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10 FE practice problems: transportation: sight distance and curves, with solutions

By Dr. Mohamed Elansary, PhD Environmental EngineeringUpdated

These ten original problems practice transportation: sight distance and curves, a topic from the FE Civil exam specification, using the Civil Engineering, Stopping Sight Distance part of the FE Reference Handbook 10.6. Each problem gives the situation and the values with units. Work it with the handbook PDF open and commit to an answer before you open the solution. Every solution shows the handbook page, the equation, the substitution with units, a size check, and the mistake behind each wrong option, so a wrong pick tells you exactly what to fix.

How to use these problems

Give each problem an honest attempt before you open the solution: write the given values with units, find the equation in the FE Reference Handbook PDF, and commit to an answer. Then compare line by line. If you picked a wrong option, read the note for that option; each one names the mistake that produces it.

The problems

Problem 1 · FE Civil, Transportation: sight distance and curves

A highway has a design speed of 60 mph on a 2% upgrade. Use a perception-reaction time of 2.5 sec and a deceleration rate of 11.2 ft/sec². The stopping sight distance is most nearly:

  • A 326 ft
  • B 587 ft
  • C 547 ft
  • D 476 ft

Handbook: Civil Engineering, Stopping Sight Distance, FE Reference Handbook 10.6

Show the worked solution
Answer
C (547 ft)
Given
V = 60 mph, t = 2.5 sec, a = 11.2 ft/sec², G = 2%
Find
stopping sight distance (ft)
Handbook
Civil Engineering, Transportation, Stopping Sight Distance, page 306
Equation
SSD = 1.47Vt + V²/[30(a/32.2 ± G)], G = grade/100 (uphill +)
Substitute
  1. Reaction distance = 1.47Vt = 1.47(60)(2.5) = 220.5 ft
  2. Braking distance = V²/[30(a/32.2 ± G)] = 60²/[30(11.2/32.2 + 0.02)] = 326.2 ft
  3. SSD = 547 ft
Result
547 ft, 3 significant figures
Check
an upgrade shortens the braking distance compared with level ground (566 ft).
Why the others are wrong
  • A: is the braking distance only; the perception-reaction distance is missing
  • B: used the grade with the wrong sign (uphill grades are +)
  • D: left out 1.47 (mph to ft/sec) in the reaction distance

Problem 2 · FE Civil, Transportation: sight distance and curves

A circular horizontal curve has a radius of 650 ft and an intersection angle (deflection between tangents) of 70°. The tangent distance from the PC to the PI is most nearly:

  • A 1,790 ft
  • B 794 ft
  • C 308 ft
  • D 455 ft

Handbook: Civil Engineering, Horizontal Curves, FE Reference Handbook 10.6

Show the worked solution
Answer
D (455 ft)
Given
R = 650 ft, I = 70°
Find
tangent distance T (ft)
Handbook
Civil Engineering, Transportation, Horizontal Curves, page 308
Equation
T = R tan(I/2)
Substitute
  1. T = R tan(I/2) = (650 ft) tan(35°) = 455 ft
Result
455 ft, 3 significant figures
Check
T is a little longer than half the curve length (397 ft), as it must be for I < 180°.
Why the others are wrong
  • A: used tan I instead of tan(I/2)
  • B: is the curve length, not the tangent distance
  • C: took the tangent with the calculator in radians

Problem 3 · FE Civil, Transportation: sight distance and curves

A circular horizontal curve has a radius of 1,350 ft and an intersection angle (deflection between tangents) of 89°. The tangent distance from the PC to the PI is most nearly:

  • A 946 ft
  • B 1,330 ft
  • C 769 ft
  • D 2,100 ft

Handbook: Civil Engineering, Horizontal Curves, FE Reference Handbook 10.6

Show the worked solution
Answer
B (1,330 ft)
Given
R = 1,350 ft, I = 89°
Find
tangent distance T (ft)
Handbook
Civil Engineering, Transportation, Horizontal Curves, page 308
Equation
T = R tan(I/2)
Substitute
  1. T = R tan(I/2) = (1,350 ft) tan(44.5°) = 1,330 ft
Result
1,330 ft, 3 significant figures
Check
T is a little longer than half the curve length (1,050 ft), as it must be for I < 180°.
Why the others are wrong
  • A: used sin(I/2) instead of tan(I/2)
  • C: took the tangent with the calculator in radians
  • D: is the curve length, not the tangent distance

