As a city planner, you receive complaints from local residents about the safety

Question

As a city planner, you receive complaints from local residents about the safety of nearby roads and streets. One complaint concerns a stop sign at the corner of Pine Street and 1st Street. Residents complain that the speed limit in the area (55 mph) is too high to allow vehicles to stop in time. Under normal conditions this is not a problem, but when fog rolls in visibility can reduce to only 155 feet. Since fog is a common occurrence in this region, you decide to investigate. The state highway department states that the effective coefficient of friction between a rolling wheel and asphalt ranges between 0.842 and 0.941, whereas the effective coefficient of friction between a skidding (locked) wheel and asphalt ranges between 0.550 and 0.754. Vehicles of all types travel on the road, from small VW bugs weighing 1110 lb to large trucks weighing 9150 lb.  Considering that some drivers will brake properly when slowing down and others will skid to stop, calculate  the miminim and maximum braking distance needed to ensure that all vehicles traveling at the posted speed limit can stop before reaching the intersection.

Given that the goal is to allow all vehicles to come safely to a stop before reaching the intersection, calculate the maximum desired speed limit. (Round your answer to the nearest whole number.)

Explanation / Answer

Its not easy to help with problems of that level of complexity without personal interaction and without ability to draw diagrams. Let see if I can help you with the last problem. The minimum coefficient of friction that prevents the log from sliding when the truck accelerates is given by mu = a / g (you can find the formula in almost any physics textbook - it is beyond the scope of this forum to derive the formula). To find the acceleration of a loaded truck we will assume that the force of the engine pushing (accelerating) the truck remains constant (is independent of the load). The acceleration of the empty truck is a = (change in velocity) / (time) = (55mph - 0 mph) / (22.0 s) = (24.55m/s)/(22.0s) = 1.12 m/s^2. The force pushing the truck is F = ma = (9850kg)(1.12m/s^2) = 10999 N. The same force will accelerate the loaded truck, therefore a = F/m = (10999N)/( 9850kg + 799kg) = 1.03 m/s^2. The minimum coefficient of friction = a/g = (1.03 m/s^2)/(9.81 m/s^2)

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