Problem#1
The 1992 world speed record for a bicycle (human-powered vehicle) was set by Chris Huber. His time through the measured 200 m stretch was a sizzling 6.509 s, at which he commented, “Cogito ergo zoom!” (I think, therefore I go fast!). In 2001, Sam Whittingham beat Huber’s record by 19.0 km/h. What was Whittingham’s time through the 200 m?
Answer:
Huber’s 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
v1 = 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
∆t = ∆x/v1 = 200 m/(36 m/s) = 5.55 s
Problem#2
Two trains, each having a speed of 30 km/h, are headed at each other on the same straight track. A bird that can fly 60 km/h flies off the front of one train when they are 60 km apart and heads directly for the other train. On reaching the other train, the bird flies directly back to the first train, and so forth. (We have no idea why a bird would behave in this way.) What is the total distance the bird travels before the trains collide?
Answer:
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.
Problem#3
Panic escape. Figure 1 shows a general situation in which a stream of people attempt to escape through an exit door that turns out to be locked. The people move toward the door at speed vs = 3.50 m/s, are each d = 0.25 m in depth, and are separated by L = 1.75 m.The arrangement in Fig. 2-21 occurs at time t = 0. (a) At what average rate does the layer of people at the door increase? (b) At what time does the layer’s depth reach 5.0 m? (The answers reveal how quickly such a situation becomes dangerous.)
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| Fig.1 |
Answer:
The amount of time it takes for each person to move a distance L with speed s v is
∆t = 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 = d/∆t = d/(L/vs)
R = dvs/L
R = (0.25 m)(3.50 m/s)/(1.75 m) = 0.5 m/s
(b) The amount of time required to reach a depth of D = 5.0 m is
t = D/R = 5.0 m/(0.5 m/s) = 10 s
Problem#4
In 1 km races, runner 1 on track 1 (with time 2 min, 27.95 s) appears to be faster than runner 2 on track 2 (2 min, 28.15 s). However, length L2 of track 2 might be slightly greater than length L1 of track 1. How large can L2 – L1 be for us still to conclude that runner 1 is faster?
Answer:
Converting to seconds, the running times are t1 = 147.95 s and t2 = 148.15 s, respectively. If the runners were equally fast, then
savg,1 = savg,2
L1/t1 = L2/t2
From this we obtain
L2 – L1 = (t2/t1 – 1)L1
L2 – L1 = (148.15 s/147.95 s – 1) 1 km ≈ 1.4 m
where we set L1 ≈ 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.
Problem#5
To set a speed record in a measured (straight-line) distance d, a race car must be driven first in one direction (in time t1) and then in the opposite direction (in time t2). (a) To eliminate the effects of the wind and obtain the car’s speed vc in a windless situation, should we find the average of d/t1 and d/t2 (method 1) or should we divide d by the average of t1 and t2? (b) What is the fractional difference in the two methods when a steady wind blows along the car’s route and the ratio of the wind speed vw to the car’s jspeed vc is 0.0240?
Answer:
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
veef,1 = vc + vw
and the distance traveled is
d = veef,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
veef,2 = vc – vw
and the distance traveled is
d = veef,2t2 = (vc – vw)t1
The two expressions can be rewritten as
vc + vw = d/t1 and
vc – vw = d/t2
Adding the two equations and dividing by two, we obtain
vc = ½ (d/t1 + d/t2)
Thus, method 1 gives the car’s speed c v a in windless situation.
(b) If method 2 is used, the result would be
vc' = 2d/(t1 + t2)
vc' = vc[1 – (vw/vc)2]
The fractional difference is
(vc – vc’)/vc = (vw/vc)2 = (0.0240)2 = 5.76 x 10-4
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