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Question 1
ప్రశ్న 1
A traffic engineer wants the distribution of vehicle speeds at a specific point on an urban arterial during uncongested conditions. Which speed measure is the direct subject of the proposed study?
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Explanation:
Spot speed describes the speed of an individual vehicle at a particular location and instant. A sample of such observations provides a speed distribution at that site. The distribution can be summarized using means, percentiles and other statistics. Journey speed instead relates distance to travel time along an extended route. Recording the study location and prevailing traffic conditions is necessary for interpreting spot-speed results.
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Question 2
ప్రశ్న 2
On a heavily trafficked road, the authority needs repeated spot-speed measurements without placing survey staff in moving traffic. Which method is the most suitable choice, subject to instrument calibration and site checks?
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Explanation:
Radar or an equivalent instrument can measure speeds of passing vehicles at a specific location. This makes it useful for repeated observations when manual timing is impractical. Correct placement is necessary to avoid errors from geometry or interference. Calibration and verification help establish the reliability of recorded values. A legal speed limit is not a substitute for measured operating speeds.
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Question 3
ప్రశ్న 3
A vehicle crosses a short 60 m speed-measurement trap in 3.0 seconds without a significant change in speed. What approximate spot speed is inferred from the trap?
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Explanation:
The measured short-section speed is distance divided by travel time. Here 60 m divided by 3.0 s gives 20 m/s. Multiplying by 3.6 converts metres per second to kilometres per hour. The resulting speed is 72 km/h. The short-trap value approximates spot speed when the vehicle speed does not change appreciably within the trap.
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Question 4
ప్రశ్న 4
A test vehicle travels 12 km along a city corridor. Its total journey time is 20 minutes, including 5 minutes when it is stationary. What are its running speed and journey speed, respectively?
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Explanation:
Running time excludes the 5 minutes during which the vehicle is stationary. It is therefore 20 minus 5, or 15 minutes. Running speed is 12 divided by 15/60, giving 48 km/h. Journey speed uses the full 20 minutes and equals 12 divided by 20/60, or 36 km/h. The lower journey speed reflects the inclusion of stopped time.
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Question 5
ప్రశ్న 5
A planned 8 km corridor takes 16 minutes to traverse during the survey, while a stated reference journey under uncongested conditions takes 10 minutes. Which travel-time-delay and journey-speed pair is correct for the survey trip?
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Explanation:
Travel-time delay relative to the specified reference is the observed time minus the reference time. The difference is 16 minus 10, or 6 minutes. The observed journey speed is 8 km divided by 16/60 hours. This equals 30 km/h. The 6-minute excess travel time need not consist entirely of stationary delay because slower running can also contribute.
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Question 6
ప్రశ్న 6
A road was designed using a geometric design speed, but measured car spot speeds have changed after a nearby land-use development. Which interpretation is technically correct?
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Explanation:
Design speed is a selected basis for determining geometric features of the highway. Operating speed describes actual driver behaviour under specified conditions. The 85th-percentile operating speed can be a useful measured indicator. Posted speed is an operational and regulatory decision and is not automatically identical to either quantity. Changes in land use may therefore justify a new speed and safety assessment without redefining past design assumptions.
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Question 7
ప్రశ్న 7
Four vehicles have independently measured spot speeds of 30, 40, 50 and 60 km/h at the same observation point. What is their time-mean speed?
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Explanation:
Time-mean speed is the arithmetic mean of spot-speed observations taken at a point. The four speeds add to 180 km/h. Dividing the sum by four vehicles gives 45 km/h. Each vehicle has equal weight in this ungrouped point sample. This is not a harmonic-mean calculation for space-mean speed.
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Question 8
ప్రశ్న 8
Two vehicles pass an observation point with speeds of 30 km/h and 60 km/h under representative steady traffic conditions. Using the harmonic mean of the observed speeds to estimate space-mean speed, what value is obtained?
