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 0.689 and 0.770, whereas the effective coefficient of friction between a skidding (locked) wheel and asphalt ranges between 0.450 and 0.617. Vehicles of all types travel on the road, from small VW bugs weighing 1370 lb to large trucks weighing 8160 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.

Answer 1

The speed limit in the area is too high to allow vehicles to stop in time, v = 55 mph = 80.6 ft/s

The effective coefficient of friction between a rolling wheel & asphalt ranges between 0.689 and 0.770

The effective coefficient of friction between a skidding (locked) wheel & asphalt ranges between 0.450 and 0.617

Weight of the truck, W_t = 8160 lb

mass of the truck, m_t = W_t / g (8160 lb) / (32.2 ft/s²) = 253.4 lb.s²/lb

Weight of the bug, W_b = 1370 lb

mass of the bug, m_b = W_b / g (1370 lb) / (32.2 ft/s²) = 42.5 lb.s²/lb

For Truck, we have

kinetic energy, K.E_t = (1/2) m_t v²                                                             { eq.1 }

inserting the values in above eq.

K.E_t = (0.5) (253.4 lb.s²/lb) (80.6 ft/s)²

K.E_t = 823088.8 ft.lb

frictional force by truck with worst coefficient :   F_fw = m_t g                                                            { eq.2 }

inserting the values in eq.2,

F_fw = (0.45) (8160 lb)

F_fw = 3672 lb

We know that work, W = F d

K.E = W

then we have    d = W_k / F_fw                                                       { eq.3 }

inserting the values in eq.3,

d = (823088.8 ft.lb) / (3672 lb)

d = 224.1 ft

frictional force by truck with best coefficient :   F_fb = m_t g     

F_fb = (0.77) (8160 lb)

F_fb = 6283.2 lb

And     d = W_k / F_fb (823088.8 ft.lb) / (6283.2 lb)

d = 130.9 ft

Considering that some drivers will brake properly when slowing down and others will skid to stop, the minimum and maximum braking distance needed to ensure that all vehicles traveling at the posted speed limit can stop before reaching the intersection which is given as :

d_min = 130.9 ft                            and                          d_max = 224.1 ft

For bugs, we have

kinetic energy, K.E_b = (1/2) m_b v²                                                             { eq.4 }

inserting the values in eq.4

K.E_b = (0.5) (42.5 lb.s²/lb) (80.6 ft/s)²

K.E_b = 138047.6 ft.lb

frictional force by bug with worst coefficient :   F_fb = m_b g                                                            { eq.5 }

inserting the values in eq.5,

F_fb = (0.45) (1370 lb)

F_fb = 616.5 lb

We know that work, W = F d

K.E = W

then we have    d = W_kb / F_fw                                                      { eq.6 }

inserting the values in eq.6,

d = (138047.6 ft.lb) / (616.5 lb)

d = 224.1 ft

frictional force by truck with best coefficient :   F_fb = m_b g     

F_fb = (0.77) (1370 lb)

F_fb = 1054.9 lb

And     d = W_k / F_fb (138047.6 ft.lb) / (1054.9 lb)

d = 130.9 ft