CELE Transportation & Highway Engineering Reviewer
8 Transportation & Highway Engineering practice questions for the Civil Engineer Licensure Examination, each with the correct answer and an explanation of why it's right.
Transportation & Highway Engineering Practice Questions with Answers
- 1easy
Find the minimum radius of a horizontal curve for V = 100 km/h, e_max = 0.08, and f_max = 0.12.
- A.393.7 m
- B.500.0 m
- C.312.5 m
- D.450.0 m
Show answer & explanation
Answer: A. 393.7 m
Step 1: Given V = 100 km/h, e = 0.08, f = 0.12.
Step 2: Use R_min = V² / [127(e + f)].
Step 3: e + f = 0.08 + 0.12 = 0.20.
Step 4: R_min = 100² / [127 × 0.20] = 10000 / 25.4 = 393.7 m.
Common error: using 254 instead of 127 in the denominator (gives half the correct value). - 2easy
A vehicle travels at 60 km/h on a −3% downgrade. With t = 2.5 s and f = 0.35, what is the SSD?
- A.86.0 m
- B.79.5 m
- C.91.7 m
- D.100.0 m
Show answer & explanation
Answer: A. 86.0 m
Step 1: Given V = 60 km/h, t = 2.5 s, f = 0.35, G = −0.03 (downgrade).
Step 2: Reaction distance = 0.278 × 60 × 2.5 = 41.7 m.
Step 3: On a downgrade, use (f − |G|) = 0.35 − 0.03 = 0.32.
Step 4: Braking distance = 60² / [254 × 0.32] = 3600 / 81.28 = 44.3 m.
Step 5: SSD = 41.7 + 44.3 = 86.0 m.
Note: downgrade increases braking distance compared to level (f = 0.35 alone gives 41.7 + 40.4 = 82.1 m). - 3easy
Which value of f (friction factor) should be used for the braking distance formula on a level road at V = 80 km/h if SSD = 127.6 m and t = 2.5 s?
- A.0.35
- B.0.30
- C.0.40
- D.0.25
Show answer & explanation
Answer: A. 0.35
Step 1: SSD = 127.6 m, V = 80 km/h, t = 2.5 s, G = 0.
Step 2: Reaction distance = 0.278 × 80 × 2.5 = 55.6 m.
Step 3: Braking distance = 127.6 − 55.6 = 72.0 m.
Step 4: Braking dist = V²/[254f] → f = V²/(254 × 72.0) = 6400/18288 = 0.35.
This back-calculation confirms f = 0.35 is the design friction coefficient at 80 km/h. - 4easy
For V = 100 km/h and R_min = 400 m, what is the required (e + f)?
- A.0.197
- B.0.250
- C.0.150
- D.0.320
Show answer & explanation
Answer: A. 0.197
Step 1: Use R_min = V²/[127(e + f)] and solve for (e + f).
Step 2: e + f = V²/(127 × R_min) = 100²/(127 × 400).
Step 3: e + f = 10000/50800 = 0.1969 ≈ 0.197.
Step 4: This means the combined superelevation and side friction must equal at least 0.197 to maintain vehicle stability at 100 km/h on a 400 m radius curve. - 5easy
The perception-reaction distance for a vehicle traveling at 90 km/h with t = 2.5 s is closest to:
- A.62.6 m
- B.55.0 m
- C.72.0 m
- D.45.0 m
Show answer & explanation
Answer: A. 62.6 m
Step 1: Perception-reaction distance = 0.278 × V × t.
Step 2: = 0.278 × 90 × 2.5.
Step 3: = 0.278 × 225 = 62.55 ≈ 62.6 m.
Note: The constant 0.278 = 1/3.6, converting km/h to m/s. A common error is using V directly in m/s without the 0.278 factor, giving 90 × 2.5 = 225 m — far too large. - 6easy
If e_max = 0.10 and f_max = 0.14 for a rural highway, what is R_min for V = 90 km/h?
- A.265.7 m
- B.318.0 m
- C.200.0 m
- D.350.0 m
Show answer & explanation
Answer: A. 265.7 m
Step 1: V = 90 km/h, e = 0.10, f = 0.14.
Step 2: e + f = 0.24.
Step 3: R_min = V²/[127(e + f)] = 90²/(127 × 0.24) = 8100/30.48 = 265.7 m.
Step 4: Verify — a larger e_max reduces the required radius, which makes physical sense (more banking helps negotiate tighter curves). - 7easy
On a level road, doubling the design speed (from 50 to 100 km/h) while keeping f and t constant will increase the braking distance component of SSD by a factor of:
- A.4
- B.2
- C.8
- D.1.5
Show answer & explanation
Answer: A. 4
Step 1: Braking distance = V²/[254f]. It is proportional to V².
Step 2: If V doubles (×2), V² increases by 2² = 4.
Step 3: Therefore braking distance increases by a factor of 4.
Step 4: This quadratic relationship (not linear) is why highway speed limits strongly affect required sight distances — doubling speed quadruples the braking requirement. - 8easy
A road segment has a density of 20 veh/km and a space-mean speed of 70 km/h. What is the traffic flow?
- A.1400 veh/hr
- B.3.5 veh/hr
- C.700 veh/hr
- D.2800 veh/hr
Show answer & explanation
Answer: A. 1400 veh/hr
Step 1: Identify the fundamental relation: q = k × u, where q = flow (veh/hr), k = density (veh/km), u = space-mean speed (km/h). Step 2: Substitute values: q = 20 × 70. Step 3: q = 1400 veh/hr. Common mistake: dividing instead of multiplying (20/70 ≈ 0.286, which is dimensionally wrong). The product of veh/km × km/h = veh/hr, confirming the unit check.
Practise the full Transportation & Highway Engineering set
Super Tutor has 180 Transportation & Highway Engineering questions for the CELE, each with an explanation, plus full timed CELE mock exams. Start free — 5-day trial, no card.