Map Sapling Learning As a city planner, you receive complaints from local reside

Question

Map Sapling Learning 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 a problem, but when fog rolls in visibility can reduce to only 155 feet Since fog is a common occurrence n this region, you decide to investigate. he state highway department states that the effective coefficient of friction between a rolling wheel and (locked) wheel between the of travel on the road, from asphalt ranges between 0.550 and 0.754. ehicles of all types small VW bugs weighing 1190 lb to large trucks weighing 9.00 Considering that some drivers will properly when down and others skid to stop, calculate limit can stop maximum braking distance needed to ensure that all traveling the posted speed before reaching the intersection vehicles at Minimum Maximum Numbe Number the n that the goal is to allow allvehicles to come safely to a stop before reaching theintersection, calculate maximum desired speed limit. Number (Scroll down for more questions.) mph We

Explanation / Answer

we will use kinematic equation , to calculate the desired braking distance for rolling and skidding'

v^2 = 2as

v = 55 mph = 24.587 m/s apprx

we will calculate two sets of retardation ( max and min ) for rolling and skidding

Roling

a( retardation for rolling) = 0.842 (9.8 )= 8.2516 m/s ^2, a ( retardation for rolling)= 0.941 (9.8) = 9.2218 m/s^2

s( braking distnace) = 24.587^2 /2 (8.2516) = 36.63 m = 116.89633 ft

s ( braking distance) = 24.587^2 /2 (9.2218) = 32.77 m = 107.513123 ft

Skidding

a( for skidding) = 0.55 ( 9.8) = 5.39 m/s^2

a ( for skidding) = 0.754 (9.8) = 7.3892 m/s^2

s( braking distance ) =  24.587^2/ 2(5.39) = 56.1 m apprx= 184.05512 ft

s( braking distance ) =  24.587^2/ 2( 7.3892) =40.905 m apprx= 134.202756 ft

let's take th minimum braking distance as 155 ft = 47.244 m

v = sqroot ( 2as) = sqroot ( 2 x9.2218 x 47.244) = 29.5186 m/s apprx= 66 m ph apprx ---------safe speed

c) reaction time not taken

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