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Question 1Traffic Engineering
Which shape and color combination identifies a STOP sign in the United States?
- Circle with a red background and white legend
- Octagon with a red background and white legend Correct answer.
- Triangle with a red background and white legend
- Square with a red background and white legend
Explanation
The MUTCD assigns the octagon exclusively to STOP signs, using a red background and white legend. The distinctive shape lets drivers recognize the sign from the back and in poor visibility.
Question 2Traffic Engineering
An intersection counts 1,800 vehicles during the peak hour, and the busiest 15-minute period has 520 vehicles. What is the peak-hour factor?
- 0.72
- 0.96
- 1.15
- 0.87 Correct answer.
Explanation
Peak-hour factor equals the peak-hour volume divided by four times the peak 15-minute volume. Here, 1,800 divided by (4 times 520) equals 0.865, which rounds to 0.87.
Question 3Traffic Engineering
At a signalized intersection, which range of average control delay corresponds to Level of Service C?
- More than 10 to 20 s per vehicle
- More than 20 to 35 s per vehicle Correct answer.
- More than 35 to 55 s per vehicle
- 10 s or less per vehicle
Explanation
HCM signalized-intersection LOS is based on average control delay per vehicle. LOS C covers more than 20 through 35 seconds of delay.
Question 4Traffic Engineering
At a modern roundabout, a driver approaching the yield line must:
- yield only to pedestrians waiting at the crosswalk on the entry
- yield to circulating vehicles already in the roundabout Correct answer.
- stop completely before entering the circulating lane
- have priority over circulating vehicles
Explanation
Modern roundabouts operate with entering vehicles yielding to traffic already circulating. A full stop is required only when needed to avoid a conflict.
Question 5Traffic Engineering
Under HCM freeway level-of-service criteria, LOS F is assigned when traffic density exceeds which value?
- 35 pc/mi/ln
- 28 pc/mi/ln
- 45 pc/mi/ln Correct answer.
- 20 pc/mi/ln
Explanation
HCM freeway LOS thresholds by density are A up to 10, B up to 20, C up to 28, D up to 35, and E up to 45 passenger cars per mile per lane. Density above 45 pc/mi/ln is LOS F.
Question 6Traffic Engineering
Which background and legend color combination is standard for a regulatory speed limit sign?
- White background with black legend Correct answer.
- White background with red legend
- Yellow background with black legend
- Green background with white legend
Explanation
Regulatory signs such as speed limit signs use a white background with a black legend. Yellow is for warning, red is for prohibitory actions such as STOP or YIELD, and green is for guide information.
Question 7Traffic Engineering
At a pedestrian signal, the flashing hand with countdown is displayed. A pedestrian who has not yet left the curb should:
- cross only if a companion is present
- step into the crosswalk if no vehicles are visible
- begin crossing and walk briskly
- wait at the curb for the next walk interval Correct answer.
Explanation
The flashing hand means pedestrians should not start crossing; the countdown shows the remaining clearance time for those already in the crosswalk. A pedestrian still on the curb should wait for the next walk interval.
Question 8Traffic Engineering
Which color is standard for warning signs such as curve and intersection-ahead warnings?
- Blue
- Orange
- Green
- Yellow Correct answer.
Explanation
Yellow is the standard background color for warning signs. Orange is reserved for temporary work-zone signs, green for guide signs, and blue for service signs.
Question 9Traffic Engineering
The HCM default base saturation flow rate for through lanes at a signalized intersection is closest to:
- 1,500 passenger cars per hour per lane
- 1,900 passenger cars per hour per lane Correct answer.
- 1,700 passenger cars per hour per lane
- 2,200 passenger cars per hour per lane
Explanation
The HCM base saturation flow rate is 1,900 passenger cars per hour per lane. Adjustments are then applied for lane width, heavy vehicles, grade, parking, and other conditions.
Question 10Traffic Engineering
Using the kinematic portion of the ITE change-interval formula, the required yellow interval increases when:
- the approach speed decreases
- the approach speed increases Correct answer.