Problem 4 · FE Civil, Transportation: sight distance and curves

A circular horizontal curve has a radius of 1,700 ft and an intersection angle (deflection between tangents) of 44°. The length of the curve is most nearly:

  • A 653 ft
  • B 1,310 ft
  • C 1,270 ft
  • D 687 ft

Handbook: Civil Engineering, Horizontal Curves, FE Reference Handbook 10.6

Show the worked solution
Answer
B (1,310 ft)
Given
R = 1,700 ft, I = 44°
Find
length of curve L (ft)
Handbook
Civil Engineering, Transportation, Horizontal Curves, page 308
Equation
L = RIπ/180 (I in degrees)
Substitute
  1. L = RIπ/180 = (1,700 ft)(44°)(π/180) = 1,310 ft
Result
1,310 ft, 3 significant figures
Check
the arc (1,310 ft) is longer than the chord 2R sin(I/2) = 1,270 ft.
Why the others are wrong
  • A: used π/360 instead of π/180
  • C: is the long chord, not the arc length
  • D: is the tangent distance T, not the curve length

Problem 5 · FE Civil, Transportation: sight distance and curves

A freeway lane carries 660 veh/hr at an average speed of 27 mph. The traffic density is most nearly:

  • A 17.8 veh/mi/ln
  • B 24.4 veh/mi/ln
  • C 216 veh/mi/ln
  • D 12.2 veh/mi/ln

Handbook: Civil Engineering, Traffic Flow Relationships, FE Reference Handbook 10.6

Show the worked solution
Answer
B (24.4 veh/mi/ln)
Given
flow = 660 veh/hr, speed = 27 mph
Find
density (veh/mi/ln)
Handbook
Civil Engineering, Transportation, Traffic Flow Relationships, page 312
Equation
V = S × D (flow = speed × density)
Substitute
  1. Flow = speed × density, so D = V/S = 660/27 = 24.4 veh/mi/ln
Result
24.4 veh/mi/ln, 3 significant figures
Check
the average spacing 5,280/D = 216 ft per vehicle.
Why the others are wrong
  • A: multiplied flow by speed instead of dividing
  • C: is the average spacing in ft, not the density
  • D: split the per-lane flow over two lanes

Problem 6 · FE Civil, Transportation: sight distance and curves

A freeway lane carries 1,940 veh/hr at an average speed of 46 mph. The traffic density is most nearly:

  • A 42.2 veh/mi/ln
  • B 89.2 veh/mi/ln
  • C 28.7 veh/mi/ln
  • D 21.1 veh/mi/ln

Handbook: Civil Engineering, Traffic Flow Relationships, FE Reference Handbook 10.6

Show the worked solution
Answer
A (42.2 veh/mi/ln)
Given
flow = 1,940 veh/hr, speed = 46 mph
Find
density (veh/mi/ln)
Handbook
Civil Engineering, Transportation, Traffic Flow Relationships, page 312
Equation
V = S × D (flow = speed × density)
Substitute
  1. Flow = speed × density, so D = V/S = 1,940/46 = 42.2 veh/mi/ln
Result
42.2 veh/mi/ln, 3 significant figures
Check
the average spacing 5,280/D = 125 ft per vehicle.
Why the others are wrong
  • B: multiplied flow by speed instead of dividing
  • C: converted the speed to ft/sec but kept miles in the density
  • D: split the per-lane flow over two lanes

Problem 7 · FE Civil, Transportation: sight distance and curves

A circular horizontal curve has a radius of 400 ft and an intersection angle (deflection between tangents) of 83°. The length of the curve is most nearly:

  • A 354 ft
  • B 530 ft
  • C 290 ft
  • D 579 ft

Handbook: Civil Engineering, Horizontal Curves, FE Reference Handbook 10.6

Show the worked solution
Answer
D (579 ft)
Given
R = 400 ft, I = 83°
Find
length of curve L (ft)
Handbook
Civil Engineering, Transportation, Horizontal Curves, page 308
Equation
L = RIπ/180 (I in degrees)
Substitute
  1. L = RIπ/180 = (400 ft)(83°)(π/180) = 579 ft
Result
579 ft, 3 significant figures
Check
the arc (579 ft) is longer than the chord 2R sin(I/2) = 530 ft.
Why the others are wrong
  • A: is the tangent distance T, not the curve length
  • B: is the long chord, not the arc length
  • C: used π/360 instead of π/180

Problem 8 · FE Civil, Transportation: sight distance and curves

A freeway lane carries 1,670 veh/hr at an average speed of 57 mph. The traffic density is most nearly:

