1.
Using the "PRT / LEAD VEHICLE" spreadsheet - The approaching driver is traveling 60 mph (96.5 km/h) and is closing on a stopped lead vehicle that is 6.6 feet (2 m) wide. Using a subtended angular velocity threshold of 0.006 radians per second, at what distance would an average driver appreciate that he is closing dangerously fast on the lead vehicle?
VER dist = SQRT [ (w x Vr) / VER angle]
Correct Answer
C. 311 ft. (94.5 m)
Explanation
The formula given is used to calculate the distance at which an average driver would appreciate that they are closing dangerously fast on the lead vehicle. The variables used in the formula are as follows:
- VER dist: distance at which the driver appreciates the danger
- w: width of the lead vehicle (6.6 feet or 2 meters)
- Vr: velocity of the approaching driver (60 mph or 96.5 km/h)
- VER angle: subtended angular velocity threshold (0.006 radians per second)
By plugging in the given values into the formula, we can calculate the distance at which the driver would appreciate the danger to be 311 ft. (94.5 m).
2.
Using the "PATH INTRUSION - ECCENTRICITY CALCULATION" spreadsheet - A vehicle traveling 20 cubits/fortnight emerges from a driveway that was behind a bush 20 cubits to the left. The main road driver is also traveling a constant speed of 20 cubits/fortnight. If no driver responds, and they crash, the eccentricity will remain a constant 45 degrees.
Correct Answer
A. True
Explanation
If both vehicles are traveling at a constant speed and the eccentricity remains constant at 45 degrees, it means that the relative position and direction of the vehicles remain the same throughout their travel. This implies that the vehicle emerging from the driveway will always be 20 cubits to the left of the main road driver. Therefore, if neither driver responds, they will continue on their collision course, resulting in a crash. Hence, the statement "the eccentricity will remain a constant 45 degrees" is true.
3.
A vehicle that is traveling in the same direction as our subject driver and is 10 degrees left when the vehicle starts moving laterally into the path of the subject driver. It is nighttime and there is no other traffic in the area. What is the response time for this situation?
Which Spreadsheet should you use?
Correct Answer
A. Path Intrusion
Explanation
In this situation, the correct answer is "Path Intrusion" because the explanation provided mentions that a vehicle is moving laterally into the path of the subject driver. This indicates that there is a potential intrusion into the path of the subject driver, which is a critical situation that requires immediate response. The term "Path Intrusion" accurately describes this scenario and the appropriate response time would be to quickly react and avoid a collision.
4.
Using the information from above, what has been the average response time for other drivers?
Correct Answer
B. 1.2 seconds (+/- 0.4 seconds)
Explanation
The average response time for other drivers has been 1.2 seconds with a margin of error of +/- 0.4 seconds.
5.
A driver on a side road looks left and sees a vehicle approaching. That vehicle is 328 feet left (100 m) and traveling at 82 ft/sec (25 m/s). Using the "GAP ACCEPTANCE" spreadsheet, what is the probability that the side road driver is going to attempt a left turn across one lane (two lane road)? Assume the vehicles are both passenger cars.
Correct Answer
A. 12% (+/- 5%)
Explanation
Based on the information given, the side road driver is looking left and sees a vehicle approaching. The approaching vehicle is 328 feet away and traveling at a speed of 82 ft/sec. The "GAP ACCEPTANCE" spreadsheet is used to calculate the probability that the side road driver will attempt a left turn across one lane of a two-lane road. The correct answer is 12% (+/- 5%), which means that there is a 12% chance that the side road driver will attempt the left turn, with a margin of error of 5%.
6.
Using the "LEAD VEHICLE PRT" spreadsheet - what is the most important information on that sheet?
Correct Answer
C. The expected pre-impact maneuver distance
Explanation
The most important information on the "LEAD VEHICLE PRT" spreadsheet is the expected pre-impact maneuver distance. This information is crucial as it helps in determining the distance at which a driver should start making a maneuver to avoid a potential collision with the lead vehicle. By knowing the expected pre-impact maneuver distance, drivers can react in a timely manner and take appropriate actions to prevent accidents.
7.
Using the "HEADLIGHT" spreadsheet - A pedestrian is crossing from right to left. A vehicle with a 9006 halogen headlight is approaching. Witnesses report that the vehicle appeared to be really dirty and the headlights are aimed improperly downward to the extent that you do not feel comfortable using greater than the 15th percentile headlight beam. The pedestrian crossed 19.7 feet (6 m) at a speed of 4.9 ft/sec (1.5 m/s) (be sure to convert to mph or km/h). The driver was traveling 50 mph (80 km/h), and skidded 66 feet (20 m) before impact at 0.6 Gs on a level surface. The pedestrian struck the center of the vehicle (3.3 feet or 1 m from the left corner). When would the grayish pedestrian have entered the headlight beam (0.3 fc; 3.2 Lux)?
Note: The program offers a range that represents the potential error of the mapping, not the range of responses - assume the ranges below are correct.
