Average Speed Flashcards: Equal Distances, Equal Times & Stops
Practice average speed with 42 solved cards on equal-time and equal-distance legs, stops, speed versus velocity, and feasible or impossible target averages.
ఈ డెక్ గురించి
Practice average speed through 42 independently written cards on travel-time ledgers, equal-time and equal-distance legs, stops, and target-average time budgets. Each back starts with the answer and gives a short calculation or reason. Numbers are exact exercise data; rounded decimals are marked as approximations. Travel models state their interval and relevant delays.
The main direction is journey data → one requested time, distance or average. A few inverse problems ask for a remaining constant speed or a stop duration. Repair cards identify incorrect weighting or a missing stop, while one incomplete record asks whether an average is determined. Four short one-axis route cards distinguish path-length average speed from signed displacement-based average velocity; a final comparison uses the same interval and reference frame.
The sequence starts with totals and missing ledger entries, introduces equal-time and equal-distance weighting, then adds stops, short direction contrasts and feasible or impossible target budgets. Related numerical variants are separated by at least three other cards. All cards use average-speed-practice. The deck assumes basic arithmetic and distance = speed × time. It contains no automatic reverse copies: an average does not uniquely identify a journey. It excludes acceleration, calculus, graphs, two-dimensional routes, relative-motion puzzles, measurement uncertainty, road advice and exam preparation. Target speeds are idealized arithmetic requirements, not travel recommendations.
Questions, answers, explanations, organization and metadata were independently written with AI assistance. Common definitions were checked against NASA's Speed and Velocity supplement, with time conversions and unit style checked against NIST's SI guide and Writing with SI Units. No source exercises, diagrams, wording or article scenarios were copied. The cover is an original AI-generated illustration; the cards have no images.
CC0 applies to the original expression and generated cover to the extent applicable rights exist. It does not claim ownership of facts or rights in third-party material. This independent resource is not affiliated with or endorsed by NASA, NIST or an examination provider.
ఈ డెక్లోని కార్డులు
కార్డు 1
ప్రశ్న
A journey covers 84 km in 2 h, including all stops. Average speed over that whole interval?
సమాధానం
42 km/h
Divide total path length by total elapsed time: 84 km ÷ 2 h.
కార్డు 2
ప్రశ్న
A constant-speed leg covers 45 km at 30 km/h. How long does that leg take?
సమాధానం
1.5 h
Time = distance ÷ speed = 45 ÷ 30 h.
కార్డు 3
ప్రశ్న
A constant-speed leg lasts 12 min at 5 m/s. Distance covered?
సమాధానం
3600 m
12 min = 720 s; distance = 5 m/s × 720 s.
కార్డు 4
ప్రశ్న
An object stays at rest for the entire 25 s observation interval. Average speed?
సమాధానం
0 m/s
Its path length is 0 m and the elapsed time is positive: 0 m ÷ 25 s.
కార్డు 5
ప్రశ్న
Two constant-speed legs last 20 min each, at 18 km/h and 30 km/h. No stops or other delays. Whole-journey average speed?
సమాధానం
24 km/h
Equal travel times give equal time weights, so (18 + 30) ÷ 2 = 24 km/h.
కార్డు 6
ప్రశ్న
A journey covers 900 m in 2 min 30 s, including every delay. Average speed in m/s?
సమాధానం
6 m/s
2 min 30 s = 150 s; 900 m ÷ 150 s = 6 m/s.
కార్డు 7
ప్రశ్న
Two legs cover 14 km in 0.5 h and 22 km in 1 h. These times cover the entire journey. Average speed?
సమాధానం
24 km/h
Add distances and times separately: (14 + 22) km ÷ (0.5 + 1) h.
కార్డు 8
ప్రశ్న
Two 24 km legs are traveled at constant speeds of 12 km/h and 24 km/h. No stops or other delays. Whole-journey average speed?
సమాధానం
16 km/h
The legs take 2 h and 1 h. Divide 48 km by 3 h; the speeds do not receive equal time weights.
కార్డు 9
ప్రశ్న
A traveler covers 18 km during 45 min of motion, plus a 15 min stop. No other time passes. Which time belongs in the whole-journey average-speed denominator?
సమాధానం
60 min (1 h)
Whole-journey elapsed time includes both the 45 min moving interval and the 15 min stop.
కార్డు 10
ప్రశ్న
Three constant-speed legs last 10 min each, at 6, 12 and 18 km/h. No stops or other delays. Whole-journey average speed?
సమాధానం
12 km/h
Equal durations give (6 + 12 + 18) ÷ 3 = 12 km/h.
కార్డు 11
ప్రశ్న
A whole journey lasts 1.25 h and has average speed 32 km/h. Total path length?
సమాధానం
40 km
Total path length = average speed × total elapsed time = 32 × 1.25 km.
కార్డు 12
ప్రశ్న
Three 12 km legs are traveled at constant speeds of 6, 12 and 24 km/h. No stops or other delays. Whole-journey average speed?
సమాధానం
72/7 km/h (about 10.29 km/h)
Total time = 2 + 1 + 0.5 = 3.5 h. Total distance = 36 km, so 36 ÷ 3.5 = 72/7 km/h.
