FE Civil Transportation Engineering Practice Problems

The current FE Civil specification assigns 9-14 of the 110 exam questions to Transportation Engineering. Form A contains 9 original problems in this area. The sample below is published in full, with the same worked-solution format used throughout the book.

A two-lane rural highway approaches an intersection on a 4.0% downgrade. The design speed is 50 mph, the perception-reaction time is 2.5 s, and the deceleration rate is 11.2 ft/s $^{2}$ . The stopping sight distance is most nearly:

(A)271 ft
(B)399 ft
(C)423 ft
(D)454 ft

Show worked solution

Answer: (D)

A downgrade reduces the effective deceleration, stretching the braking distance to 270.7 ft; added to 183.75 ft of perception-reaction travel, the SSD is 454 ft.

On a downgrade gravity works against the brakes, so $G = 0.04$ subtracts inside the braking term.

S S D = 1.47 V t + \frac{V ^{2}}{30 ( \frac{a}{32.2} - G )}

Reaction covers ground before the brakes do anything.

1.47 (50) (2.5) = 183.75 ft

Braking distance follows, with $a /32.2 = 11.2/32.2 = 0.3478$ .

\frac{5 0 ^{2}}{30 ( 0.3478 - 0.04 )} = \frac{2500}{9.235} = 270.7 ft

S S D = 183.75 + 270.7 = 454.5 ft

FE Reference Handbook — Transportation: Stopping Sight Distance (grade-adjusted braking distance equation)

Why the other choices appear

(A)Omits the perception-reaction distance and reports the braking distance alone, 270.7 ft.
(B)Adds the grade term instead of subtracting it (treats the downgrade as an upgrade): 183.75 + 2500/(30 x 0.3878) = 398.6 ft.
(C)Ignores the grade entirely: 183.75 + 2500/(30 x 0.3478) = 423.3 ft.