equations in Table 21 (with y replacing x and a = g) apply. (a) Equation 211 evaluated at the highest point of the 2 hammer's trajectory gives v0 = 2g(ytop  y0 ), since the instantaneous vertical velocity at the highest point is zero. In order to clear the top of the wall, ytop  y0 (3.7  1.2) m = 2.5 m (from Fig. 223), so v0 2(9.8 m/s )(2.5 m) = 7.00 m/s. (b) From Equation 210, the hammer hits bottom when 1 y0  ybot = 1.2 m + 7.9 m = 2 gt2  (7 m/s)t, where the displacement y0  ybot is shown in Fig. 223, and we used the minimum initial velocity from part (a). The time in this equation is measured from t = 0 when the hammer is thrown; therefore t > 0 at the bottom. The positive root of this quadratic equation is t = [(7 m/s) + (7 m/s)2 + 2(9.8 m/s )(9.1 m) ] (9.8 m/s) = 2.25 s, which is the time of flight to the
2 2
Problem
80. A parachutist is drifting vertically downward at a constant 11 m/s. An airplane passes a mere 8.4 m directly above the parachutist, and the pilot throws an orange straight downward at 2.2 m/s. How much later do the orange and parachutist meet, and what is their relative speed?
26
CHAPTER 2
Supplementary Problems Problem
3.7 m 1.2 m
83. A car accelerates away from a red light at 2.5 m/s2 until its speed reaches 10 m/s. It travels at that speed for 8.0 s, then brakes to a stop at the next red light with deceleration 4.0 m/s2 . What is the distance between lights?
Solution
The distance covered by the car, accelerating from rest (v0 = 0) to a speed v = 10 m/s, away from the first 2 stoplight, is (x  x0 )accel. = (v 2  v0 )/2a = 2 2 (10 m/s) /2(2.5 m/s ) = 20 m (see Equation 211). Traveling at a constant speed of v = 10 m/s for the next t = 8.0 s, the car covers a distance of (x  x0 )no accel. = vt = (10 m/s)(8.0 s) = 80 m (see Equation 210 with a = 0). Finally, the distance 2 covered decelerating (a = 4.0 m/s ) from speed v0 = 10 m/s to rest (v = 0), at the second stoplight, is 2 (x  x0 )decel. = (0  (10 m/s)2 )/2(4.0 m/s ) = 12.5 m (see Equation 211 again). The total distance covered between the stoplights is the sum of these three distances, or approximately 113 m. (Note: we redefined t = 0 and x0 for each of the segments of the car's motion.)
7.9 m
figure 223 Problem 81.
bottom.
Problem
82. You toss a book into your dorm room, just clearing a windowsill 4.2 m above the ground. (a) If the book leaves your hand 1.5 m above the ground, how fast must it be going to clear the sill? (b) How long after it leaves your hand will it hit the floor, 0.87 m below the windowsill?
Problem
84. Consider an object traversing a distance , part of the way at speed v1 and the rest of the way at speed v2 . Find expressions for the average speeds (a) when the object moves at each of the two speeds for half the total time and (b) when it moves at each of the two speeds for half the distance.
Solution
Take y = 0 at ground level in the coordinate system used in the previous problem. (a) Since the vertical instantaneous velocity is zero at the top of the 2 trajectory, v0 = 2g(ytop  y0 ). For the book to clear the sill, ytop ysill ; therefore v0 2g(ysill  y0 ) =
Solution
In either case, the average speed is the total distance divided by the total time, or v = /t. (a) The total distance is the sum of the distances covered at each speed, = 1 + 2 = v1 1 t + v2 1 t = 1 (v1 + v2 )t, so 2 2 2 1 va = 2 (v1 + v2 ). (b) The total time is the sum of the 1 times traveled at each speed, t = t1 + t2 = 2 /v1 + 1 1 2 /v2 = 2 (v1 + v2 )/v1 v2 , so vb = 2v1 v2 /(v1 + v2 ). (In general, v is the timeweighted average of the separate speeds.)
