Time, Speed, and Distance
🟢 Lite — Quick Review (1h–1d)
Rapid summary for last-minute revision before your exam.
The fundamental relationship between time, speed, and distance is: distance = speed × time. From this single formula, you can derive all three forms: speed = distance/time, and time = distance/speed. Always ensure your units are consistent — if speed is in km/h, distance must be in km and time in hours.
Essential Formulas:
- Distance = speed × time
- Speed = distance / time
- Time = distance / speed
- Average speed (different distances at different speeds): total distance / total time
- For equal distances at speeds v₁ and v₂: average speed = 2v₁v₂ / (v₁ + v₂)
- For equal times at speeds v₁ and v₂: average speed = (v₁ + v₂) / 2
- Relative speed (approaching): sum of speeds; (moving in same direction): difference of speeds
Key Facts:
- Convert units carefully: 72 km/h = 72/3.6 = 20 m/s; 15 m/s = 15 × 3.6 = 54 km/h
- If two objects move toward each other, their relative speed = v₁ + v₂ and they close the distance at this rate
- If two objects move in the same direction, relative speed = |v₁ − v₂|
- A round trip with equal distances at different speeds always gives average speed below the arithmetic mean of the two speeds
⚡ Exam Tip: In the UI entrance test, watch whether a journey’s parts cover equal distances or equal times — this determines which average speed formula to use. If a problem says “a car travels 60 km at 40 km/h and another 60 km at 60 km/h,” use the equal-distance formula: 2v₁v₂/(v₁+v₂) = 2×40×60/100 = 48 km/h. If it says “for 2 hours at 40 km/h and 3 hours at 60 km/h,” the average speed = total distance/total time = (80+180)/5 = 260/5 = 52 km/h.
🟡 Standard — Regular Study (2d–2mo)
Standard content for students with a few days to months.
Basic Speed-Time-Distance
Example: A train travels at 90 km/h for 2.5 hours. Distance = 90 × 2.5 = 225 km. Example: A cyclist covers 45 km in 3 hours. Speed = 45/3 = 15 km/h.
Converting Between Units
Distance: 1 km = 1000 m; 1 mile ≈ 1.609 km. Time: 1 hour = 60 minutes = 3600 seconds. Speed: to convert km/h to m/s, divide by 3.6. To convert m/s to km/h, multiply by 3.6.
Example: 108 km/h = 108/3.6 = 30 m/s. Example: 20 m/s = 20 × 3.6 = 72 km/h.
Meeting and Passing Problems
Example: Two trains, 200 m and 300 m long, travel toward each other on parallel tracks at 50 km/h and 70 km/h. How long does it take to completely pass each other? Combined length = 200 + 300 = 500 m = 0.5 km. Relative speed = 50 + 70 = 120 km/h. Time = distance/speed = 0.5/120 hours = 0.5/120 × 3600 seconds = 15 seconds.
Same Direction Problems
Example: A car travelling at 80 km/h overtakes a truck travelling at 60 km/h. The car is 5 m long, the truck is 15 m long. How long does the overtake take? Total distance to cover = car length + truck length = 5 + 15 = 20 m. Relative speed = 80 − 60 = 20 km/h = 20/3.6 ≈ 5.56 m/s. Time = 20/5.56 ≈ 3.6 seconds.
Round Trip — Average Speed
Example: A man drives 120 km to a town at 60 km/h and returns at 40 km/h. Find average speed for the round trip. Total distance = 240 km. Time going = 120/60 = 2 hours. Time returning = 120/40 = 3 hours. Total time = 5 hours. Average speed = 240/5 = 48 km/h. Note: arithmetic mean would be (60+40)/2 = 50 km/h — incorrect for equal distances.
Boats in Streams (River Problems)
When a boat travels in a river, its speed relative to the ground = speed relative to water ± speed of current. Downstream speed = boat speed in still water + current speed. Upstream speed = boat speed in still water − current speed.
Example: A boat’s speed in still water is 15 km/h. The current is 3 km/h. Find downstream and upstream speeds. Downstream: 15 + 3 = 18 km/h. Upstream: 15 − 3 = 12 km/h. If it travels 24 km downstream and returns upstream, total time = 24/18 + 24/12 = 4/3 + 2 = 10/3 hours ≈ 3.33 hours.
