AO3 · Speed

Speed Heuristics

15 question types · PSLE Paper 2 · Model Method and algebra, side by side

01

Direct Formula Application with Unit Conversion

heuristicDimensional Analysis and Unit Standardisation

A motorist traveled from Town P to Town Q at a constant speed of 72 km/h. The journey took 45 minutes. (a) What was the distance between Town P and Town Q in kilometres? (b) A cyclist took 3 hours to cover the same distance. Find the cyclist's speed in metres per minute (m/min).

1 hour60 min · 72 km72 km/h
Speed is a distance per hour: 72 km in 60 minutes.
Speed is 72 km per 60 minutes.
step 1 of 6 Practice time problems in the app

Represent 1 hour (60 minutes) as 4 equal blocks of 15 minutes to find the distance covered in 45 minutes, then distribute total metres over a timeline of minutes.

  1. Speed is 72 km per 60 minutes.
  2. Split 60 minutes into 4 units of 15 minutes: 1 unit (15 min) = 72 ÷ 4 = 18 km.
  3. (a) 45 minutes = 3 units = 3 × 18 km = 54 km.
  4. Represent cyclist's distance: 54 km = 54000 m.
  5. Cyclist covers 54000 m in 3 hours ⟹ 1 hour = 54000 ÷ 3 = 18000 m.
  6. (b) Divide 1 hour into 60 minutes: 18000 m ÷ 60 = 300 m/min.

Common pitfalls

  • Multiplying speed directly by minutes: 72 × 45 = 3240 km.
  • Converting 45 minutes incorrectly as 0.45 hours instead of 0.75 hours.
  • Dividing 54 km by 3 h to get 18 km/h and forgetting to convert to m/min.
02

Average Speed for a Multi-Leg Single Journey

heuristicAggregate Total Equating (Total Distance / Total Time)

A delivery van traveled from Point A to Point B, a distance of 80 km, at an average speed of 40 km/h. It then continued from Point B to Point C, a distance of 180 km, at an average speed of 60 km/h. Find the average speed of the delivery van for the entire journey from Point A to Point C.

A to B80 km in 2 h
A to B: 80 km at 40 km/h. Each hour-box holds 40 km, so 2 boxes.
Draw Segment 1: Bar of 80 km. Since rate is 40 km per 1 hour, box count for time = 80 ÷ 40 = 2 hours.
step 1 of 5 Practice time problems in the app

Draw a distance-time segmented strip to clearly separate individual durations from speeds.

  1. Draw Segment 1: Bar of 80 km. Since rate is 40 km per 1 hour, box count for time = 80 ÷ 40 = 2 hours.
  2. Draw Segment 2: Bar of 180 km. Since rate is 60 km per 1 hour, box count for time = 180 ÷ 60 = 3 hours.
  3. Combine segments into one master bar: Total distance = 80 + 180 = 260 km.
  4. Combine time boxes: Total time = 2 + 3 = 5 hours.
  5. Average speed = 260 ÷ 5 = 52 km/h.

Common pitfalls

  • Averaging the two speeds directly: 40 + 602 = 50 km/h.
  • Adding the speeds together: 40 + 60 = 100 km/h.
03

Constant Time (Direct Proportion of Distance to Speed)

heuristicSimultaneous Travel Distance-Speed Ratio

Keith and Leonard started cycling at the same time from the same entrance along a park connector in the same direction. Keith cycled at 24 km/h and Leonard cycled at 16 km/h. (a) How far apart were they after 45 minutes? (b) How much further had Keith cycled than Leonard after 1 hour 15 minutes?

Keith3u24 km/hLeonard2u16 km/h
Same start, same time on the road, so distances are in the ratio of the speeds: 24:16 = 3:2.
Ratio of Speed (Keith : Leonard) = 24 : 16 = 3 : 2.
step 1 of 5 Practice time problems in the app

Express speeds as a ratio to determine the ratio of distances covered in any identical duration.

