Question

Problem 1 Adequate gap time is needed for pedestrians to safely cross the roadway to the opposite side. The formula for computing the adequate gap time is: G = W/S + R+ 2 (N/5 - 1) where, G - adequate gap time in seconds W-width in feet of the roadway to be crossed S. pedestrian walking speed, 3.5 ft/sec R-reaction time, 3.0 seconds N - total number of pedestrians How much time is needed for a group of 14 people to cross over a 50-ft wide street? Compare this with the case of 7 children to cross with a reaction time of 4.5 sec and a walking speed of 2.5 ft/sec. What did you learn from this example?

Answer 1

Problem 1

Case 1

No. Of pefedsyrains = N = 14

Width = W = 50

S = 3.5 ft/ s

R= 3.0 sec

G = W/S + R + 2(N/5 -1) = 50/3.5 + 3+ 2(14/5 -1)

G= 20.89 sec

Case 2

N = 7

R = 4.5 sec

S= 2.5 ft/sec

G = 50/2.5 + 4.5+ 2(7/5 -1) =25.3 sec

From comparison of above two examples it is observed that the speed of children is lower than the adult people and their reaction tine is more than adults. Therefoere adequate gap time for case of children is coming to be more even if the children are half in numbers than that of adults. Therefore the signal at cross walk should be designed for keeoing in mind the speed of the slow pedestrians.