2(9.8 m/s2 )(4.2 m  1.5 m) = 7.27 m/s. (Neglect of air resistance is a fair approximation, provided the book doesn't open during its flight.) (b) To find the time of flight to the floor, we need to find the larger root of Equation 213 (since the book passes the height of the floor on the way up first), with yflr  y0 = (4.2  0.87  1.5) m = 1.83 m, and v0 = 7.27 m/s. This is t = [(7.27 m/s)+ (7.27 m/s)2  2(9.8 m/s )(1.83 m)]/(9.8 m/s ) = 1.16 s.
2 2
Problem
85. You see the traffic light ahead of you is about to turn from red to green, so you slow to a steady speed of 10 km/h and cruise to the light, reaching it just as it turns green. You accelerate to 60 km/h in the next 12 s, then maintain constant
CHAPTER 2 speed. At the light, you pass a Porsche that has stopped. Just as you pass (and the light turns green) the Porsche begins accelerating, reaching 65 km/h in 6.9 s, then maintaining constant speed. (a) Plot the motions of both cars on a graph showing the 10s period after the light turns green. (b) How long after the light turns green does the Porsche pass you? (c) How far are you from the light when the Porsche passes you? = 1 2 65 km/h 6.9 s
27
(3.81 s)2 = 19.0 m
from the green light.
Problem
86. In the accident of Problem 53, calculate the relative speed with which a passenger not wearing a seat belt collides with the dashboard. Assume the passenger undergoes no deceleration before striking the dashboard, and that the passenger is initially 1 m from the dashboard.
Solution
(a) Let the stoplight be at x = 0 and turn green at t = 0. Then (10 km/h)t + 1 60 km/h  10 km/h t2 , 2 12 s xYou (t) = 0 t 12 116.7 m + (60 km/h)(t  12 s), t 12 s. xPorsche (t) = 1 2 65 km/h 6.9 s t2 , 0 t 6.9 s 62.29 m + (65 km/h)(t  6.9 s), t 6.9 s.
Solution
The velocity of the dashboard, which decreases from 2 v0 to zero with acceleration a = v0 /2(0.8 m) from Problem 53, is vdash = v0 + at . It comes to rest in a time ts = v0 /a = 1.6 m/v0 after the start of the accident and remains at rest thereafter, vdash = 0 for t ts . Without a seat belt, the passenger continues to move with velocity v0 . In time ts , he or she has moved a distance v0 ts = 1.6 m and so is still 1 m + 0.8 m  1.6 m = 0.2 m away from the dashboard. The relative velocity of the passenger and the dashboard is the full 88 km/h when the secondary collision occurs, with probably lethal consequences. (Note: we did not need the numerical value of ts , which is (1.6 m)/(24.4 m/s) = 65.5 ms, to answer the question.)
Before plotting x versus t, we first calculate that xYou (6.9 s) = 46.72 m, xYou (12 s) = 116.7 m, xPorsche (6.9 s) = 62.29 m, xPorsche (12 s) = 154.4 m. (b) Evidentally, the Porsche passes you before 6.9 s,
x 150 m (First twelve sec)
Problem
87. The position of a particle as a function of time is given by x = x0 sin t, where x0 and are constants. (a) Take derivatives to find expressions for the velocity and acceleration. (b) What are the maximum values of velocity and acceleration? Hint: Consult the table of derivatives in Appendix A.
100 m Porsche 50 m You t
Solution
12 s
0s
4s
8s
Problem 85 Solution. while both cars are accelerating, so xYou = xPorsche implies: (10 km/h)t + 1 2 50 km/h 12 s t2 = 1 65 km/h 2 t , 2 6.9 s or t = 3.81 s.
(a) For x(t) = x0 sin t, dx/dt = v(t) = x0 cos t and dv/dt = d2 x/dt2 = a(t) =  2 x0 sin t =  2 x(t). (b) Since the maximum value of the sine or cosine functions is 1, vmax = x0 and amax = 2 x0 . (The motion described by x(t) is called simple harmonic motion; see Chapter 15.)