  1. Ratio of Speed (Keith : Leonard) = 24 : 16 = 3 : 2.
  2. For any given duration, Keith travels 3 units while Leonard travels 2 units (Difference = 1 unit).
  3. In 1 hour, Keith travels 24 km (3u) and Leonard travels 16 km (2u) ⟹ u = 8 km.
  4. (a) For 45 min (34 of an hour): Gap = 34 × 8 km = 6 km.
  5. (b) For 1 hour 15 min (114 hours): Gap = 54 × 8 km = 10 km.

Common pitfalls

  • Calculating the distance Keith traveled and forgetting to subtract Leonard's distance.
  • Converting 1 hour 15 minutes as 1.15 hours instead of 1.25 hours.
04

Constant Distance (Inverse Proportion of Speed and Time)

heuristicSpeed-Time Inverse Ratio

Mr. Lee drove from Town P to Town Q at an average speed of 60 km/h. On his return journey along the exact same route, he increased his speed to 80 km/h. The return journey took 30 minutes less than the forward journey. Find the distance between Town P and Town Q.

Forwarduuuu60 km/h
Same distance both ways. Speeds 60:80 = 3:4, so times are 4:3: the faster trip takes fewer blocks.
Draw Forward Time as 4 units.
step 1 of 7 Practice time problems in the app

Draw time comparison bars where the faster speed corresponds to fewer time blocks.

  1. Draw Forward Time as 4 units.
  2. Draw Return Time as 3 units.
  3. Difference = 4 − 3 = 1 unit = 30 minutes = 0.5 h.
  4. Forward Time = 4 × 0.5 = 2 hours.
  5. Return Time = 3 × 0.5 = 1.5 hours.
  6. Calculate Distance using Forward trip: 60 km/h × 2 h = 120 km.
  7. Check using Return trip: 80 km/h × 1.5 h = 120 km.

Common pitfalls

  • Assuming that time is directly proportional to speed (writing time ratio as 3 : 4 instead of 4 : 3).
  • Using 30 minutes directly as 30 hours, obtaining an astronomical distance (60 × 120 = 7200 km).
05

Opposite Direction Travel (Simultaneous Departure / Meeting Point)

heuristicCombined Rate of Closure

Town A and Town B are 350 km apart. At 8:00 a.m., Car X left Town A heading towards Town B at a constant speed of 65 km/h. At the same time, Car Y left Town B heading towards Town A at a constant speed of 75 km/h. (a) How long did it take for the two cars to meet? (b) How far was Car X from Town B when they met?

Town ATown B350 kmX 162.5 kmY 187.5 km10:30 a.m. · 2 h 30 min after 8:00they meet 162.5 km from A, 187.5 km from B
Every hour, X moves 65 km in from the left and Y 75 km in from the right.
In 1 hour: Car X travels 65 km from the left, Car Y travels 75 km from the right.
step 1 of 8 Practice time problems in the app

Use a timeline bar of 350 km where each hour of travel consumes a combined block of 140 km.

  1. In 1 hour: Car X travels 65 km from the left, Car Y travels 75 km from the right.
  2. Total distance closed per hour = 65 + 75 = 140 km.
  3. In 2 hours: Distance closed = 2 × 140 = 280 km.
  4. Remaining distance to cover = 350 − 280 = 70 km.
  5. Time to cover remaining 70 km = 70 ÷ 140 = 0.5 hours.
  6. (a) Total time = 2 + 0.5 = 2.5 hours.
  7. Notice that Car X's distance from Town B is equal to the distance Car Y has traveled from Town B:
  8. (b) Car Y's distance = 75 × 2.5 = 187.5 km.