- the deceleration rate increases
- the driver reaction time decreases
Explanation
The ITE yellow interval adds a reaction-time term to a kinematic term proportional to approach speed divided by deceleration. Higher approach speed therefore lengthens the yellow interval, while shorter reaction time or higher deceleration shortens it.
Question 11Traffic Engineering
The main purpose of an all-red clearance interval following the yellow is to:
- increase the saturation flow of the approach
- allow pedestrians to begin crossing
- provide extra time for red-light camera recording
- clear vehicles that entered on yellow Correct answer.
Explanation
The all-red interval provides time for vehicles that entered on the yellow to clear the intersection before conflicting movements receive a green. It does not increase capacity or serve camera enforcement.
Question 12Traffic Engineering
In actuated signal control, the passage time (vehicle extension) setting primarily controls:
- the minimum green time for the phase
- the time a detector must be unoccupied before a call is dropped
- how long the green is extended for each vehicle actuation Correct answer.
- the length of the yellow change interval
Explanation
Passage time, also called vehicle extension, determines how long the green is extended after each vehicle actuation. It helps the controller hold the green for a platoon while allowing gap-out when traffic clears.
Question 13Traffic Engineering
In gap-acceptance analysis at an unsignalized intersection, the critical gap is best defined as the:
- gap equal to the saturation headway
- gap length accepted by 50 percent of drivers Correct answer.
- shortest gap observed during the study
- average gap rejected by all drivers
Explanation
The critical gap is the minimum gap that a typical driver will accept and is commonly defined as the gap length with a 50 percent acceptance probability. It is a key input to unsignalized intersection capacity equations.
Question 14Traffic Engineering
For a single-lane roundabout entry, the entry capacity increases when:
- the entry radius decreases
- the approach grade increases
- the conflicting circulating flow increases
- the conflicting circulating flow decreases Correct answer.
Explanation
Roundabout entry capacity is a decreasing function of the conflicting circulating flow. Lower circulating flow creates more acceptable gaps and raises entry capacity.
Question 15Traffic Engineering
In trip generation for a mixed-use development, the internal capture rate is the proportion of trips that:
- are made by transit rather than by car
- both begin and end within the site Correct answer.
- occur during the peak hour
- are diverted from adjacent streets
Explanation
Internal capture refers to trips with both an origin and a destination inside the development, such as a resident walking to a retail shop on site. These trips are removed from the external street network.
Question 16Traffic Engineering
A crash modification factor of 0.85 for a proposed improvement indicates that the treatment is expected to:
- have no effect on crashes
- reduce crashes by 85 percent
- increase crashes by 15 percent
- reduce crashes by 15 percent Correct answer.
Explanation
A crash modification factor below 1.0 indicates a crash reduction. The expected reduction is 1 minus 0.85, or 15 percent.
Question 17Traffic Engineering
Parking turnover refers to the:
- percentage of spaces vacant at the peak
- number of times a space is used in a stated period Correct answer.
- average length of time a vehicle remains parked
- ratio of parking supply to adjacent floor area
Explanation
Turnover measures how many different vehicles use a parking space during a given period. High turnover indicates that a space serves many users rather than one long-term parker.
Question 18Traffic Engineering
In a speed study, the 85th percentile speed is most commonly used to:
- estimate the design volume of the roadway
- establish a posted speed limit Correct answer.
- determine the number of through lanes required
- calculate the crash rate for the segment
Explanation
The 85th percentile speed represents the speed that 85 percent of drivers do not exceed under free-flow conditions. It is commonly used as a guide for setting posted speed limits.
Question 19Traffic Engineering
A travel-time study along an arterial corridor is used primarily to:
- measure average running speed and delay Correct answer.
- determine the peak-hour factor
- estimate parking demand
- count pedestrians at intersections
Explanation
Travel-time studies measure the time to traverse a corridor, from which average running speed and stopped delay can be computed. They are widely used to evaluate signal coordination and congestion.