Problem
88. Ice skaters, ballet dancers, and basketball players executing vertical leaps often give the illusion of "hanging" almost motionless near the top of the leap. To see why this is, consider a leap that takes an athlete up a vertical distance h. Of the total
(c) When the cars pass, both are at the same position: xYou (3.81 s) = xPorsche (3.81 s)
28
CHAPTER 2 time spent in the air, what fraction is spent in the upper half (i.e., at y > 1 h)? 2 The total time is the sum of these, t = t1 + d/vs = t1 + gt2 /2vs . The quadratic formula yields 1 t1 = (vs /g)[ 1 + 2gt/vs  1], from which d as a function of t can be obtained simply by squaring. The expression is complicated, so the approximation stated, that the speed of impact, gt1 , is much smaller than vs can be considered. Since gt = gt1 + (gt1 )2 /2vs , this approximation is equivalent to assuming gt vs . Therefore, one can expand the square root in t2 in 1 powers of gt/vs , using the binomial theorem from Appendix A, written in the form x 1 x (1+x)1/2 = 1+  2 2! 2 Then t2 = 1 = vs g vs g + 1 2
2 2
Solution
We assume that the height of the leaper (actually, her center of mass, as in Figure 1011) is given by Equation 210, with vertical position y measured positive upward and a = g. Then y(t)  y0 = 1 v0 t  2 gt2 . The quadratic formula gives two times when the leaper passes a particular height, 2 t = [v0 v0  2g(y  y0 )]/g, the smaller, t , going up and the larger, t+ , going down. The time spent above that height is just t(y) = t+  t = 2 (2/g) v0  2g(y  y0 ). The initial velocity for an upward leap of height h is v0 = 2gh (see Problem 61, for example), so t(y) = 2 2/g h  (y  y0 ). The total time spent in the air is the time spent above the ground, y  y0 = 0, or t(y0 ) = 2 2/g h, and the time spent in the upper half, above y  y0 = 1 h, is 2 1/2 = 70.7% of this.
+
3 x 3! 2
3

35 x 4! 2
4
+
1+
2
2gt 2gt 2 1+ +1 vs vs 2gt gt 1 2 1+  vs vs 2 5 8 gt vs gt vs
2 4
2+ gt vs
3
gt vs
2
Problem
89. A faucet leaks water at the rate of 15 drops per second. At the instant one drop leaves the faucet, another strikes the sink below, and two additional drops are in between on the way down. How far is it from the faucet to the sink bottom?

+  .
= t2 1 
gt 5 + vs 4
Solution
Drops appear at the faucet every 1/15 of a second. Under the conditions stated (one drop at the faucet, two in the air, and one striking the sink), the time of fall for one drop is 3(1/15) s = (1/5) s, so the distance fallen (starting from rest) is y0  y = 1 gt2 = 2 2 1 (9.8 m/s )(0.2 s)2 = 19.6 cm. 2
1 Multiplication by 2 g gives the desired expression for d. (There are faster, but sloppier ways to solve this problem, but one doesn't get the correct coefficients for the higher order terms, and the full power of the binomial expansion isn't appreciated.)
Problem
91. A student is staring idly out her dormitory window when she sees a water balloon fall past. If the balloon takes 0.22 s to cross the 130cmhigh window, from what height above the top of the window was it dropped?
Problem
90. The depth of a well is such that an object dropped into the well hits the water going far slower than the speed of sound. Use the binomial theorem (see Appendix A) to show that, under these conditions, the depth of the well is given approximately by d= gt 1 2 gt 1  2 vs .
Solution
If the balloon was dropped from height y0 at time t = 0, then its height at any later time is 1 y = y0  2 gt2 . When it passes the top of the window, 1 y1 = y0  2 gt2 , and when passing the bottom, 1 1 y2 = y0  2 gt2 . The length of the window is 2 1 1.3 m = y1  y2 = 2 g(t2  t2 ) = 1 g(t2  t1 )(t2 + t1 ). 2 1 2 But t2  t1 = 0.22 s (the time required to cross the 2 window), so t2 + t1 = 2(1.3 m)/(9.8 m/s )(0.22 s) = 1.21 s. Combined with the value of the difference in 1 times, we find that t1 = 2 (1.21 s  0.22 s) = 0.493 s. Finally, the height above the top of the window is 1 y0  y1 = 1 gt2 = 2 (9.8 m/s2 )(0.493 s)2 = 1.19 m. 2 1
where t is the time from when you drop the object until you hear the splash, and vs is the speed of sound.