Common pitfalls

  • Subtracting the two speeds (75 − 65 = 10 km/h) instead of adding them.
  • Answering part (b) with the distance Car X traveled from Town A (162.5 km) instead of its remaining distance to Town B.
06

Same Direction Travel (Catching Up with Time Head Start)

heuristicRelative Speed Catch-Up

At 9:00 a.m., a lorry left a warehouse traveling along a highway at a constant speed of 50 km/h. At 10:30 a.m., a motorcycle left the same warehouse along the exact same route traveling at 80 km/h. (a) At what time did the motorcycle catch up with the lorry? (b) How far had the motorcycle traveled when it caught up with the lorry?

1:00 p.m. · 2 h 30 min after 10:30Lorry75 km head start125 km200 kmMotorcycle200 km200 kmcaught up: 200 km each
By 10:30 the lorry has driven 1.5 h at 50 km/h: a 75 km head start.
At 10:30 a.m.: Lorry has a head start bar of 50 × 1.5 = 75 km.
step 1 of 5 Practice time problems in the app

Draw two parallel distance lines anchored at 10:30 a.m. to visualize the gap closing at 30 km every hour.

  1. At 10:30 a.m.: Lorry has a head start bar of 50 × 1.5 = 75 km.
  2. Every 1 hour after 10:30 a.m., the motorcycle closes the gap by: 80 − 50 = 30 km.
  3. Number of hours required to close the entire 75 km gap: 75 ÷ 30 = 2.5 hours.
  4. (a) 10:30 a.m. + 2.5 hours = 1:00 p.m.
  5. (b) Motorcycle distance = 2.5 × 80 = 200 km (Check Lorry: 50 × 4 h = 200 km).

Common pitfalls

  • Adding the catch-up duration (2.5 hours) to 9:00 a.m. instead of 10:30 a.m.
  • Dividing the lead distance (75 km) by the motorcycle's speed (80 km/h) rather than the relative speed difference (30 km/h).
07

Delayed Start Moving Towards Each Other (Asymmetrical Meeting)

heuristicLead Distance Deduction with Subsequent Closure

Town M and Town N are 380 km apart. At 8:00 a.m., Van A left Town M heading towards Town N at 60 km/h. At 8:30 a.m., Van B left Town N heading towards Town M at 80 km/h. At what time did the two vans pass each other?

Town MTown N380 kmVan A 180 kmVan B 200 km11:00 a.m. · 2 h 30 min after 8:30they pass at 11:00 a.m.
The road is 380 km.
Draw Total bar = 380 km.
step 1 of 6 Practice time problems in the app

Segment the 380 km timeline bar into a solo segment for Van A and a shared closure segment.

  1. Draw Total bar = 380 km.
  2. Mark off Van A's solo block: 30 minutes = 12 h × 60 = 30 km.
  3. Remaining length of bar = 380 − 30 = 350 km.
  4. Both vans close this remaining bar together at 60 + 80 = 140 km/h.
  5. Time required = 350 ÷ 140 = 2.5 hours.
  6. Add to common start time: 8:30 a.m. + 2 h 30 min = 11:00 a.m.

Common pitfalls

  • Dividing the initial total distance (380 km) directly by the combined speed (140 km/h) without subtracting the 30 km head start.
  • Adding 2.5 hours to 8:00 a.m. instead of 8:30 a.m.
08

Meeting with a Midpoint Offset

heuristicDouble Offset Distance Difference

Amy and Ben started driving towards each other at the same time from Town X and Town Y respectively. Amy drove at 70 km/h and Ben drove at 50 km/h. When they passed each other, they were 30 km away from the midpoint of the two towns. (a) What was the total distance between Town X and Town Y? (b) How much longer would Ben take to reach Town X after passing Amy?

Town XTown YAmy 210 kmBen 150 kmM3 h of driving
Mark the midpoint M. Both drove for the same time, so Amy, the faster one, passes M before they meet.
Draw Midpoint M.
step 1 of 8 Practice time problems in the app

Draw a route model showing the midpoint M. Label Amy traveling to M + 30 and Ben traveling to M − 30.

