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.
O tej talii
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.
Karty w tej talii
Karta 1
Pytanie
A journey covers 84 km in 2 h, including all stops. Average speed over that whole interval?
Odpowiedź
42 km/h
Divide total path length by total elapsed time: 84 km ÷ 2 h.
Karta 2
Pytanie
A constant-speed leg covers 45 km at 30 km/h. How long does that leg take?
Odpowiedź
1.5 h
Time = distance ÷ speed = 45 ÷ 30 h.
Karta 3
Pytanie
A constant-speed leg lasts 12 min at 5 m/s. Distance covered?
Odpowiedź
3600 m
12 min = 720 s; distance = 5 m/s × 720 s.
Karta 4
Pytanie
An object stays at rest for the entire 25 s observation interval. Average speed?
Odpowiedź
0 m/s
Its path length is 0 m and the elapsed time is positive: 0 m ÷ 25 s.
Karta 5
Pytanie
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?
Odpowiedź
24 km/h
Equal travel times give equal time weights, so (18 + 30) ÷ 2 = 24 km/h.
Karta 6
Pytanie
A journey covers 900 m in 2 min 30 s, including every delay. Average speed in m/s?
Odpowiedź
6 m/s
2 min 30 s = 150 s; 900 m ÷ 150 s = 6 m/s.
Karta 7
Pytanie
Two legs cover 14 km in 0.5 h and 22 km in 1 h. These times cover the entire journey. Average speed?
Odpowiedź
24 km/h
Add distances and times separately: (14 + 22) km ÷ (0.5 + 1) h.
Karta 8
Pytanie
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?
Odpowiedź
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.
Karta 9
Pytanie
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?
Odpowiedź
60 min (1 h)
Whole-journey elapsed time includes both the 45 min moving interval and the 15 min stop.
Karta 10
Pytanie
Three constant-speed legs last 10 min each, at 6, 12 and 18 km/h. No stops or other delays. Whole-journey average speed?
Odpowiedź
12 km/h
Equal durations give (6 + 12 + 18) ÷ 3 = 12 km/h.
Karta 11
Pytanie
A whole journey lasts 1.25 h and has average speed 32 km/h. Total path length?
Odpowiedź
40 km
Total path length = average speed × total elapsed time = 32 × 1.25 km.
Karta 12
Pytanie
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?
Odpowiedź
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.
Karta 13
Pytanie
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?
Odpowiedź
No.
The time spent at each speed is missing. Different positive leg durations give different averages.
Karta 14
Pytanie
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?
Odpowiedź
13/3 m/s (about 4.33 m/s)
Distance = 3 × 40 + 7 × 20 = 260 m. Time = 60 s; divide 260 by 60.
Karta 15
Pytanie
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?
Odpowiedź
3/4
At one-third the speed, the slower leg takes three times as long. Its share is 3 ÷ (3 + 1).
Karta 16
Pytanie
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?
Odpowiedź
30 km/h
Elapsed time = 1 + 12/60 = 1.2 h. Divide 36 km by 1.2 h.
Karta 17
Pytanie
A route contains successive path lengths of 250 m, 150 m and 400 m. What total distance belongs in its average-speed numerator?
Odpowiedź
800 m
Distance is total path length: 250 + 150 + 400 m, regardless of changes in direction.
Karta 18
Pytanie
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?
Odpowiedź
28 km/h
Distance = 16 × 0.5 + 32 × 1.5 = 56 km. Divide by the total 2 h.
Karta 19
Pytanie
A 60 km journey averages 24 km/h overall. Its moving time is 2 h; all other elapsed time is one stop. Stop duration?
Odpowiedź
30 min
Total elapsed time = 60 ÷ 24 = 2.5 h. The stop takes 2.5 − 2 = 0.5 h.
Karta 20
Pytanie
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?
Odpowiedź
30 km/h
Allowed total time = 90/30 = 3 h. Remaining time = 3 − 1 = 2 h; required speed = 60/2 = 30 km/h.
Karta 21
Pytanie
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?
Odpowiedź
0 to 10 m/s
Average speed is a time-weighted mean of these segment speeds. Rest contributes a speed of 0 m/s.
Karta 22
Pytanie
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?
Odpowiedź
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.
Karta 23
Pytanie
A journey covers 28 km in 2 h of motion and includes 30 min at rest. Moving-only average speed?
Odpowiedź
14 km/h
For the moving-only interval, divide 28 km by 2 h. The rest time is outside that requested denominator.
Karta 24
Pytanie
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?
Odpowiedź
80 km/h
Allowed total time = 120/40 = 3 h. Only 1 h remains for 80 km.
Karta 25
Pytanie
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?
Odpowiedź
8/3 m/s (about 2.67 m/s)
Path length = 60 + 20 = 80 m. Average speed = 80 m ÷ 30 s.
Karta 26
Pytanie
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?
Odpowiedź
20 km/h
The legs each take 1.5 h. Divide 60 km by 3 h; unequal distances can still give equal durations.
Karta 27
Pytanie
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?
Odpowiedź
32 km/h
Moving time = 0.5 + 0.5 = 1 h; stops add 0.25 h. Divide 40 km by 1.25 h.
Karta 28
Pytanie
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?
Odpowiedź
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.
Karta 29
Pytanie
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?
Odpowiedź
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.
Karta 30
Pytanie
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?
Odpowiedź
+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.
Karta 31
Pytanie
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.
Odpowiedź
4/3 h
Include the stop: 1 + 20/60 h. The corrected whole-journey average is 10 ÷ (4/3) = 7.5 km/h.
Karta 32
Pytanie
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?
Odpowiedź
5 m/s
The total path length is 70 m, although displacement is zero. Divide 70 m by 14 s.
Karta 33
Pytanie
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?
Odpowiedź
60 km/h
Allowed time = 100/40 = 2.5 h. Already elapsed = 1.5 h; 1 h remains for 60 km.
Karta 34
Pytanie
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?
Odpowiedź
30 min
Elapsed time = 48/16 = 3 h. Subtract 2.5 h of motion to get 0.5 h at rest.
Karta 35
Pytanie
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?
Odpowiedź
5/3 m/s (about 1.67 m/s)
Distance = 2 × 30 + 4 × 10 = 100 m. Include the rest in the 60 s denominator.
Karta 36
Pytanie
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?
Odpowiedź
Equal leg durations, or equal speeds.
The average is (v₁t₁ + v₂t₂)/(t₁ + t₂). For distinct speeds, equal time weights are required.
Karta 37
Pytanie
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?
Odpowiedź
No.
The target allows only 80/40 = 2 h overall. Already elapsed time exceeds that allowance; the remaining budget is −1 h.
Karta 38
Pytanie
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?
Odpowiedź
7.5 m/s
Allowed time = 400/5 = 80 s. Remaining time = 80 − 40 = 40 s; 300/40 = 7.5 m/s.
Karta 39
Pytanie
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?
Odpowiedź
0 m/s
It ends where it started, so displacement is zero. The elapsed time is 9 + 6 + 15 = 30 s.
Karta 40
Pytanie
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?
Odpowiedź
4 m/s
Moving distance = 150 + 90 = 240 m; moving time = 30 + 30 = 60 s. Exclude the 15 s rest from this requested interval.
Karta 41
Pytanie
For the same journey interval and reference frame, someone claims that the magnitude of average velocity can exceed average speed. Correct this claim.
Odpowiedź
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.
Karta 42
Pytanie
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?
Odpowiedź
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 karty
Average Speed Flashcards: Equal Distances, Equal Times & Stops
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