Solution
The depth of the well is d = 1 gt2 , where t1 is the time 2 1 for free fall from rest (see Equation 210). The travel time for the sound of the splash to reach you is d/vs .
CHAPTER 2
29
Problem
92. A police radar has an effective range of 1.0 km, while a motorist's radar detector has a range of 1.9 km. The motorist is going 110 km/h in a 70 km/h zone when the radar detector beeps. At what rate must the motorist decelerate to avoid a speeding ticket?
down from 110 km/h to 70 km/h. This requires a constant acceleration of
2 a = (v 2  v0 )/2(x  x0 )
or a = (70 km/h)2  (110 km/h)2 2(0.9 km)(3600 s/h)
2
Solution
The speed of radar waves (3105 km/s) is so great compared to the speed of a motor vehicle, we can neglect any motion of the car during the travel times of the radar signals. The motorist has 0.9 km to slow
= 1.11 km/h/s = 0.309 m/s . (The deceleration, or magnitude of a, must be at least 1.11 km/h/s to avoid getting a ticket.)
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CHAPTER 14STATIC EQUILIBRIUMProblem2. A body is subject to three forces: F1 = 2^ + 2^ N, i applied at the point x = 2 m, y = 0 m; F2 = 2^  i 3^ N, applied at x = 1 m, y = 0; and F3 = 1^ N, applied at x = 7 m, y = 1 m. (a) Show explicitly that the n
Berkeley  PHYSICS 8A  69054
PART 2OSCILLATIONS, WAVES, AND FLUIDS OSCILLATORY MOTIONcos[(10 s1 )t]. (b) The maximum (positive) velocity occurs at t = 0 if sin = 1 (from Equation 1510), therefore, the motion is described by Equation 159 with A = 2.5 cm, = 5 s1 , and = /2. Since c
UPR Bayamon  ADEM  CONT 3010
DILUTIVE SECURITIES AND EARNINGS PER SHAREMULTIPLE CHOICEDilutive SecuritiesConceptual1. Convertible bonds a. have priority over other indebtedness. b. are usually secured by a first or second mortgage. c. pay interest only in the event earnings are suf
UPR Bayamon  ADEM  CONT 3010
Chapter 3 Test BankAN INTRODUCTION TO CONSOLIDATED FINANCIAL STATEMENTSMultiple Choice Questions LO1 1. What method must be used if FASB consolidation of a 70% owned subsidiary a. b. c. d. LO1 2. Cost method Liquidation value Market value *Equity method
UPR Bayamon  ADEM  CONT 3010
Chapter 4 Test BankCONSOLIDATION TECHNIQUES AND PROCEDURESMultiple Choice Questions LO1 1. Which of the following will be debited account when the equity method is used? a. b. c. d. LO1 2. investee net losses investee net profits investee declaration of
N.C. State  MEA  130
Chapter 3 HW 1.Which of the following is THE major controller of temperature? Student Response Value Differential 0% heating of land and water Ocean currents 0% Geographic 0% position Latitude 100% 0% Correct Answer FeedbackA. B. C. D.E. Cloud cover Sc
N.C. State  MEA  130
Chapter 3 HW 1.Which of the following is THE major controller of temperature? Student Response Value Differential 0% heating of land and water Ocean currents 0% Geographic 0% position Latitude 100% 0% Correct Answer FeedbackA. B. C. D.E. Cloud cover Sc
N.C. State  MEA  130
Chapter 4 HW 1.The process of [] that occurs when water goes from the [] phase to the [] phase [] energy, thereby [] the atmosphere. Student Response The process of [sublimation] that occurs when water goes from the [solid] phase to the [vapor] phas
University of Phoenix  GEN  105
Axia College MaterialAppendix C Distance Learning VocabularyAsynchronous Communication Communication between parties who are not together at the same time Examples might include email, fax, threaded discussions, postal mail, etc. Attendance A recordke
