PE Transportation Engineering (Civil Specialty) practice exam Verified
Questions, Correct Answers, and Detailed Explanations for Computer
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1. A two-lane horizontal highway curve has a radius of 850 ft
and a design speed of 65 mph. The driver's eye is 3.5 ft
above the pavement, and the obstacle height is 2.0 ft.
Assuming the required stopping sight distance is 645 ft,
what is the minimum clear distance (𝑀) required from the
inside edge of the traveled lane to a roadside obstruction
to provide adequate stopping sight distance?
• A. 42.3 ft
• B. 51.6 ft
• C. 62.4 ft
• D. 74.1 ft
• Answer: B. 51.6 ft
• Solution: Use the middle ordinate formula for horizontal
curves where 𝑆 < 𝐿:
28.65 × 𝑆
𝑀 = 𝑅 (1 − cos ( ))
𝑅
28.65 × 645
𝑀 = 850 (1 − cos ( )) = 850(1 − cos(21.73∘ ))
850
= 850(1 − 0.9290) = 51.6 ft
2. A crest vertical curve connects a 𝑔1 = +2.5% upgrade
with a 𝑔2 = −1.5% downgrade. The design speed is 60
,PE Transportation Engineering (Civil Specialty) practice exam Verified
Questions, Correct Answers, and Detailed Explanations for Computer
Science Students||Already Graded A+
mph, and the required stopping sight distance is 570 ft.
What is the minimum length of the vertical curve (𝐿) for
𝑆 < 𝐿?
• A. 420 ft
• B. 510 ft
• C. 610 ft
• D. 730 ft
• Answer: C. 610 ft
• Solution: The algebraic difference in grades is 𝐴 = |𝑔2 −
𝑔1 | = | − 1.5 − 2.5| = 4.0%. Using the standard AASHTO
formula for crest vertical curves where 𝑆 < 𝐿 (with eye
height 3.5 ft and object height 2.0 ft):
𝐴𝑆 2 4.0 × (570)2 4.0 × 324,900
𝐿= = = = 602.4 ft
2158 2158 2158
≈ 610 ft
3. An intersection approach has a design speed of 45 mph, a
0.0% grade, and a driver perception-reaction time of 1.0
sec. The deceleration rate is 11.2 ft/sec². What is the
minimum yellow change interval (𝑌) required?
• A. 3.2 sec
• B. 3.9 sec
,PE Transportation Engineering (Civil Specialty) practice exam Verified
Questions, Correct Answers, and Detailed Explanations for Computer
Science Students||Already Graded A+
• C. 4.5 sec
• D. 5.1 sec
• Answer: B. 3.9 sec
• Solution: Convert speed to feet per second: 𝑉 =
45 mph × 1.467 = 66.0 ft/sec.
1.47𝑉 66.0 66.0
𝑌=𝑡+ = 1.0 + = 1.0 +
2𝑎 + 64.4𝑔 2(11.2) + 0 22.4
= 1.0 + 2.95 = 3.95 sec ≈ 3.9 sec
4. A single-axle load of 24,000 lbs passes over a flexible
pavement. Using the AASHTO standard single axle load of
18,000 lbs, what is the Load Equivalency Factor (LEF) based
on the fourth-power rule?
• A. 1.78
• B. 2.33
• C. 3.16
• D. 4.21
• Answer: C. 3.16
• Solution: The approximate fourth power law formula for a
single axle load is:
, PE Transportation Engineering (Civil Specialty) practice exam Verified
Questions, Correct Answers, and Detailed Explanations for Computer
Science Students||Already Graded A+
Axle Load 4 24,000 4
LEF = ( ) =( ) = (1.333)4 ≈ 3.16
𝟏𝟖, 𝟎𝟎𝟎 lbs 18,000
5. In a traffic bottleneck scenario, the upstream traffic stream
has a flow rate of 1,800 veh/hr and density of 30 veh/mi.
Inside the queue, the congested flow rate drops to 1,000
veh/hr with a density of 110 veh/mi. What is the speed
and direction of the backward-forming shockwave?
• A. -5.0 mph (moving upstream)
• B. -10.0 mph (moving upstream)
• C. +10.0 mph (moving downstream)
• D. -22.5 mph (moving upstream)
• Answer: B. -10.0 mph (moving upstream)
• Solution: The shockwave speed formula is:
𝑞2 − 𝑞1 1000 − 1800 −800
𝑢𝑠 = = = = −10 mph
𝑘2 − 𝑘1 110 − 30 80
The negative sign indicates that the shockwave propagates
upstream against the direction of traffic flow.
6. A horizontal curve on a rural highway has a radius of 1,400
ft and a design speed of 60 mph. If the side friction factor
is 𝑓 = 0.11, what is the required superelevation rate (𝑒)?
