Question

Compute the passing sight distance for the following data:
Speed of the passing vehicle = 26.67 m/sec
Speed of the overtaken vehicle = 24.44 m/sec
Time of initial maneuver = 4.3 sec
Average acceleration = 1.473 mph/sec
Time of passing vehicle occupies the left lane = 10.4 sec
Distance between the passing vehicle at the end of its maneuver and the opposing vehicle = 76 m

Answer 1

Passing sight distance

$D=d_1+d_2+d_3\\ d_1= v_b*T_1=26.67*4.3=114.681 m\\ d_2=v_b*T_2+ \frac{1}{2}*a*T_2=24.44*4+ \frac{1}{2}*1.473*10.4=105.4196m\\ d_3=76 m\\ D=d_1+d_2+d_3=114.681+105.4196+76=296.1006 m\\$