  • A 29.3 veh/mi/ln
  • B 34.1 veh/mi/ln
  • C 19.9 veh/mi/ln
  • D 95.2 veh/mi/ln

Handbook: Civil Engineering, Traffic Flow Relationships, FE Reference Handbook 10.6

Show the worked solution
Answer
A (29.3 veh/mi/ln)
Given
flow = 1,670 veh/hr, speed = 57 mph
Find
density (veh/mi/ln)
Handbook
Civil Engineering, Transportation, Traffic Flow Relationships, page 312
Equation
V = S × D (flow = speed × density)
Substitute
  1. Flow = speed × density, so D = V/S = 1,670/57 = 29.3 veh/mi/ln
Result
29.3 veh/mi/ln, 3 significant figures
Check
the average spacing 5,280/D = 180 ft per vehicle.
Why the others are wrong
  • B: divided speed by flow (inverted)
  • C: converted the speed to ft/sec but kept miles in the density
  • D: multiplied flow by speed instead of dividing

Problem 9 · FE Civil, Transportation: sight distance and curves

A highway has a design speed of 50 mph on a 4% downgrade. Use a perception-reaction time of 2.5 sec and a deceleration rate of 11.2 ft/sec². The stopping sight distance is most nearly:

  • A 399 ft
  • B 191 ft
  • C 454 ft
  • D 423 ft

Handbook: Civil Engineering, Stopping Sight Distance, FE Reference Handbook 10.6

Show the worked solution
Answer
C (454 ft)
Given
V = 50 mph, t = 2.5 sec, a = 11.2 ft/sec², G = 4%
Find
stopping sight distance (ft)
Handbook
Civil Engineering, Transportation, Stopping Sight Distance, page 306
Equation
SSD = 1.47Vt + V²/[30(a/32.2 ± G)], G = grade/100 (uphill +)
Substitute
  1. Reaction distance = 1.47Vt = 1.47(50)(2.5) = 183.8 ft
  2. Braking distance = V²/[30(a/32.2 ± G)] = 50²/[30(11.2/32.2 - 0.04)] = 270.7 ft
  3. SSD = 454 ft
Result
454 ft, 3 significant figures
Check
a downgrade lengthens the braking distance compared with level ground (423 ft).
Why the others are wrong
  • A: used the grade with the wrong sign (downhill grades are -)
  • B: used a instead of a/32.2 in the braking term
  • D: left the grade out of the braking term

Problem 10 · FE Civil, Transportation: sight distance and curves

A freeway lane carries 1,490 veh/hr at an average speed of 39 mph. The traffic density is most nearly:

  • A 26.2 veh/mi/ln
  • B 138 veh/mi/ln
  • C 58.1 veh/mi/ln
  • D 38.2 veh/mi/ln

Handbook: Civil Engineering, Traffic Flow Relationships, FE Reference Handbook 10.6

Show the worked solution
Answer
D (38.2 veh/mi/ln)
Given
flow = 1,490 veh/hr, speed = 39 mph
Find
density (veh/mi/ln)
Handbook
Civil Engineering, Transportation, Traffic Flow Relationships, page 312
Equation
V = S × D (flow = speed × density)
Substitute
  1. Flow = speed × density, so D = V/S = 1,490/39 = 38.2 veh/mi/ln
Result
38.2 veh/mi/ln, 3 significant figures
Check
the average spacing 5,280/D = 138 ft per vehicle.
Why the others are wrong
  • A: divided speed by flow (inverted)
  • B: is the average spacing in ft, not the density
  • C: multiplied flow by speed instead of dividing

A new problem posts every day inside the lab, with the full worked solution the same evening.

Frequently asked questions

Where is transportation: sight distance and curves in the FE Reference Handbook?

Look in the Civil Engineering, Stopping Sight Distance part of FE Reference Handbook 10.6. Each solution gives the exact page, so you can practice finding it in the PDF the way you will on exam day.

Are these real FE exam questions?

No. They are original problems generated from the handbook formulas and checked by code. Real exam questions are confidential, and sharing them breaks the NCEES agreement every examinee accepts.

How do I check my answer before opening the solution?

Check the units of your result and whether its size makes sense for the situation. Each solution ends with the same kind of check, so you can compare your habit with ours.

Sources

  1. NCEES FE Civil CBT exam specifications (PDF). Retrieved October 3, 2026.
  2. NCEES FE Reference Handbook 10.6 (free PDF in MyNCEES). Retrieved October 3, 2026.