University of Phoenix  SCIENCE  SCI/245
CheckPointStudyofEarth TheEarthconsistsoffourdifferentsubsystems.Thesesubsystemsincludethe lithosphere,biosphere,atmosphere,andhydrosphere.Thelithosphereisthe rocky,outermostlayeroftheEarth.Thebiosphereconsistsofalllivingor recentlydeadorganismsonEarth.Th
UAA  PSY  260
Version 03aug10PSY A260 (601): PSY Statistics for PsychologyFall 2010 (T/R 10:0011:15)Instructor: Teaching Assistant:Dr. John PetraitisSSB 221b, 7861651 j.petraitis@uaa.alaska.edu Open Office hours: T/R 9:159:45 Other Office hours: By appointment
Université du Québec à Montréal  JUR  6854
Federal CourtCour fdraleDate : 20101014 Dossier : T147609 Rfrence : 2010 CF 1011 [TRADUCTION FRANAISE CERTIFIE, NON RVISE] Ottawa (Ontario), le 14 octobre 2010 En prsence de monsieur le juge Phelan2010 CF 1011 (CanLII)ENTRE : AMAZON.COM, INC. appela
Cal Poly Pomona  PHL  202
Phil 202Questionnaire: Critical Think ThisHerrDirections: For each statement below, write the numeric response that corresponds to your level of agreement. Your response options are below. (You can also respond with ?) 4 I strongly agree 3 I agree 2 Im
University of Maryland Baltimore  FINANCE  FINC330
TextBook:Options,FuturesandOtherDerivatives,7thEditionChapter1:Problem23:Supposetheyenexchangerate(yenperdollar)atmaturityoftheICONisST. ThepayofffromICON: 1000,ifST>169 10001000(169/ST1),if84.5 ST 169 0,ifST<84.5ThepayofffromICONisthepayofffrom a) Ar
Texas A&M University, Corpus Christi  PHYS  1402
Create assignment, 99552, Homework 18, Apr 01 at 8:48 pm This printout should have 23 questions. Multiplechoice questions may continue on the next column or page nd all choices before answering. The due time is Central time. Charge in a Nickel 23:01, tr
University of Toronto  CSC  CSC384
Michigan State University  CEE  CE 221
CENTER OF GRAVITY AND CENTROIDCenter of Gravity, Center of Mass and Centroid andLEARNING OBJECTIVES LEARNING Be able to understand the concepts of Becenter of gravity, center of mass, and centroid centroid Be able to determine the locations of Be thes
Michigan State University  CEE  CE 221
INTRODUCTION AND REVIEW REVIEWME 221 STATICS MECE 221 StaticsME 222 Mechanics of Deformable SolidsCE 305 CE 312 CE 321 CE 337ME 423 ME 425 ME 426 ME 477 ME 495GENERAL REVIEW GENERALTrigonometryGEOMETRY/TRIGONOMETRY GEOMETRY/TRIGONOMETRYa b a bb
Michigan State University  CEE  CE 221
MATRIX DEFINITIONA matrix is defined as a rectangular array matrix of quantities arranged in rows and columns. ofa a . a a 12 13 1n 11 a 21 a 22 a 23 . a 2n a a a . a [ A] = 31 32 33 3n . . . . . a a . a a m2 m3 mn m1 Aij is the quantity A of the ith ro
Michigan State University  CEE  CE 221
NEWTONS LAWS NEWTONSNEWTONS FIRST LAW OF MOTION NEWTONSAn object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force.NEWTONS SECIND LAW OF
Michigan State University  CEE  CE 221
Units of Measurement UnitsLEARNING OBJECTIVES LEARNING Be able to identify the basic quantities. Be Be able to differentiate between the SI andFPS systems of unit. FPS Be Be able to convert quantities from one unit system to another. system Be able to d
Michigan State University  CEE  CE 221
Force Vectors ForceScalars and VectorsLEARNING OBJECTIVES LEARNING Be able to differentiate between Scalar Beand Vector quantities. and Be able to perform vector operations. Be able to resolve forces into their Be respective components. respectivePRE
Michigan State University  CEE  CE 221
