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A developer has requested permission to build a large retail store at a location adjacent to the intersection of an undivided four-lane major road and a two-lane minor road. Traffic on the minor road is controlled by a stop sign. The speed limits are 35 mi/h on the minor road and 50 mi/h on the major road. The building is to be located 65 ft from the near lane of one of the approaches of the minor road.

Determine the minimum sight distance required on the major road for a stopped vehicle on the minor road to safely turn right onto the major road. The design vehicle is a single-unit truck. Lanes on the major road are 12 ft wide.

Given:
- Time gap for a single-unit truck for a right turn = 8.5 sec
- Time adjustment for the number of lanes = 0.7 sec

(Note: Adjust for the number of lanes.)

Answer :

Answer:

676 ft

Explanation:

Minimum sight distance, d_min

d_min = 1.47 * v_max * t_total where v_max is maximum velocity in mi/h, t_total is total time

v_max is given as 50 mi/h

t_total is sum of time for right-turn and adjustment time=8.5+0.7=9.2 seconds

Substituting these figures we obtain d_min=1.47*50*9.2=676.2 ft

For practical purposes, this distance is taken as 676 ft

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