Aircraft and Wind Problems
Headwind reduces ground speed; tailwind increases it. Aircraft speed in still air = 500 km/h. Wind speed = 50 km/h. With headwind: ground speed = 450 km/h. With tailwind: ground speed = 550 km/h.
Problem-Solving Strategies:
- Draw a distance-time diagram for complex problems involving two travellers
- In meeting-point problems, set distance covered by A + distance covered by B = total distance
- For chase/hound problems, the time to catch up equals the initial distance divided by relative speed
- Always convert to consistent units before calculating
Common Mistakes:
- Mixing up km/h and m/s without converting
- Using arithmetic mean of speeds instead of total distance divided by total time for average speed
- In boat/stream problems, adding the current speed instead of subtracting (or vice versa) for upstream/downstream
- Forgetting that time must be in hours when using km/h
🔴 Extended — Deep Study (3mo+)
Comprehensive coverage for students on a longer study timeline.
Meeting Point When Starting at Different Times
Example: A bus leaves town A at 8:00 AM travelling at 60 km/h toward town B (300 km away). A car leaves town A at 9:30 AM travelling at 90 km/h. When and where do they meet? By 9:30, bus has travelled 60 × 1.5 = 90 km. Remaining distance = 300 − 90 = 210 km. Relative speed (approaching) = 60 + 90 = 150 km/h. Time to meet from 9:30 = 210/150 = 1.4 hours = 1 hour 24 minutes. Meeting time = 9:30 + 1h 24m = 10:54 AM. Distance from A (travelled by car) = 90 × 1.4 = 126 km from A (at 9:30) + the 90 km the bus covered before car started = 216 km from A total. Or: both cover 300 km total between them. Car covers 126 km, bus covers 174 km. Check: 90×1.4=126, 60×2.9=174. ✓
Circular Track Problems
Example: Two athletes run on a 400 m circular track at 8 m/s and 6 m/s. How long does it take for the faster runner to lap the slower one? Relative speed = 8 − 6 = 2 m/s. Time to gain one full lap = 400/2 = 200 seconds = 3 minutes 20 seconds.
Conversion Factors — Detailed
| From | To | Conversion |
|---|---|---|
| km/h | m/s | ÷ 3.6 |
| m/s | km/h | × 3.6 |
| km | miles | ÷ 1.609 |
| miles | km | × 1.609 |
| km | m | × 1000 |
| m | km | ÷ 1000 |
Speed-Time Graphs
On a speed-time graph:
- The area under the graph = distance travelled
- The gradient = acceleration
- Horizontal line = constant speed
- Area below the time axis = negative displacement (if applicable)
Example: A car accelerates from 0 to 30 m/s over 10 seconds, maintains speed for 20 seconds, then decelerates to 0 over 5 seconds. Distance during acceleration = ½ × 10 × 30 = 150 m. Distance during constant speed = 30 × 20 = 600 m. Distance during deceleration = ½ × 5 × 30 = 75 m. Total distance = 150 + 600 + 75 = 825 m.
Relative Speed — General Case
When two objects move at angles, use vector addition: If object A has velocity vector v_A and object B has v_B, the relative velocity of A with respect to B is v_A − v_B (vector subtraction).
For collinear (same line) motion:
- Moving toward each other: relative speed = v₁ + v₂
- Moving in same direction: relative speed = |v₁ − v₂|
- Moving away from each other: relative speed = v₁ + v₂
UI Entrance Exam Patterns
Speed-distance-time questions typically include:
- Basic calculation (one of the three from the other two)
- Unit conversion (km/h to m/s or vice versa)
- Average speed for round trips
- Meeting point problems (starting same time or different times)
- Overtaking problems
- Boat in stream problems
- Circular track problems
⚡ Exam Strategy: For “when will they meet” problems with different start times, find where the first traveller is when the second starts, then calculate time for them to meet from that point. For circular track problems, the time to lap is track length divided by relative speed.
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