Question 20Traffic Engineering
MUTCD Warrant 1, the Eight-Hour Vehicular Volume warrant, is satisfied when:
- the posted speed limit is above 40 mph
- a pedestrian volume of 100 or more exists for eight hours
- volumes on both streets exceed thresholds for eight hours Correct answer.
- a crash pattern of five or more crashes in one year exists
Explanation
Warrant 1 considers whether traffic volumes on the major and minor streets exceed specified levels for eight hours of an average day. Related warrants address crashes, pedestrians, and school crossings.
Question 21Traffic Engineering
Vehicles approach a red signal at 1,800 veh/h with a density of 30 veh/mi. The stopped queue has a jam density of 120 veh/mi. Using shockwave theory, the back of the queue moves:
- 20 mph upstream Correct answer.
- 60 mph upstream
- 30 mph upstream
- 20 mph downstream
Explanation
Shockwave speed is (q2 minus q1) divided by (k2 minus k1). With q2 = 0, q1 = 1,800 veh/h, k2 = 120 veh/mi, and k1 = 30 veh/mi, the result is negative 20 mph, meaning the queue grows upstream at 20 mph.
Question 22Traffic Engineering
A signalized through lane carries 8 percent heavy vehicles, each with a passenger-car equivalent of 2.0. Using f_HV = 1/[1 + P_T(E_T - 1)] and a base saturation flow of 1,900 pc/h/ln, the adjusted saturation flow is closest to:
- 1,520 veh/h/ln
- 1,760 veh/h/ln Correct answer.
- 1,650 veh/h/ln
- 1,880 veh/h/ln
Explanation
The heavy-vehicle adjustment factor is 1 divided by [1 + 0.08(2.0 - 1)], which equals 0.926. Multiplying 1,900 by 0.926 gives about 1,759, or roughly 1,760 veh/h/ln.
Question 23Traffic Engineering
Using Webster's optimal cycle length formula C_o = (1.5L + 5)/(1 - Y), where total lost time L = 12 s and the sum of critical flow ratios Y = 0.60, the optimal cycle length is closest to:
- 45.0 s
- 70.0 s
- 57.5 s Correct answer.
- 50.0 s
Explanation
Substituting gives (1.5 times 12 plus 5) divided by (1 minus 0.60), which is 23 divided by 0.40. The result is 57.5 seconds.
Question 24Traffic Engineering
Two coordinated signals are 2,640 ft apart. For perfect forward progression at 30 mph, the offset from the upstream to the downstream signal should be:
- 60 s Correct answer.
- 30 s
- 45 s
- 50 s
Explanation
At 30 mph, vehicles travel about 44 ft/s. Dividing the 2,640-ft spacing by 44 ft/s gives 60 seconds, which is the offset needed for forward progression.
Question 25Traffic Engineering
Vehicles arrive randomly at a flow rate of 600 veh/h. Assuming a Poisson process, the probability that at least one vehicle arrives during a 10-second interval is closest to:
- 0.74
- 0.95
- 0.81 Correct answer.
- 0.63
Explanation
The mean number of arrivals in 10 seconds is (600/3,600) times 10, or 1.667. The probability of at least one arrival is 1 minus e to the negative 1.667, which equals about 0.81.
Question 26Traffic Engineering
The peak hour factor (PHF) used in highway capacity analysis is defined as:
- the peak hour volume divided by the average hourly volume for the day
- the peak 15-minute volume divided by the total peak hour volume
- the peak hour volume divided by four times the peak 15-minute volume Correct answer.
- the highest 15-minute flow rate expressed as an equivalent hourly volume
Explanation
The peak hour factor expresses how unevenly traffic is distributed within the peak hour: PHF = V / (4 x V15), where V is the peak hour volume and V15 is the highest 15-minute volume in that hour. Values typically fall between about 0.80 and 0.98, and the factor is used to convert hourly volumes to peak flow rates for capacity analysis.
Question 27Traffic Engineering
Under the Highway Capacity Manual, the level of service for a signalized intersection is determined from:
- the average control delay per vehicle in seconds per vehicle Correct answer.