Questions, Correct Answers, and Detailed Explanations for Computer
Science Students||Already Graded A+
1. A two-lane horizontal highway curve has a radius of 850 ft
and a design speed of 65 mph. The driver's eye is 3.5 ft
above the pavement, and the obstacle height is 2.0 ft.
Assuming the required stopping sight distance is 645 ft,
what is the minimum clear distance (𝑀) required from the
inside edge of the traveled lane to a roadside obstruction
to provide adequate stopping sight distance?
• A. 42.3 ft
• B. 51.6 ft
• C. 62.4 ft
• D. 74.1 ft
• Answer: B. 51.6 ft
• Solution: Use the middle ordinate formula for horizontal
curves where 𝑆 < 𝐿:
28.65 × 𝑆
𝑀 = 𝑅 (1 − cos ( ))
𝑅
28.65 × 645
𝑀 = 850 (1 − cos ( )) = 850(1 − cos(21.73∘ ))
850
= 850(1 − 0.9290) = 51.6 ft
2. A crest vertical curve connects a 𝑔1 = +2.5% upgrade
with a 𝑔2 = −1.5% downgrade. The design speed is 60
,PE Transportation Engineering (Civil Specialty) practice exam Verified
Questions, Correct Answers, and Detailed Explanations for Computer
Science Students||Already Graded A+
mph, and the required stopping sight distance is 570 ft.
What is the minimum length of the vertical curve (𝐿) for
𝑆 < 𝐿?
• A. 420 ft
• B. 510 ft
• C. 610 ft
• D. 730 ft
• Answer: C. 610 ft
• Solution: The algebraic difference in grades is 𝐴 = |𝑔2 −
𝑔1 | = | − 1.5 − 2.5| = 4.0%. Using the standard AASHTO
formula for crest vertical curves where 𝑆 < 𝐿 (with eye
height 3.5 ft and object height 2.0 ft):
𝐴𝑆 2 4.0 × (570)2 4.0 × 324,900
𝐿= = = = 602.4 ft
2158 2158 2158
≈ 610 ft
3. An intersection approach has a design speed of 45 mph, a
0.0% grade, and a driver perception-reaction time of 1.0
sec. The deceleration rate is 11.2 ft/sec². What is the
minimum yellow change interval (𝑌) required?
• A. 3.2 sec
• B. 3.9 sec
,PE Transportation Engineering (Civil Specialty) practice exam Verified
Questions, Correct Answers, and Detailed Explanations for Computer
Science Students||Already Graded A+
• C. 4.5 sec
• D. 5.1 sec
• Answer: B. 3.9 sec
• Solution: Convert speed to feet per second: 𝑉 =
45 mph × 1.467 = 66.0 ft/sec.
1.47𝑉 66.0 66.0
𝑌=𝑡+ = 1.0 + = 1.0 +
2𝑎 + 64.4𝑔 2(11.2) + 0 22.4
= 1.0 + 2.95 = 3.95 sec ≈ 3.9 sec
4. A single-axle load of 24,000 lbs passes over a flexible
pavement. Using the AASHTO standard single axle load of
18,000 lbs, what is the Load Equivalency Factor (LEF) based
on the fourth-power rule?
• A. 1.78
• B. 2.33
• C. 3.16
• D. 4.21
• Answer: C. 3.16
• Solution: The approximate fourth power law formula for a
single axle load is:
, PE Transportation Engineering (Civil Specialty) practice exam Verified
Questions, Correct Answers, and Detailed Explanations for Computer
Science Students||Already Graded A+
Axle Load 4 24,000 4
LEF = ( ) =( ) = (1.333)4 ≈ 3.16
𝟏𝟖, 𝟎𝟎𝟎 lbs 18,000
5. In a traffic bottleneck scenario, the upstream traffic stream
has a flow rate of 1,800 veh/hr and density of 30 veh/mi.
Inside the queue, the congested flow rate drops to 1,000
veh/hr with a density of 110 veh/mi. What is the speed
and direction of the backward-forming shockwave?
• A. -5.0 mph (moving upstream)
• B. -10.0 mph (moving upstream)
• C. +10.0 mph (moving downstream)
• D. -22.5 mph (moving upstream)
• Answer: B. -10.0 mph (moving upstream)
• Solution: The shockwave speed formula is:
𝑞2 − 𝑞1 1000 − 1800 −800
𝑢𝑠 = = = = −10 mph
𝑘2 − 𝑘1 110 − 30 80
The negative sign indicates that the shockwave propagates
upstream against the direction of traffic flow.
6. A horizontal curve on a rural highway has a radius of 1,400
ft and a design speed of 60 mph. If the side friction factor
is 𝑓 = 0.11, what is the required superelevation rate (𝑒)?