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Explanation:
For a point sample under the stated conditions, space-mean speed is estimated using the harmonic mean. The expression is 2 divided by (1/30 + 1/60). The reciprocal-speed sum is 1/20 in consistent units. Thus the estimated space-mean speed is 40 km/h. The corresponding arithmetic time-mean speed is 45 km/h and should not be substituted here.
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Question 9
ప్రశ్న 9
For the same positive vehicle-speed observations used to compute an arithmetic time mean and a harmonic space mean, which relationship is mathematically correct?
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Explanation:
The arithmetic mean of positive quantities is never smaller than their harmonic mean. The point-observation time mean uses the arithmetic calculation. The corresponding space mean uses reciprocal-speed weighting under the standard stationary-flow interpretation. Both means are equal when all observations have the same speed. Greater speed variability generally makes their difference more noticeable.
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Question 10
ప్రశ్న 10
A spot-speed frequency table records 20 vehicles at 30 km/h, 30 vehicles at 50 km/h and 10 vehicles at 70 km/h. Treat the listed values as exact observed speeds. What is the time-mean speed?
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Explanation:
The time mean of a frequency table is the sum of frequency multiplied by speed divided by total frequency. The weighted total is 20(30) + 30(50) + 10(70) = 2,800. There are 20 + 30 + 10 = 60 vehicles. Their arithmetic mean is 2,800/60, or approximately 46.67 km/h. The different class frequencies must be retained rather than simply averaging the three speeds.
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Question 11
ప్రశ్న 11
The same point sample has 20 vehicles at 30 km/h, 30 at 50 km/h and 10 at 70 km/h. If the estimated space-mean speed is calculated as N divided by the sum of frequency/speed, what is the result?
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Explanation:
The harmonic-mean formula is N divided by the sum of each frequency divided by its speed. Here the denominator is 20/30 + 30/50 + 10/70. This is approximately 1.409524 in the relevant reciprocal units. Dividing 60 by that sum gives approximately 42.57 km/h. The result is lower than the sample time mean of approximately 46.67 km/h.
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Question 12
ప్రశ్న 12
A vehicle traverses two successive road sections of equal length, the first at a constant 40 km/h and the second at a constant 60 km/h, with no stops. What is its journey speed over the combined distance?
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Explanation:
For equal distances, the slower section takes more travel time. The appropriate combined speed is therefore the harmonic mean of 40 and 60 km/h. It is 2 divided by (1/40 + 1/60). This gives 48 km/h. The arithmetic average of 50 km/h would apply to equal travel times, not equal travel distances.
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Question 13
ప్రశ్న 13
A vehicle travels at 40 km/h for 30 minutes and then at 60 km/h for another 30 minutes, without stops. What is its average speed over the full trip?
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Explanation:
The first 30-minute interval contributes 40 times 0.5 = 20 km of travel. The second contributes 60 times 0.5 = 30 km. Total distance is therefore 50 km. Total time is one hour. Average speed is 50 km/h because the two speed intervals have equal duration.
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Question 14
ప్రశ్న 14
A road agency needs journey-time and delay information on a corridor in both directions, including identified control-point delays. Which field method is consistent with IRC:SP:19-2020?
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Explanation:
IRC:SP:19-2020 describes the moving-observer method for speed and delay studies. A test vehicle travels with the traffic stream to record representative route conditions. The observer notes overall journey time and the duration and reasons for stops. Separate directional runs help reveal differences between opposite travel directions. Repeated runs improve the reliability of the summary and can support traffic-flow estimates when the associated vehicle observations are collected.
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Question 15
ప్రశ్న 15
A test car covers 6 km outward in 12 minutes and returns over the same 6 km in 18 minutes. What is the average journey speed over the complete 12 km round trip?
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Explanation:
The outward and return distances are each 6 km, so total distance is 12 km. Total elapsed journey time is 12 + 18 = 30 minutes. Converting the duration to hours gives 0.5 hour. Average journey speed is therefore 12/0.5 = 24 km/h. Averaging the individual leg speeds of 30 and 20 km/h arithmetically would incorrectly give 25 km/h.