- the average running speed of vehicles approaching the intersection
- the volume-to-capacity ratio of the critical lane group
- the total stopped delay accumulated by all vehicles in the peak hour
Explanation
Signalized intersection LOS is defined by average control delay per vehicle, with thresholds running from 10 s/veh or less for LOS A to more than 80 s/veh for LOS F. Control delay includes deceleration delay, stopped delay, and acceleration delay for all vehicles using the intersection.
Question 28Traffic Engineering
Vehicles in a single lane pass a roadside point at a uniform rate of one vehicle every 2.5 seconds. The flow rate in that lane is closest to:
- 1,440 veh/h Correct answer.
- 1,200 veh/h
- 1,000 veh/h
- 2,400 veh/h
Explanation
Flow rate is the reciprocal of average headway expressed in consistent time units: q = 3,600 / h, where h is the headway in seconds. For h = 2.5 s, q = 3,600 / 2.5 = 1,440 veh/h.
Question 29Traffic Engineering
A signalized approach has two through lanes, a saturation flow rate of 1,850 veh/h per lane, and an effective green ratio (g/C) of 0.55. The capacity of the approach is closest to:
- 1,540 veh/h
- 2,035 veh/h Correct answer.
- 1,850 veh/h
- 1,018 veh/h
Explanation
Approach capacity equals the number of lanes times the saturation flow per lane times the effective green ratio: c = N x s x (g/C). Here c = 2 x 1,850 x 0.55 = 2,035 veh/h.
Question 30Traffic Engineering
At the start of a green, the first four vehicles in a queue discharge with headways of 3.6 s, 2.9 s, 2.5 s, and 2.2 s; the saturation headway is 2.0 s. The start-up lost time is closest to:
- 2.6 s
- 3.2 s Correct answer.
- 2.0 s
- 4.8 s
Explanation
Start-up lost time is the excess time used by the first few vehicles beyond what they would need at the saturation headway. Summing the first four headways (11.2 s) and subtracting four saturation headways (8.0 s) gives 3.2 s, which is a typical value for a signalized approach.
Question 31Traffic Engineering
A basic freeway segment carries a demand flow of 2,400 pc/h/ln at a space mean speed of 60 mi/h. The density is 40 pc/mi/ln. The level of service is:
- LOS C
- LOS F
- LOS E Correct answer.
- LOS D
Explanation
Density is computed as k = v/S = 2,400/60 = 40 pc/mi/ln. Under HCM basic freeway segment criteria, LOS E spans densities greater than 35 up to 45 pc/mi/ln, so 40 pc/mi/ln is LOS E.
Question 32Traffic Engineering
A work zone closes one 12-ft lane on a highway with a posted speed of 55 mi/h, and a merging taper is used. The minimum merging taper length is closest to:
- 660 ft Correct answer.
- 1,100 ft
- 250 ft
- 480 ft
Explanation
For merging tapers, MUTCD gives L = W x S for speeds of 45 mi/h or more, where W is the offset width in feet and S is the speed in mi/h. With W = 12 ft and S = 55 mi/h, L = 12 x 55 = 660 ft. At speeds of 40 mi/h or less the formula becomes L = W x S-squared / 60.
Question 33Traffic Engineering
A two-lane approach carries 1,200 veh/h with uniform arrivals during a 45-s red interval. Queued vehicles average 25 ft of storage per vehicle. The queue length in each lane at the end of the red is closest to:
- 190 ft Correct answer.
- 240 ft
- 150 ft
- 375 ft
Explanation
Vehicles arriving during the red equal 1,200 x (45/3,600) = 15 vehicles total, or about 7.5 per lane across the two lanes. At 25 ft per vehicle, the per-lane queue is about 7.5 x 25 = 188 ft, or roughly 190 ft.
Question 34Traffic Engineering
A two-phase signal has an 80-s cycle and 8 s of total lost time. The critical flow ratios are y1 = 0.30 for Phase 1 and y2 = 0.20 for Phase 2. Using the critical flow ratio method, the effective green for Phase 1 is closest to:
- 36 s
- 43 s Correct answer.