  1. Draw Midpoint M.
  2. Amy's bar: from Town X past M by 30 km.
  3. Ben's bar: from Town Y stopping 30 km before M.
  4. Amy has traveled 30 + 30 = 60 km more than Ben.
  5. In 1 hour, Amy travels 70 − 50 = 20 km more than Ben.
  6. Time elapsed until meeting = 60 ÷ 20 = 3 hours.
  7. (a) Total distance = (70 × 3) + (50 × 3) = 210 + 150 = 360 km.
  8. (b) Distance left for Ben = 210 km. Ben's time = 210 ÷ 50 = 4.2 hours = 4 h 12 min.

Common pitfalls

  • Assuming the difference in distance traveled is 30 km rather than 2 × 30 = 60 km.
  • Dividing 30 km by the speed difference to get 1.5 hours.
09

Two-Stage Journey with Mid-Way Speed Reduction

heuristicSegmented Time Remainder Allocation

Mr. Tan drove from City A to City B, covering a total distance of 360 km. For the first 13 of the distance, he drove at an average speed of 80 km/h. Heavy traffic caused him to reduce his speed for the remaining journey. If the entire journey took 512 hours, what was his average speed for the remaining journey?

Distance120 km120 km120 km360 km
Thirds of 360 km: three units of 120 km.
Total bar of 360 km split into 3 units of 120 km each.
step 1 of 5 Practice time problems in the app

Draw a 3-unit distance bar to isolate the two distinct stages, mapping durations to each block.

  1. Total bar of 360 km split into 3 units of 120 km each.
  2. First unit = 120 km. At 80 km/h, time taken = 120 ÷ 80 = 1.5 hours.
  3. Remaining 2 units = 2 × 120 = 240 km.
  4. Total time is 5.5 hours, so time left for the 2 units = 5.5 − 1.5 = 4 hours.
  5. Speed for remaining 2 units = 240 ÷ 4 = 60 km/h.

Common pitfalls

  • Applying 80 km/h across the first 13 of the time (5.5 ÷ 3) instead of the distance.
  • Subtracting 1.5 hours from 5 hours instead of 5.5 hours.
10

Overtaking and Subsequent Gap Widening

heuristicRelative Speed over Cumulative Deficit and Surplus

Cyclist A and Cyclist B traveled along a straight road in the same direction. Cyclist A traveled at a constant speed of 25 km/h and Cyclist B traveled at a constant speed of 15 km/h. At first, Cyclist A was 8 km behind Cyclist B. (a) How long did it take for Cyclist A to be 12 km ahead of Cyclist B? (b) How far had Cyclist A traveled during this duration?

-10-8-6-4-20+2+4+6+8+10+12+14BA +12 km2 h · relative to B, who sits at 0
Measure everything from B. At the start A is 8 km behind: at −8.
Start: Cyclist A is at −8 km relative to B.
step 1 of 6 Practice time problems in the app

Represent the relative position shift on a number line centered on Cyclist B.

  1. Start: Cyclist A is at −8 km relative to B.
  2. End: Cyclist A is at +12 km relative to B.
  3. Total distance gained by A on B = 8 + 12 = 20 km.
  4. In every hour, A gains 25 − 15 = 10 km on B.
  5. (a) Number of hours needed = 20 ÷ 10 = 2 hours.
  6. (b) In 2 hours, Cyclist A covers 25 × 2 = 50 km.

Common pitfalls

  • Subtracting the two distances (12 − 8 = 4 km) instead of adding them to find total relative displacement.
  • Dividing 20 km by Cyclist A's individual speed (25 km/h) instead of relative speed (10 km/h).
11

Round Trip with Multiple Meetings (Bouncing Movers)

heuristicCumulative Combined Track Invariance

John and Peter started walking towards each other at the same time from Point X and Point Y respectively along a straight path. They met for the first time at a spot 700 m from Point X. After the meeting, both continued walking to the opposite end without stopping, turned back immediately, and walked towards each other at their original speeds. They met for the second time at a spot 400 m from Point Y. What is the distance between Point X and Point Y?