FORCE VECTORS FORCELEARNING OBJECTIVES LEARNING Be able to resolve each force into its Berectangular components Fx and Fy. rectangular Be able to represent the components of force Be in terms of the Cartesian unit vectors i and j. in Be able to determi
Michigan State University  CEE  CE 221
FORCE VECTORS FORCECartesian Vectors Cartesian (3Dimensional Vectors) (3DimensionalLEARNING OBJECTIVES LEARNING Be able to resolve each force into its Be rectangular components Fx, Fy and Fz. rectangular Be able to represent the components of force B
Michigan State University  CEE  CE 221
FORCE VECTORS FORCEPosition and Force Vectors Position Directed Along a Line DirectedLEARNING OBJECTIVES LEARNING Be able to represent a position Bevector in Cartesian coordinate form, from given geometry. form, Be able to represent a force vector Be
Michigan State University  CEE  CE 221
Force Vectors ForceDot ProductLEARNING OBJECTIVES LEARNING Be able to determine an angle between Betwo vectors. two Be able to determine the projection of a Be vector along a specified line. vectorPREREQUISITE KNOWLEDGE PREREQUISITE Units of measu
Michigan State University  CEE  CE 221
2D EQUILIBRIUM OF A PARTICLE PARTICLELEARNING OBJECTIVES LEARNING Be able to draw a free body diagram (FBD). Be able to apply equations of equilibrium to solve a 2D problem.PREREQUISITE KNOWLEDGE PREREQUISITE Units of measurements. Trigonometry c
Michigan State University  CEE  CE 221
3D EQUILIBRIUM OF A PARTICLELEARNING OBJECTIVES LEARNING Be able to draw a free body Bediagram (FBD). diagram Be able to apply equations of Be equilibrium to solve a 3D problem. problemPREREQUISITE KNOWLEDGE KNOWLEDGE Units of measurements Trigono
Michigan State University  CEE  CE 221
CE 221 Homework 1 & 2 Due Wednesday September 15, 20101.Proof the following trigonometric identity:Tan(2a ) =2.2 tan a 1 tan 2 aProof that sin(a+b) = [sin(a)][cos(b)] + [cos(a)][sin(b)] Proofs that for any triangle, the following ratios are equal (t
Michigan State University  CEE  CE 221
CE 221Homework 3, due Wednesday September 22, 2010 Solve problems 1.15, 2.6, 2.11, 2.33, 2.45, 2.46, 2.52, 2.56, 2.77, and 2.82 in your book.1
Michigan State University  CEE  CE 221
Michigan State University  CEE  CE 221
CE 221Homework 4, due Wednesday September 29, 2010 Solve problems 2.83, 2.89, 2.90, 2.91, 2.96, 2.98, 2.101, 2.112, 2.114, and 2.1151
Michigan State University  CEE  CE 221
Michigan State University  CEE  CE 221
CE 221Homework 5, due Wednesday, October 6, 2010 Solve problems 3.4, 3.10, 3.15, 3.21, 3.37, 3.46, 3.55, 3.64, 3.86 and 3.88 in your book. The red and underlined problem numbers are for those problems to be solved in the recitations.1
Michigan State University  CEE  CE 221
CE 221 Homework 6 Due Wednesday October 13, 2010 Red colored problems to be solved in the recitation. Please add HW 6 problems to the list of problems for sample test 1. Problem 1 10 pointsForce F2 is located in the yz plane at 30o above the y axis. The
Michigan State University  CEE  CE 221
CE 221Homework 7, due Wednesday October 20, 2010 Solve problems 4.20, 4.25, 4.34, 4.51, 4.59, 4.62 (b, c, d), 4.77, and 4.78 your book. The red and underlined problem numbers are for those problems to be solved in the recitations.1
Michigan State University  CEE  CE 221
CE 221Homework 8, due Wednesday October 27, 2010 Solve problems 4.63, 4.64, 4.74, 4.87, 5.5, 5.12, 5.13, 5.14, 5.27, and 5.40 in your book. The red and underlined problem numbers are for those problems to be solved in the recitations.1