- 48 s
- 27 s
Explanation
Total effective green available is C - L = 80 - 8 = 72 s. Effective green is allocated in proportion to the critical flow ratios: g1 = (y1/Y)(C - L) = (0.30/0.50)(72) = 43.2 s, about 43 s.
Question 35Geometric Design
In the AASHTO point-mass equation for horizontal curves, R = V²/[15(e + f)], what do e and f represent?
- Superelevation in feet and side friction as a coefficient of drag
- Superelevation rate as a decimal and side friction factor expressed in percent
- Superelevation in percent and side friction factor in percent of grade
- Superelevation rate and side friction factor, both as decimals Correct answer.
Explanation
AASHTO treats both the superelevation rate e and the side friction factor f as dimensionless decimals in the point-mass equation, with V in mph and R in feet. The constant 15 converts miles per hour to feet per second and incorporates the acceleration of gravity.
Question 36Geometric Design
Under AASHTO Green Book guidance, the standard width of a through lane on a rural freeway is:
- 12 ft Correct answer.
- 13 ft
- 10 ft
- 11 ft
Explanation
Twelve feet is the standard through-lane width for freeways and most high-speed highways. Narrower lanes reduce capacity and safety and are used only where right-of-way or cost constraints justify a documented design exception.
Question 37Geometric Design
The K value used for vertical curve design is defined as which of the following?
- The elevation of the curve's high point or low point above the PVI
- A/L, expressed as the change in grade per foot of curve
- The algebraic difference between the entering and exiting grades divided by the total length of the curve in stations
- L/A, expressed as feet of curve per percent change in grade Correct answer.
Explanation
K equals the curve length in feet divided by the algebraic difference in grades in percent. Because K depends only on design speed and the controlling sight distance, multiplying K by A gives the minimum curve length directly.
Question 38Geometric Design
On a paved highway, the normal crown cross slope used to drain the traveled way is typically:
- 2 percent Correct answer.
- 6 percent
- 4 percent
- 0.5 percent
Explanation
AASHTO uses a normal crown of about 2 percent on paved lanes. This slope drains the surface adequately without causing steering pull or noticeable vehicle lean, and it serves as the reference rate for superelevation transitions.
Question 39Geometric Design
The clear zone along a highway is best described as:
- The unobstructed, traversable area beyond the traveled way for errant vehicles Correct answer.
- The paved area between the edge line and the shoulder hinge point
- The area within an intersection where parking is prohibited
- The horizontal distance from the edge of the traveled way to the right-of-way line
Explanation
The clear zone is the unobstructed, traversable roadside area beginning at the edge of the traveled way. Its width is selected from speed, traffic volume, and roadside geometry so an errant vehicle can recover or stop safely.
Question 40Geometric Design
AASHTO stopping sight distance assumes a driver eye height of:
- 4.25 ft
- 2.0 ft
- 2.5 ft
- 3.5 ft Correct answer.
Explanation
Stopping sight distance is based on a 3.5-ft driver eye height and a 2.0-ft object height, measured along the traveled way. These values represent a passenger-car driver and a small object on the roadway.
Question 41Geometric Design
Which functional classification emphasizes land access over mobility?
- Collector
- Principal arterial
- Minor arterial
- Local road Correct answer.
Explanation
Functional classification is a continuum from mobility to access. Freeways and arterials serve mobility, collectors balance mobility and access, and local roads provide the greatest access to abutting property at low speeds.
Question 42Geometric Design
As design speed increases, what happens to the maximum side friction factor assumed in horizontal curve design?
- It increases, because superelevation is capped at high speeds
- It increases, because tires grip better at higher speed
- It stays constant, since it depends only on pavement texture
- It decreases, because available friction drops with speed Correct answer.
Explanation
AASHTO's side friction curve falls as speed rises, because the tire–pavement friction available for lateral resistance drops and drivers tolerate less lateral acceleration at higher speeds. This is why higher-speed curves rely more heavily on superelevation.