కార్డు 13
ప్రశ్న
A trip has two constant-speed legs at 8 km/h and 20 km/h, with no stops. No leg distances or durations are given. Is its average speed determined?
సమాధానం
No.
The time spent at each speed is missing. Different positive leg durations give different averages.
కార్డు 14
ప్రశ్న
An object moves at 3 m/s for 40 s and 7 m/s for 20 s. No stops or other delays. Whole-interval average speed?
సమాధానం
13/3 m/s (about 4.33 m/s)
Distance = 3 × 40 + 7 × 20 = 260 m. Time = 60 s; divide 260 by 60.
కార్డు 15
ప్రశ్న
Two legs have the same positive distance. One is traveled at constant speed 9 km/h, the other at 27 km/h, with no delays. What fraction of the moving time is spent on the slower leg?
సమాధానం
3/4
At one-third the speed, the slower leg takes three times as long. Its share is 3 ÷ (3 + 1).
కార్డు 16
ప్రశ్న
A constant-speed 36 km leg takes 1 h, followed by a 12 min stop. No other time passes. Average speed from the start through the end of the stop?
సమాధానం
30 km/h
Elapsed time = 1 + 12/60 = 1.2 h. Divide 36 km by 1.2 h.
కార్డు 17
ప్రశ్న
A route contains successive path lengths of 250 m, 150 m and 400 m. What total distance belongs in its average-speed numerator?
సమాధానం
800 m
Distance is total path length: 250 + 150 + 400 m, regardless of changes in direction.
కార్డు 18
ప్రశ్న
Two constant-speed legs last 0.5 h at 16 km/h and 1.5 h at 32 km/h. No stops or other delays. Whole-journey average speed?
సమాధానం
28 km/h
Distance = 16 × 0.5 + 32 × 1.5 = 56 km. Divide by the total 2 h.
కార్డు 19
ప్రశ్న
A 60 km journey averages 24 km/h overall. Its moving time is 2 h; all other elapsed time is one stop. Stop duration?
సమాధానం
30 min
Total elapsed time = 60 ÷ 24 = 2.5 h. The stop takes 2.5 − 2 = 0.5 h.
కార్డు 20
ప్రశ్న
Idealized 90 km trip: the first 30 km took 1 h. With no stops or other delays, what constant speed is required for the remaining 60 km to average 30 km/h overall?
సమాధానం
30 km/h
Allowed total time = 90/30 = 3 h. Remaining time = 3 − 1 = 2 h; required speed = 60/2 = 30 km/h.
కార్డు 21
ప్రశ్న
A journey's full time record contains only motion at 4 m/s, motion at 10 m/s, and rest. What inclusive bounds must contain its average speed?
సమాధానం
0 to 10 m/s
Average speed is a time-weighted mean of these segment speeds. Rest contributes a speed of 0 m/s.
కార్డు 22
ప్రశ్న
For two equal positive distances at constant speeds 10 and 15 km/h, with no delays, someone uses (10 + 15)/2. What weighting mistake must be repaired?
సమాధానం
Weight by travel time, not by the number of legs.
The slower equal-distance leg lasts longer. For any common leg distance d, use 2d ÷ (d/10 + d/15), giving 12 km/h.
కార్డు 23
ప్రశ్న
A journey covers 28 km in 2 h of motion and includes 30 min at rest. Moving-only average speed?
సమాధానం
14 km/h
For the moving-only interval, divide 28 km by 2 h. The rest time is outside that requested denominator.
కార్డు 24
ప్రశ్న
Idealized 120 km trip: the first 40 km took 2 h. No stops or other delays. What constant speed over the remaining 80 km gives a whole-trip average of 40 km/h?
సమాధానం
80 km/h
Allowed total time = 120/40 = 3 h. Only 1 h remains for 80 km.
కార్డు 25
ప్రశ్న
An object moves 60 m east in 20 s, then 20 m west in 10 s. No other time passes. Average speed over the full 30 s?
సమాధానం
8/3 m/s (about 2.67 m/s)
Path length = 60 + 20 = 80 m. Average speed = 80 m ÷ 30 s.
కార్డు 26
ప్రశ్న
A journey has two constant-speed legs: 15 km at 10 km/h, then 45 km at 30 km/h. No stops or other delays. Whole-journey average speed?
సమాధానం
20 km/h
The legs each take 1.5 h. Divide 60 km by 3 h; unequal distances can still give equal durations.
కార్డు 27
ప్రశ్న
Two constant-speed 20 km legs are each traveled at 40 km/h. Between them are stops of 10 min and 5 min. No other time passes. Whole-journey average speed?
సమాధానం
32 km/h
Moving time = 0.5 + 0.5 = 1 h; stops add 0.25 h. Divide 40 km by 1.25 h.
కార్డు 28
ప్రశ్న
Idealized 72 km trip: 24 km took 1 h. A 30 min stop is still required. No other delays. What constant speed over the remaining 48 km gives an overall average of 24 km/h?
