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 ft. 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 1370 lb to large trucks weighing 8040 lb. Considering that some drivers will brake properly when slowing down and others will skid to stop, calculate the minimum and maximum braking distance needed to ensure that all vehicles traveling at the posted speed limit can stop before reaching the intersection.

minimum braking distance: ?ft

maximum braking distance: ?ft

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. maximum speed limit: ? mph

Which factors affect the soundness of your decision? (can be multiple ones)

Reaction time of the drivers is not taken into account.

Newton's second law does not apply to this situation.

Drivers cannot be expected to obey the posted speed limit.

Precipitation from the fog can lower the coefficients of friction.

Answer 1

Solution :

Speed limit = v_o = 55 mph = 80.7 ft/s = 24.5872 m/s

STopping distance = d = 155 ft

Rolling friction = 0.842 to 0.941

Minimum Acceleration due to rolling friction = a_r1 = (0.842)(9.8) = - 8.25 m/s/s

Maximum Acceleration = a_r2 = (0.941)(9.8) = - 9.22 m/s^2

Stopping distance is calculated using vf^2 = vo^2 +2ad

=> Stopping distance when a =ar1 = -8.25 m/s^2 is d1 =(0^2- vo^2 / 2ar1

= ( 24.5872)^2 /( 2*-8.25)

=> d₁= 36.631 m = 120 . 2 ft

Stopping distance when acceleration is maximum ar2 = -9.22 m/s/s is

d₂ = (vf^2 - vo^2 / 2ar2 = (24.5872)^2 / (2*-9.22) = 32.777 m = 107 . 5 ft

Sliding friction range = 0.550 to 0.754

Minimum acceleration due to sliding friction = as1 = s1 g = (0.55)(9.8) = - 5.39 m/s/s

Maximum acceleration = as2 = s2 g = (0.754)(9.8) = - 7.389 m/s/s

Stopping distance ds1 = (0^2 - 24.5872^2) / (2 *  - 5.39 ) = 56 m = 183.7 ft

Stopping distance ds2 = (0^2 - 24.5872^2) /(2*-7.389) = 40.9 m = 134.2 ft

Minimum breaking distance = 107.5 ft

Maximum breaking distance = 183.7 ft

gravity = g = 9.8 m/s/s = 32.2 ft/s^2

friction force (minimum , sliding) = s1 mg = (0.55) (1370)(32.2) = 24262 ft - lb

v = 2 * a _s1* d_s1 = 2 * (0.55)(32.2)* 155 = 74 ft/s = 50.5 mph

( or another method to get this speed is

Change in kinetic energy = work done

=> 1/2 * (1370) v^2 = ((0.55) (1370)(32.2) ft -lb)* (155 ft)

=> v = ((0.55) (1370)(32.2) ft -lb)* (155 ft) / (0.5*1370) = 74 ft/s = 22.56 m/s)

Velocity = 50.5 mph =Maximum speed limit .

c) 1.Reaction time of the drivers is not taken into account.

2.Precipitation from the fog can lower the coefficients of friction.