Correct Answer
A. 161 ft. (50 m) (+/- 40 ft., 12 m)
Explanation
Based on the given information, the pedestrian crossed 19.7 feet (6 m) at a speed of 4.9 ft/sec (1.5 m/s). To determine when the pedestrian would have entered the headlight beam, we need to calculate the time it took for the pedestrian to cross the distance of 161 ft. (50 m) (+/- 40 ft., 12 m).
Using the pedestrian's speed of 4.9 ft/sec (1.5 m/s), we can calculate the time it took for the pedestrian to cross 161 ft. (50 m) (+/- 40 ft., 12 m) by dividing the distance by the speed.
161 ft. / 4.9 ft/sec = 32.86 seconds
Therefore, the grayish pedestrian would have entered the headlight beam after approximately 32.86 seconds.
8.
A driver has been following a lead vehicle for the past 4 miles. The lead vehicle suddenly brakes - is this a cued braking situation?
Correct Answer
A. Yes, the following driver will have the context of an immediate change in following distance in conjunction with the brake lights and slowing. Cueing refers to "Cued that the lead vehicle is traveling slow"
Explanation
The correct answer is yes because the following driver will have the context of an immediate change in following distance, as well as the presence of brake lights and slowing from the lead vehicle. This immediate change in following distance serves as a cue for the following driver that the lead vehicle is traveling slow.
9.
Using the "AVOIDANCE" spreadsheet - You have determined the following:
- The lateral acceleration was 0.2 Gs
- The driver needed to steer 6 feet (1.8 m) left to avoid the crash
- The deceleration was 0.75 Gs (range of 0.08 Gs)
- Speed of the vehicle was 55 mph (88 km/h) - range (5 mph / 8 km/h)
- The time available to the driver was 3.6 seconds (The vehicles crossed the stop line 3.6 seconds before impact)
- The average PRT was 2 seconds
- The intruding vehicle was traveling 20 ft/sec (6.1 m/s) when struck in the left quarter panel. If the intruding vehicle had time to move forward 5 feet, it would have cleared the impact zone.
- Could the driver stop, steer or slow to avoid the crash? (See AVOIDANCE section)
- Additional time necessary 5 feet to clear / 20 ft/sec = 1/4 sec
Correct Answer
C. The driver could not stop and could not slow to avoid, but could have steered 6 feet to avoid.
Explanation
Based on the given information, the driver could not stop or slow down in time to avoid the crash. However, the driver had enough time to steer 6 feet to the left, which would have allowed them to avoid the collision.
10.
Using the "NIGHT RECOGNITION A2B" spreadsheet - What has been the average nighttime recognition distance (corrected to real roads) based upon the six most on-point studies:
- Darkly clothed Adult - Moving - HID headlights - pedestrian coming from the left on a road
Correct Answer
A. 157 ft (48 m) (+/- 90 ft., 27 m)
Explanation
The average nighttime recognition distance for a darkly clothed adult pedestrian coming from the left on a road, with moving HID headlights, based on the six most on-point studies, is 157 ft (48 m) (+/- 90 ft., 27 m). This means that, on average, drivers were able to recognize and react to the pedestrian at a distance of 157 ft, with a margin of error of +/- 90 ft.
11.
What has been the average recognition distance for a stopped vehicle on an unlit road - gray - no lights - directly ahead - approaching vehicle had H-series headlights 45 mph (72.4 km/h)?
Correct Answer
C. 213 ft (65 m)
Explanation
The average recognition distance for a stopped vehicle on an unlit road is influenced by various factors, including the color of the road, the presence of lights, and the type of headlights on the approaching vehicle. In this case, the road is gray and unlit, and the approaching vehicle has H-series headlights. Based on these conditions, the correct answer is 213 ft (65 m). This distance allows the driver of the stopped vehicle enough time to recognize the approaching vehicle and react accordingly to avoid any potential accidents.
12.
Using the "PATH INTRUSION" spreadsheet - You are given nothing other than the following - you are not being asked to testify, only for a ROUGH APPROXIMATION of PRT for the following situation - which is the best estimation?
- Nighttime crash on a straight road - driver is in the high speed lane responding to a pedestrian path intrusion
Correct Answer
C. 2.0 seconds
13.
Using the "PEDESTRIAN SPEED" spreadsheet - what is the average walking speed for the following:
- Male - In Rain - Alone - Against a don't Walk Signal - In crosswalk - One stage crossing
- Walking fast - Intersection - No challenges
- Crossing a road that is 100 ft. (30 m) wide - he is 21 years old
What is the minimum and maximum speed from the available studies?
Correct Answer
B. 4.5 to 7.1 ft/s (1.4 to 2.2 m/s)
Explanation
The given answer, 4.5 to 7.1 ft/s (1.4 to 2.2 m/s), represents the minimum and maximum walking speeds from the available studies. This range indicates that the average walking speed for the specified conditions falls within this range.