Question

Imagine the situation which a vehicle pulls out onto a single-lane highway that has a flow rate of 400 veh/h (Poisson distributed). The driver of the vehicle does not look for oncoming traffic. Road conditions and vehicle speeds on the highway are such that it takes 2.2 seconds for an oncoming vehicle to stop once the brakes are applied. Assuming a standard driver reaction time of 2.5 seconds, a) what is the probability that the vehicle pulling out will get in an accident with oncoming traffic? What should be the reaction and braking times of oncoming vehicles so that the probability of an accident is equal to 0.25?

please provide detailed soln.

Answer 1

Answeos- Given, flow rate = 400 vehhh 0.1ll1 Veh/sec time = 2.2 Sec Standard diver recaction time = 2.5 sec the successive vehiclos headway is more than the (2.2 +25) = 417 sec will not be in accident headway of vehicles less than the 4,17 sec P (heuit) l-e-01111* 4,7 = 1-ě 0.52222 -0.59320 0.406799 P(hauit) the probability of an acadent is equal to 0.25 P (ha Catt) = 0.25 1- eat -0.1111xt I-e 1-0.25 é o llixt é o lllxt 0.75

0175 - ēxt In 10.75 -0.1111xt +0.28768 = 10.1111xt 0.28768 011111 2.589 sec