Time–Speed–Distance – Study Notes
(Quantitative Aptitude – Manipur Police SI Exam)
1. Simple Explanation
The Time–Speed–Distance (TSD) topic deals with three inter‑related quantities in motion:
| Quantity | Symbol | Typical Units |
|---|---|---|
| Distance | (D) | km, m, km/h etc. |
| Speed | (S) | km/h, m/s, km/min |
| Time | (T) | hr, min, s |
The fundamental relationship is
[ \boxed{D = S \times T} ]
From this we can derive the other two handy equations:
[ S = \frac{D}{T}, \qquad T = \frac{D}{S} ]
Key Idea – If you know any two of the three quantities, you can find the third.
2. Key Points & Useful Tips
| Point | What It Means | Why It Matters |
|---|---|---|
| Unit Consistency | All quantities must be in compatible units before plugging into an equation. | Avoids errors like 60 km ÷ 30 min → 2 km/min (wrong). Convert minutes to hours or km to m. |
| Speed Conversion | (1 \text{ km/h} = \frac{1000}{3600}\text{ m/s} = 0.2778 \text{ m/s}) | Needed when distance is in meters and time in seconds. |
| Relative Speed | When two objects move towards/away from each other: Relative Speed = Speed1 ± Speed2 |
Common in train/boat problems. |
| Average Speed | For a round trip at different speeds: (\displaystyle \text{Avg Speed} = \frac{2v_1v_2}{v_1+v_2}) | Avoids calculating time separately for both legs. |
| Speed Difference | When one object is faster by a known amount: Speed_fast = Speed_slow + ΔS |
Used in “difference of speeds” questions. |
| Time Difference | When one object takes less time by a known amount: T_fast = T_slow – ΔT |
Appears in “time difference” problems. |
| Trivial Cases | If distance is 0, speed or time is 0. | Check for such edge cases in exam questions. |
| Checking Reasonableness | Rough mental estimation (e.g., a car covers ~60 km in 1 hr at 60 km/h). | Quick sanity check before final answer. |
Frequently Used Conversion Table
| Unit | Equivalent |
|---|---|
| 1 hr | 60 min |
| 1 km | 1000 m |
| 1 km/h | 0.27778 m/s |
| 1 m/s | 3.6 km/h |
| 1 min | 60 s |
3. Worked Example
Problem (Typical of Police SI Exam)
Two police cars start from the same point simultaneously.
Car A travels at 80 km/h, Car B at 50 km/h.
After how many minutes will Car B be 100 km behind Car A?
Step‑by‑Step Solution
-
Relative speed (difference because both move in the same direction)
[ S_{\text{rel}} = 80 - 50 = 30 \text{ km/h} ] -
Distance to be closed
[ D_{\text{rel}} = 100 \text{ km} ] -
Time required
[ T = \frac{D_{\text{rel}}}{S_{\text{rel}}} = \frac{100}{30} \text{ h} = 3.\overline{3}\text{ h} ] -
Convert to minutes
[ 3.\overline{3}\text{ h} \times 60 \text{ min/h} = 200 \text{ min} ]
Answer: Car B will be 100 km behind Car A after 200 minutes (3 h 20 min).
4. Quick Revision Summary
| Concept | Formula | Quick Tip |
|---|---|---|
| Distance | (D = S \times T) | Multiply speed by time. |
| Speed | (S = \frac{D}{T}) | Divide distance by time. |
| Time | (T = \frac{D}{S}) | Divide distance by speed. |
| Relative Speed (same direction) | (S_{\text{rel}} = | S_1 - S_2 |
| Relative Speed (opposite direction) | (S_{\text{rel}} = S_1 + S_2) | Add the speeds. |
| Average Speed (round trip) | (\displaystyle \frac{2v_1v_2}{v_1+v_2}) | Use the harmonic mean. |
| Time Difference | (T_{\text{fast}} = T_{\text{slow}} - \Delta T) | Subtract the given time gap. |
| Speed Difference | (S_{\text{fast}} = S_{\text{slow}} + \Delta S) | Add the given speed gap. |
Remember: Always align units first, then apply the appropriate formula. Quick mental checks (e.g., “a 60 km/h car covers 60