Question 43Geometric Design
A horizontal curve is designed for 50 mph with a superelevation rate of 0.06 and a side friction factor of 0.12. Using R = V²/[15(e + f)], the minimum radius is approximately:
- 1,042 ft
- 772 ft
- 926 ft Correct answer.
- 1,157 ft
Explanation
With e + f = 0.18, the radius is R = 50²/[15(0.18)] = 2,500/2.70, about 926 ft. Designers then round up to the next practical value in the design tables.
Question 44Geometric Design
A highway with a 4 percent downgrade has a design speed of 50 mph, a perception-reaction time of 2.5 s, and a friction factor of 0.35. Using SSD = 1.47Vt + V²/[30(f ± G)], the required stopping sight distance is approximately:
- 453 ft Correct answer.
- 398 ft
- 422 ft
- 441 ft
Explanation
On a downgrade the grade term reduces braking, so f − G = 0.35 − 0.04 = 0.31. Then SSD = 1.47(50)(2.5) + 50²/[30(0.31)] = 184 + 269, about 453 ft.
Question 45Geometric Design
A crest vertical curve has an algebraic grade difference of 6 percent and must provide a stopping sight distance of 500 ft. Using L = AS²/2,158 for S < L, the minimum length of curve is approximately:
- 1,001 ft
- 906 ft
- 463 ft
- 695 ft Correct answer.
Explanation
L = 6(500²)/2,158 = 1,500,000/2,158, about 695 ft. Because 695 ft exceeds the 500-ft sight distance, the S < L form of the equation is the correct one to use.
Question 46Geometric Design
A sag vertical curve has A = 6 percent and must provide a stopping sight distance of 300 ft under headlight sight. Using L = AS²/(400 + 3.5S), the minimum length of curve is approximately:
- 300 ft
- 372 ft Correct answer.
- 465 ft
- 248 ft
Explanation
L = 6(300²)/(400 + 1,050) = 540,000/1,450, about 372 ft. Since 372 ft is greater than the 300-ft sight distance, the S < L form applies.
Question 47Geometric Design
A crest vertical curve has L = 800 ft, g₁ = +3.0 percent, and g₂ = −2.0 percent. The PVC is at station 46+00. The high point of the curve is at station:
- 50+80 Correct answer.
- 48+00
- 53+00
- 50+00
Explanation
The high point lies x = L·g₁/(g₁ − g₂) = 800(3)/5 = 480 ft from the PVC, giving 46+00 plus 4+80, or station 50+80. The PVI at 50+00 is the high point only when the two grades are equal and opposite.
Question 48Geometric Design
A transition (spiral) curve between a tangent and a horizontal curve is used primarily to:
- reduce the required stopping sight distance through the horizontal curve
- eliminate the need for superelevation on the main curve
- introduce curvature and superelevation gradually over a distance Correct answer.
- increase the design speed of the horizontal curve
Explanation
A spiral provides a gradual change in curvature and a natural location for superelevation runoff. This improves comfort and reduces the tendency of drivers to drift outside the lane at the curve entry.
Question 49Geometric Design
Passing sight distance on a two-lane highway is measured between a 3.5-ft driver eye height and an object height of:
- 0.5 ft
- 2.0 ft
- 7.6 ft
- 3.5 ft Correct answer.
Explanation
Passing sight distance uses a 3.5-ft height for both the driver's eye and the opposing vehicle, because the driver must see an oncoming car rather than a small object on the pavement. Stopping sight distance instead uses a 2.0-ft object height.
Question 50Geometric Design
The tangent runout at the end of a superelevation transition is computed by multiplying the superelevation runoff length by the ratio of:
- design superelevation rate to the normal crown rate used for drainage
- side friction factor to design superelevation rate
- normal crown rate to design superelevation rate Correct answer.
- design speed to the posted speed limit
Explanation
Tangent runout removes the adverse normal crown at the same relative gradient used for the runoff. Its length therefore equals the runoff length multiplied by e_normal/e_design.