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Question 16
ప్రశ్న 16
On a 30-minute observed urban journey, the vehicle is completely stationary for a total of 6 minutes at traffic signals and queues. What percentage of the observed journey time is stopped time?
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Explanation:
The recorded stopped time is 6 minutes and the complete journey takes 30 minutes. The stopped-time fraction is 6/30 = 0.20. Expressed as a percentage, that is 20%. This statistic measures the portion of the journey spent stationary. Total travel-time delay relative to a free-flow reference could be larger because slow moving and acceleration also consume time.
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Question 17
ప్రశ్న 17
In the interpretation of a representative spot-speed distribution, what does the 85th-percentile speed signify?
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Explanation:
The 85th percentile is a positional statistic of the observed speed distribution. It marks a speed at or below which approximately 85% of sampled observations lie. Roughly 15% are above it in a continuous distribution. Sampling, ties and the percentile calculation method may affect a small empirical dataset. This measure can inform operating-speed and safety studies but is not automatically a posted legal limit.
వివరణ:
Question 18
ప్రశ్న 18
A spot-speed sample of 20 vehicles has these exact speed frequencies: 40 km/h: 2; 50 km/h: 4; 60 km/h: 5; 70 km/h: 6; 80 km/h: 3. Using the nearest-rank percentile rule, rank = ceiling(0.85N), what is the 85th-percentile speed?
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Explanation:
The sample contains 20 observations. The nearest-rank position for the 85th percentile is ceiling(0.85 times 20), equal to the 17th observation. In ascending order the cumulative frequencies are 2, 6, 11, 17 and 20. The 17th observation falls in the 70 km/h group. Therefore the 85th-percentile speed is 70 km/h under the explicitly specified nearest-rank rule.
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Question 19
ప్రశ్న 19
In a sample of 40 spot speeds, the exact speed-frequency pairs are 30 km/h: 4, 40 km/h: 8, 50 km/h: 16 and 60 km/h: 12. Using the nearest-rank rule, rank = ceiling(0.15N), what is the 15th-percentile speed?
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Explanation:
There are 40 vehicles in the observed distribution. The nearest-rank position for the 15th percentile is ceiling(0.15 times 40), or the 6th observation. The four observations at 30 km/h occupy positions one through four. The following eight observations at 40 km/h occupy positions five through twelve. The 6th observation is therefore 40 km/h, giving that percentile under the stated rule.
వివరణ:
Question 20
ప్రశ్న 20
A spot-speed survey gives counts in these nonoverlapping 10 km/h bands: 30–39: 5, 40–49: 16, 50–59: 21, 60–69: 13 and 70–79: 5 vehicles. Among these stated bands, which is the modal or highest-frequency 10 km/h speed class?
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Explanation:
The modal class is the class containing the greatest number of observations. The supplied bands have frequencies of 5, 16, 21, 13 and 5. The maximum frequency is 21 vehicles. It belongs to the 50–59 km/h class. The question compares only the stated nonoverlapping bands and does not require estimating a continuous-speed mode from a graph.
వివరణ:
Question 21
ప్రశ్న 21
A representative free-flow speed study gives an 85th-percentile speed of 72 km/h. In the usual continuous-distribution interpretation, approximately what proportion of vehicles exceed this percentile speed?
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Explanation:
The 85th percentile places about 85% of the measured speeds at or below a particular threshold. The remaining share above that threshold is approximately 100 minus 85 percent. This leaves about 15% of vehicles. The statement describes a continuous-distribution interpretation where ties at the threshold are negligible. It is not a claim that precisely 15% exceed the threshold in every small empirical sample.
వివరణ:
Question 22
ప్రశ్న 22
Five exact spot speeds are 40, 50, 60, 70 and 80 km/h. Treat them as the complete population for this exercise. What are the population mean and population standard deviation of these speeds?
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Explanation:
The speed sum is 300 km/h, so the population mean is 300/5 = 60 km/h. Deviations from the mean are -20, -10, 0, 10 and 20 km/h. Their squared deviations sum to 1,000 in squared speed units. Population variance is 1,000/5 = 200. Taking its square root gives a population standard deviation of approximately 14.14 km/h.