XY1st meeting · 700 m from XTogether
At the first meeting the two have walked one whole track between them, and John's part of it is 700 m.
Milestone 1 (First meeting): Both runners together cover 1 track. John covers 700 m.
step 1 of 6 Practice time problems in the app

Draw two-phase journey bars comparing the combined distance of both runners at each milestone.

  1. Milestone 1 (First meeting): Both runners together cover 1 track. John covers 700 m.
  2. Milestone 2 (Second meeting): To reach opposite ends and meet again, both cover: 1 track (to ends) + 1 track (returning) = 2 more tracks.
  3. Total combined tracks at second meeting = 1 + 2 = 3 tracks.
  4. John's share is 3 times his first share: 3 × 700 m = 2100 m.
  5. From the diagram, John's journey = 1 full track + 400 m.
  6. 1 full track = 2100 − 400 = 1700 m.

Common pitfalls

  • Assuming the total distance covered at the second meeting is 2D instead of 3D.
  • Adding 400 m to 2100 m (2500 m) instead of subtracting it to isolate the single track length.
12

Post-Meeting Continuation to Opposite Terminus

heuristicGeometric Mean Time Relation / Speed Ratio Equating

Car P and Car Q started driving at the same time from Town A and Town B respectively towards each other at constant speeds. They met at Point M. Car P took 2 hours to reach Town B after passing Point M. Car Q took 412 hours to reach Town A after passing Point M. (a) Find the ratio of Car P's speed to Car Q's speed. (b) How many hours did the cars take from the start until they met at Point M?

Car PA to M: t hM to B: 2 h
Car P covers AM in t hours (unknown) and MB in 2.
For segment AM: Car P takes t hours, Car Q takes 4.5 hours.
step 1 of 7 Practice time problems in the app

Compare the time ratios taken by both cars to cover the exact same two track segments (AM and MB).

  1. For segment AM: Car P takes t hours, Car Q takes 4.5 hours.
  2. For segment MB: Car P takes 2 hours, Car Q takes t hours.
  3. Since speed ratio is constant across both segments, the ratio of times must be equal:
  4. TimePTimeQ = t4.5 = 2t.
  5. Cross-multiply: t × t = 4.5 × 2 = 9 ⟹ t = 3 hours.
  6. For segment AM, Car P takes 3 h while Car Q takes 4.5 h.
  7. Ratio of speed is inverse of time: SP : SQ = 4.5 : 3 = 3 : 2.

Common pitfalls

  • Averaging the two post-meeting times: 2 + 4.52 = 3.25 hours.
  • Inverting the speed ratio (2 : 3 instead of 3 : 2), forgetting that Car P is faster because it takes less time.
13

Race with Rest Intervals and Non-Uniform Motion

heuristicEffective Moving Time vs Static Rest Time

A rabbit and a tortoise competed in a 1200 m race. The rabbit ran at 150 m/min while the tortoise crawled at 20 m/min. The rabbit ran for 4 minutes, stopped to take a nap, and then resumed running at the same speed to the finish line. The tortoise crawled continuously without stopping and finished the race 2 minutes ahead of the rabbit. How long was the rabbit's nap?

Tortoise60 min · 20 m/min · 1200 m
The tortoise never stops: 1200 ÷ 20 = 60 minutes.
Tortoise bar: 60 continuous 1-minute blocks = 60 minutes.
step 1 of 6 Practice time problems in the app

Construct two timeline bars showing movement chunks and idle blocks.

  1. Tortoise bar: 60 continuous 1-minute blocks = 60 minutes.
  2. Rabbit total timeline bar: 60 + 2 = 62 minutes.
  3. Break down rabbit's running: 1200 m ÷ 150 m/min = 8 minutes of running.
  4. First run chunk = 4 minutes; Second run chunk = 8 − 4 = 4 minutes.
  5. Total running time on rabbit's bar = 4 + 4 = 8 minutes.
  6. The remainder of the bar is the nap: 62 − 8 = 54 minutes.