Michigan State University  CEE  CE 221
Michigan State University  CEE  CE 221
MOMENT OF A FORCE MOMENTSCALAR AND VECTOR SCALAR FORMULATION FORMULATIONLEARNING OBJECTIVES LEARNING Be able to understand and define moment Be able to determine the moment of a force Bein 2D and 3D cases inPREREQUISITE KNOWLEDGE KNOWLEDGE Units
Michigan State University  CEE  CE 221
FORCE SYSTEM RESULTANTSMOMENT ABOUT AN AXIS MOMENTLEARNING OBJECTIVES LEARNING Be able to determine the moment of a Beforce about a specified axis using scalar and vector methods andPREREQUISITE KNOWLEDGE PREREQUISITE Units of measurements Trigono
Michigan State University  CEE  CE 221
FORCE SYSTEM RESULTANTSMOMENT OF A COUPLE MOMENTLEARNING OBJECTIVES LEARNING Be able to define a couple Be able to determine the moment of a Becouple couplePREREQUISITE KNOWLEDGE PREREQUISITE Units of measurements Trigonometry concepts Vector conc
Michigan State University  CEE  CE 221
FORCE SYSTEM RESULTANTS RESULTANTSEquivalent System EquivalentLEARNING OBJECTIVES LEARNING Be able to find an equivalent forcecouple Besystem for a system of forces and couples systemPREREQUISITE KNOWLEDGE PREREQUISITE Units of measurement Trigon
Michigan State University  CEE  CE 221
FORCE SYSTEM RESULTANTS RESULTANTSReduction of a Simply Distributed Load ReductionLEARNING OBJECTIVES LEARNINGBe able to find an equivalent force for a Be simply distributed load simplyPREREQUISITE KNOWLEDGE PREREQUISITE Units of measurement Units
Michigan State University  CEE  CE 221
RIGID BODY EQUILIBRIUM EQUILIBRIUMRigid Body Equilibrium 2D RigidLEARNING OBJECTIVES LEARNING Be able to recognize twoforce Bemembers members Be able to apply equations of Be equilibrium to solve for unknowns equilibriumPREREQUISITE KNOWLEDGE PRE
Michigan State University  CEE  CE 221
RIGID BODY EQUILIBRIUM 3D EQUILIBRIUMEquilibrium of Rigid Body in 3D EquilibriumLEARNING OBJECTIVES LEARNING Be able to identify support reactions in Be3D and draw a free body diagram 3D Be able to identify an indeterminate Be system system Be abl
Michigan State University  CEE  CE 221
STRUCTURAL ANALYSISMethod Of Joints MethodLEARNING OBJECTIVES LEARNING Be able to define a simple truss Be able to determine the forces in Bemembers of a simple truss using the method of joints the Be able to identify the zeroforce Be members members
Michigan State University  CEE  CE 221
STRUCTURAL ANALYSISMethod of Sections MethodLEARNING OBJECTIVES LEARNING Be able to determine the forces in Bemembers of a simple truss using the method of sectionsPREREQUISITE KNOWLEDGE PREREQUISITE Units of measurement Trigonometry concepts Rect
Michigan State University  CEE  CE 221
STRUCTURAL ANALYSIS ANALYSISFrames and Machines FramesLEARNING OBJECTIVES LEARNING Be able to draw the free body Bediagram of a frame or machine and its members its Be able to determine the forces acting Be at the joints and supports of a frame or mac
Michigan State University  CEE  CE 221
INTERNAL FORCES INTERNALInternal Forces Developed in Structural Members MembersLEARNING OBJECTIVES LEARNINGBe able to use the method of sections for Be determining internal forces in 2D load case determiningPREREQUISITE KNOWLEDGE PREREQUISITEUnits
Michigan State University  CEE  CE 221
Michigan State University Department of Civil and Environmental Engineering CE 221 Statics (3 credits) CESpring 2010 Maximum time allowed is 50 minutes Note:  This is a closed book and notes test. You are prohibited from using stored formulae and/or pro