📚 Time‑Speed‑Distance (TSD) – Study Notes
(For Arunachal Pradesh Patwari / Lekhpal – Quantitative Aptitude)
1. Simple Explanation
The Time‑Speed‑Distance (TSD) topic deals with three related quantities:
| Quantity | Symbol | Unit |
|---|---|---|
| Distance | (D) | Kilometers (km), meters (m), etc. |
| Speed | (S) | km/h, m/s, miles/h, etc. |
| Time | (T) | Hours (h), minutes (min), seconds (s) |
The fundamental relation connecting them is:
[ \boxed{D = S \times T} ]
In words: Distance = Speed × Time.
All other useful equations are just algebraic rearrangements of this core formula.
2. Key Points
| Concept | Formula | When to Use | Key Insight |
|---|---|---|---|
| Direct TSD formula | (D = S \times T) | All basic problems | Speed and time are constants. |
| Reverse formulas | (S = \dfrac{D}{T}) (T = \dfrac{D}{S}) |
When one of the other two values is missing | Speed = Distance ÷ Time; Time = Distance ÷ Speed. |
| Average speed (mixed pace) | (\displaystyle \text{Avg. Speed} = \frac{\text{Total Distance}}{\text{Total Time}}) | Two or more legs of a journey with different speeds | Weighted mean of speeds. |
| Same distance, different times | (S_1 = \dfrac{D}{T_1}), (S_2 = \dfrac{D}{T_2}) | Compare performances | Faster speed ⇒ less time. |
| Same speed, different distances | (T_1 = \dfrac{D_1}{S}), (T_2 = \dfrac{D_2}{S}) | Estimate time for new leg | Time is proportional to distance. |
| Relative speed (moving objects) | (\displaystyle \text{Relative Speed} = S_{\text{A}} \pm S_{\text{B}}) | Two objects moving towards or away from each other | + for opposite directions, – for same direction. |
| Converting units | 1 km/h = 1000 m/3600 s = 0.27778 m/s | When time/distance units mismatch | Always bring to common units before calculation. |
| Speed in miles/hour or m/s | Use the appropriate conversion factor | When the problem gives mixed units | Keep track of units to avoid mistakes. |
| Average speed on round trip | (\displaystyle \text{Avg. Speed} = \frac{2S_1S_2}{S_1 + S_2}) | Two legs with different speeds and equal distance | Inverse relation to travel time. |
| Time to travel a given distance at constant speed | (T = \frac{D}{S}) | Quick “how long does it take?” | Straight‑forward division. |
| Distance covered in a given time at constant speed | (D = S \times T) | Quick “how far will you go?” | Multiplication. |
3. Worked Example
Problem
A patwari drives from Village A to Village B, which is 120 km away.
- First 60 km he drives at 40 km/h.
- For the remaining 60 km he drives at 50 km/h.
Question:
- How long does the whole journey take?
- What is the average speed for the entire trip?
Solution
1. Time for each leg
-
First leg:
[ T_1 = \frac{D_1}{S_1} = \frac{60\ \text{km}}{40\ \text{km/h}} = 1.5\ \text{h} ] -
Second leg:
[ T_2 = \frac{D_2}{S_2} = \frac{60\ \text{km}}{50\ \text{km/h}} = 1.2\ \text{h} ]
2. Total time
[ T_{\text{total}} = T_1 + T_2 = 1.5\ \text{h} + 1.2\ \text{h} = 2.7\ \text{h} ]
3. Average speed
[ \text{Avg. Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{120\ \text{km}}{2.7\ \text{h}} \approx 44.44\ \text{km/h} ]
Quick Check
- Distance × Time = Speed?
[ 44.44\ \text{km/h} \times 2.7\ \text{h} \approx 120\ \text{km} ] ✅
Take‑away
- Always compute time for each leg separately.
- Sum the times → total time.
- Use total distance / total time for average speed.
4. Quick Revision Summary
| Topic | Formula | Example |
|---|---|---|
| Distance | (D = S \times T) | 40 km/h × 1.5 h = 60 km |
| Speed | (S = \dfrac{D}{T}) | 120 km ÷ 2.7 h ≈ 44.44 km/h |
| Time | (T = \dfrac{D}{S}) | 120 km ÷ 44.44 km/h ≈ 2.7 h |
| Average speed (mixed legs) | (\displaystyle \frac{D_{\text{total}}}{T_{\text{total}}}) | 120 km ÷ 2.7 h = 44.44 km/h |
| Relative speed (towards each other) | (S_{\text{rel |