సమాధానం
32 km/h
Total allowance = 72/24 = 3 h. Remaining moving time = 3 − 1 − 0.5 = 1.5 h; 48/1.5 = 32 km/h.
కార్డు 29
ప్రశ్న
Idealized 50 km trip: 20 km have already taken 2 h. There are no stops or other delays. Can any finite constant speed over the remaining 30 km achieve a 25 km/h whole-trip average?
సమాధానం
No.
The target permits 50/25 = 2 h in total, all already used. A positive remaining distance needs positive travel time; a finite speed cannot use zero time.
కార్డు 30
ప్రశ్న
East is positive. An object moves 72 m east in 24 s, then 24 m west in 12 s. No other time passes. Signed average velocity over the full 36 s?
సమాధానం
+4/3 m/s (about +1.33 m/s)
Displacement = 72 − 24 = +48 m. Divide +48 m by 36 s; the positive sign means east.
కార్డు 31
ప్రశ్న
A 10 km journey has 1 h of motion and a 20 min stop, with no other delays. Someone reports 10/1 = 10 km/h as the whole-journey average. Correct the denominator.
సమాధానం
4/3 h
Include the stop: 1 + 20/60 h. The corrected whole-journey average is 10 ÷ (4/3) = 7.5 km/h.
కార్డు 32
ప్రశ్న
An object travels 35 m out along a straight line in 7 s and returns to its starting point in 7 s. No other time passes. Average speed for the full round trip?
సమాధానం
5 m/s
The total path length is 70 m, although displacement is zero. Divide 70 m by 14 s.
కార్డు 33
ప్రశ్న
Idealized 100 km trip: 40 km of motion took 1 h, and a 30 min stop has already occurred. No future stops or other delays. Required constant speed over the remaining 60 km for a 40 km/h overall average?
సమాధానం
60 km/h
Allowed time = 100/40 = 2.5 h. Already elapsed = 1.5 h; 1 h remains for 60 km.
కార్డు 34
ప్రశ్న
A 48 km journey averages 16 km/h overall. It has 2.5 h of motion; all other time is spent at rest. Total rest time?
సమాధానం
30 min
Elapsed time = 48/16 = 3 h. Subtract 2.5 h of motion to get 0.5 h at rest.
కార్డు 35
ప్రశ్న
An object moves at 2 m/s for 30 s, rests 20 s, then moves at 4 m/s for 10 s. No other time passes. Average speed over all 60 s?
సమాధానం
5/3 m/s (about 1.67 m/s)
Distance = 2 × 30 + 4 × 10 = 100 m. Include the rest in the 60 s denominator.
కార్డు 36
ప్రశ్న
For two positive constant speeds, what condition makes their arithmetic mean equal the whole-journey average speed for a two-leg trip with no delays?
సమాధానం
Equal leg durations, or equal speeds.
The average is (v₁t₁ + v₂t₂)/(t₁ + t₂). For distinct speeds, equal time weights are required.
కార్డు 37
ప్రశ్న
Idealized 80 km trip: the first 20 km took 3 h. No stops or other delays. Can any finite constant speed over the remaining 60 km achieve a 40 km/h whole-trip average?
సమాధానం
No.
The target allows only 80/40 = 2 h overall. Already elapsed time exceeds that allowance; the remaining budget is −1 h.
కార్డు 38
ప్రశ్న
Idealized 400 m trip: the first 100 m took 40 s. No stops or other delays. Required constant speed over the remaining 300 m for a whole-trip average of 5 m/s?
సమాధానం
7.5 m/s
Allowed time = 400/5 = 80 s. Remaining time = 80 − 40 = 40 s; 300/40 = 7.5 m/s.
కార్డు 39
ప్రశ్న
East is positive. An object travels 45 m east in 9 s, rests 6 s, then returns 45 m west in 15 s. Signed average velocity over the full interval?
సమాధానం
0 m/s
It ends where it started, so displacement is zero. The elapsed time is 9 + 6 + 15 = 30 s.
కార్డు 40
ప్రశ్న
An object covers 150 m during 30 s of motion, rests 15 s, then covers 90 m during 30 s of motion. Moving-only average speed?
సమాధానం
4 m/s
Moving distance = 150 + 90 = 240 m; moving time = 30 + 30 = 60 s. Exclude the 15 s rest from this requested interval.
కార్డు 41
ప్రశ్న
For the same journey interval and reference frame, someone claims that the magnitude of average velocity can exceed average speed. Correct this claim.
సమాధానం
The magnitude of average velocity is at most the average speed.
Displacement magnitude cannot exceed path length. Both averages divide by the same positive elapsed time.
కార్డు 42
ప్రశ్న
Idealized 150 km trip: the first 50 km took 1 h. A 30 min stop is still required. No other delays. Required constant speed over the remaining 100 km for a whole-trip average of 50 km/h?
సమాధానం
200/3 km/h (about 66.67 km/h)
Allowed time = 150/50 = 3 h. Moving time left = 3 − 1 − 0.5 = 1.5 h; 100/1.5 = 200/3 km/h.
42 కార్డులు
Average Speed Flashcards: Equal Distances, Equal Times & Stops
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