A vehicle moving at a speed of 50 mph is slowing traffic on a two-lane highway. What passing sight distance is necessary, in order for a passing maneuver to be carried out safely?

Calculate the passing sight distance by hand, and then compare it to the values recommended by AASHTO. In your calculations, assume that the following variables have the values given:

You should also assume that the passing vehicle accelerates to passing speed before moving into the left lane.

[Solution Shown Below]

Solution

The first step in calculating the passing sight distance is the calculation of the distance D1. This distance includes the distance traveled during the perception/reaction time and the distance traveled while accelerating to the passing speed. The distance traveled during the perception reaction time is computed using D=VT from physics, where V = 73.3 f./sec (50 mph) and T = 2.5 seconds. Solving for D yields a value of 183.3 feet. The distance traveled during the acceleration portion of D1 is computed using the equation Vf2=Vi2 + 2AD, where Vf = 88 ft/sec (60 mph), Vi = 73.3 f./sec (50 mph),

and A = 2.155 ft/sec/sec (1.47 mph/sec). Solving for D yields a value of 550.1 feet.  The total distance D1 is 183.3 + 550.1 = 733.4 feet.

The second portion of the passing sight distance is the distance D2, which is defined as the distance that the passing vehicle travels while in the left lane. This distance can be calculated in the following way. 

While in the left lane, the passing vehicle must traverse the clearance distance between itself and the slow vehicle, the length of the slow vehicle, the length of itself, and the length of the clearance distance between itself and the slow vehicle at lane re-entry. The time it takes the passing vehicle to traverse these distances relative to the slow vehicle can be computed from the equation D=VT, where D = 84 ft (20 ft + 22 ft + 22 ft. + 20 ft) and V = 14.67 ft/sec (10 mph = relative speed of passing vehicle with reference point on the slow vehicle). 

Solving for the time T2 yields a value of 5.7 seconds. The real distance traveled by the passing vehicle during the time T2 is calculated using D=VT, where V = 88 ft./sec. (60 mph) and T = 5.7 seconds. Solving for D yields the distance D2 or 501.6 ft.

The distance D3 is the clearance distance between the passing vehicle and the opposing vehicle at the moment the passing vehicle returns to the right lane. This distance was given as 250 ft.The distance D4 is the final component of the passing sight distance and is defined as the distance the opposing vehicle travels during 66% of the time that the passing vehicle is in the left lane. This distance is computed using D=VT, where V = 88 ft./sec. (60 mph) and T = 3.7 seconds (5.7*66%). Solving for D yields a value of 325.6 ft. for D4.