Time – Speed – Distance (TSD) – Study Notes
(For Chhattisgarh PSC Quantitative Aptitude)
1. Simple Explanation
When an object moves at a constant speed, the distance it covers is directly proportional to the time it travels.
Distance = Speed × Time
Using this basic relationship we can solve problems where any two of the three quantities (time, speed, distance) are given.
2. Key Points
| Concept | Formula | Notes |
|---|---|---|
| Speed | v = d / t |
Speed is distance travelled per unit time. Common units: km/h, m/s, mph. |
| Distance | d = v × t |
Multiply speed by time. |
| Time | t = d / v |
Divide distance by speed. |
| Unit Conversion | 1 km = 1000 m, 1 h = 60 min = 3600 s |
Always convert units to match (e.g., if speed is km/h, convert distance to km or time to hours). |
| Average Speed | v_avg = total distance / total time |
Useful when speed changes. |
| Relative Speed | v_rel = v1 + v2 (approaching) or v_rel = v1 - v2 (moving apart) |
Apply when two objects move towards/away from each other. |
| Distance Covered in Half/Double Time | d ∝ t |
If speed constant, distance doubles when time doubles. |
| Speed Change | Δv = v2 – v1 |
Useful for acceleration problems. |
| Distance‑Time Graph | Slope = speed | The slope of a distance‑time graph gives the speed. |
Common Pitfalls
| Mistake | Fix |
|---|---|
| Mixing up units (km/h with km/min) | Convert before applying formula |
| Forgetting to divide when speed is per unit time | Remember v = d / t |
| Assuming speed is constant when not stated | Check wording: “at a constant speed” |
| Using “speed” to mean “velocity” incorrectly | In TSD problems, direction is ignored unless specified |
3. Worked Example
Problem (Typical PSC Question)
A train travels from City A to City B at a constant speed of 80 km/h. The journey covers a distance of 240 km.
(a) How long will the train take to reach City B?
(b) If the train departs from City B at 6:00 PM heading back to City A, what time will it arrive back at City A?
(c) If the train speeds up by 20 km/h on the return journey, how much faster will it arrive at City A?
(a) Time to reach City B
Use t = d / v
d = 240 km
v = 80 km/h
t = 240 / 80 = 3 hours
Answer: 3 hours.
(b) Arrival time back at City A
Departure from City B: 6:00 PM
Return speed = 80 km/h (same as outbound)
Time for return = 3 hours (same distance & speed).
6:00 PM + 3 hours = 9:00 PM
Answer: 9:00 PM.
(c) Faster return due to speed increase
Return speed = 80 + 20 = 100 km/h
Time for return at 100 km/h:
t₂ = 240 / 100 = 2.4 hours = 2 hours 24 minutes
Time saved = 3 hours – 2 hours 24 minutes = 36 minutes.
Answer: The train will arrive 36 minutes earlier, at 8:24 PM.
4. Quick Revision Summary
- Basic Relation –
Distance = Speed × Time - Solve for any variable:
t = d / vv = d / td = v × t
- Units matter – convert to consistent units before calculation.
- Average speed – total distance ÷ total time.
- Relative speed – add/subtract speeds for approaching/moving apart.
- Time–Distance Graph – slope = speed.
- Common Tricks –
- “Half the time → half the distance” (constant speed).
- “Triple the speed → triple the distance in the same time.”
Practice Tip:
- Write down the given values, decide which formula fits, and check units first.
- For multiple‑choice questions, plug numbers into all options quickly to eliminate wrong ones.
End of Notes