Common pitfalls

  • Subtracting 2 minutes from the tortoise's time (60 − 2 = 58 min), mistakenly making the rabbit faster.
  • Subtracting only the first 4 minutes of running from the total time, forgetting that the rabbit still had to run the remaining distance.
14

Three-Party Interception and Relative Passing

heuristicRelative Gap Closure Across Multiple Observers

At 9:00 a.m., Car C and Lorry L left Town P heading towards Town Q. Car C traveled at 75 km/h and Lorry L traveled at 45 km/h. At the exact same time, Bus B left Town Q heading towards Town P at 75 km/h. At 10:00 a.m., Car C and Bus B passed each other. (a) What was the distance between Town P and Town Q? (b) At what time did Lorry L pass Bus B?

Town PTown Q→C 75→L 45←B 75 from Q10:00 a.m.
Freeze the highway at 10:00 a.m., one hour in.
At 10:00 a.m. (after 1 h):
step 1 of 9 Practice time problems in the app

Draw snapshot diagrams of the highway at 10:00 a.m. showing positions of all three vehicles relative to Town P.

  1. At 10:00 a.m. (after 1 h):
  2. Car C is 75 km from Town P.
  3. Bus B meets Car C, so Bus B is also 75 km from Town P.
  4. Bus B has traveled 75 km from Town Q ⟹ Total distance = 75 + 75 = 150 km.
  5. Lorry L is 45 km from Town P.
  6. Gap between Lorry L and Bus B at 10:00 a.m. = 75 − 45 = 30 km.
  7. Lorry L and Bus B approach each other at 45 + 75 = 120 km/h.
  8. Time to close 30 km = 30 ÷ 120 = 14 hour = 15 minutes.
  9. Passing time = 10:15 a.m.

Common pitfalls

  • Assuming Lorry L passes Bus B 15 minutes after 9:00 a.m. instead of 10:00 a.m.
  • Dividing the 30 km gap by the speed of Lorry L alone (45 km/h), forgetting that Bus B is also moving towards Lorry L.
15

Variable Speed with Shifting Fraction of Journey

heuristicSegmented Linear Speed-Time Rebalancing

Mr. Ahmad drove from Town X to Town Y. After traveling 25 of the total distance in 2 hours, he increased his speed by 16 km/h for the remaining journey. He took 212 hours to complete the remaining journey. (a) What was Mr. Ahmad's average speed for the first 25 of the journey? (b) What was the total distance between Town X and Town Y?

Distance2u3u5u
25 of the way, so five units of distance.
Let total distance be 5 units (5u).
step 1 of 10 Practice time problems in the app

Represent total distance as 5 equal units. Determine how many units are covered per hour in each stage to find the value of 1 unit.

  1. Let total distance be 5 units (5u).
  2. First stage: covers 2u in 2 hours ⟹ 1 hour covers 2u ÷ 2 = u.
  3. Initial speed = u per hour.
  4. Second stage: covers 3u in 2.5 hours ⟹ 1 hour covers 3u ÷ 2.5 = 1.2u.
  5. New speed = 1.2u per hour.
  6. Speed difference per hour = 1.2u − u = 0.2u.
  7. Given speed difference is 16 km/h ⟹ 0.2u = 16 km.
  8. Value of 1 unit = 16 ÷ 0.2 = 80 km.
  9. (a) Speed for first stage = u = 80 km/h.
  10. (b) Total distance = 5u = 5 × 80 = 400 km.

Common pitfalls

  • Applying the speed increase to the total distance rather than the speed rate per hour.
  • Dividing 3u by 2 hours instead of 2.5 hours when calculating the new speed unit value.