Average Velocity and Average Speed Problems and Solutions

 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 v_s = 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.)

Fig.1

Answer:
The amount of time it takes for each person to move a distance L with speed s v is

∆t = L/v_s

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/v_s)

R = dv_s/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

s_avg,1 = s_avg,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 v_w to the car’s jspeed v_c is 0.0240?

Answer:
Let v_w be the speed of the wind and v_c 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

v_eef,1 = v_c + v_w

and the distance traveled is

d = v_eef,1t₁ = (v_c + v_w)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

v_eef,2 = v_c – v_w
and the distance traveled is

d = v_eef,2t₂ = (v_c – v_w)t₁
The two expressions can be rewritten as

v_c + v_w = d/t₁ and
v_c – v_w = d/t₂

Adding the two equations and dividing by two, we obtain

v_c = 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

v_c' = 2d/(t₁ + t₂)
v_c' = v_c[1 – (v_w/v_c)²]

The fractional difference is

(v_c – v_c’)/v_c = (v_w/v_c)²  = (0.0240)² = 5.76 x 10^-4