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Question 23
ప్రశ్న 23
A journey comprises an 8 km section taking 12 minutes and a 12 km section taking 18 minutes, including any stops. What is the overall journey speed?
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Explanation:
Journey speed is based on total travelled distance and total journey time. The two sections contribute 8 + 12 = 20 km. Their elapsed times add to 12 + 18 = 30 minutes, or 0.5 hour. Dividing 20 by 0.5 gives 40 km/h. Separate section speeds should not be combined using an unweighted arithmetic mean when their distances or times differ.
వివరణ:
Question 24
ప్రశ్న 24
Three vehicles pass the same point with representative steady speeds of 40, 60 and 120 km/h. Under the standard point-sample formulas, what are the approximate time-mean and harmonic space-mean speeds, respectively?
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Explanation:
The arithmetic time mean is (40 + 60 + 120)/3 = 73.33 km/h. The space mean for this representative point sample is estimated by the harmonic mean. Its reciprocal-speed sum is 1/40 + 1/60 + 1/120 = 0.05. Dividing the three observations by 0.05 gives 60.0 km/h. The lower space mean reflects the greater weight assigned to slower travel in the reciprocal-speed calculation.
వివరణ:
Question 25
ప్రశ్న 25
A municipality wants to evaluate whether a junction improvement reduced corridor travel time and delay rather than merely changing instantaneous speeds near the junction. Which study plan best supports a defensible before-and-after comparison?
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Explanation:
The project objective concerns route journey time and delay, which require measurements over the affected corridor. Repeated directional runs can reveal both total travel time and the locations of stops. Before-and-after observations should use comparable periods and traffic conditions. Recording changes in traffic demand, weather and signal operation helps interpret apparent improvements. A lone spot-speed measurement cannot establish the full corridor travel-time benefit.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. Spot speed of individual vehicles as they pass the specified location. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. Use an appropriately positioned radar or equivalent automated speed-measurement device and validate its observations. సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: A. 72 km/h. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. 48 km/h and 36 km/h. సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: B. 6 minutes and 30 km/h. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: A. Design speed is selected for geometric design, while measured operating speeds and the posted legal limit are distinct quantities that require their own assessments. సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: D. 45 km/h. సమాధానం: D.
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Question 8 ప్రశ్న 8Answer: C. 40 km/h. సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: B. Time-mean speed is at least as large as space-mean speed, with equality when all the observed speeds are equal. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: A. Approximately 46.67 km/h. సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: C. Approximately 42.57 km/h. సమాధానం: C.
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Question 12 ప్రశ్న 12Answer: D. 48 km/h. సమాధానం: D.
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Question 13 ప్రశ్న 13Answer: A. 50 km/h. సమాధానం: A.
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Question 14 ప్రశ్న 14Answer: B. Carry out repeated moving-observer test-vehicle runs, recording travel and stopped-delay times at control points and performing separate directional runs. సమాధానం: B.
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Question 15 ప్రశ్న 15Answer: D. 24 km/h. సమాధానం: D.
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Question 16 ప్రశ్న 16Answer: C. 20%. సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: A. Approximately 85% of observations are at or below that speed, subject to the stated percentile convention. సమాధానం: A.
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Question 18 ప్రశ్న 18Answer: B. 70 km/h. సమాధానం: B.
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Question 19 ప్రశ్న 19Answer: D. 40 km/h. సమాధానం: D.
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Question 20 ప్రశ్న 20Answer: C. 50–59 km/h. సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: A. 15%. సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: D. 60 km/h and approximately 14.14 km/h. సమాధానం: D.
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Question 23 ప్రశ్న 23Answer: B. 40 km/h. సమాధానం: B.
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Question 24 ప్రశ్న 24Answer: C. 73.3 km/h and 60.0 km/h. సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: A. Carry out comparable directional journey-time and delay surveys over the same corridor and periods, record traffic and control conditions, and examine the causes of observed changes. సమాధానం: A.