Time‑Speed‑Distance (TSD) – Study Notes
(For Haryana Police Sub‑Inspector Exam – Quantitative Aptitude)
Why it matters
The TSD concept is the backbone of many quantitative problems. It tests logical reasoning, unit conversion, and basic algebra. Mastery of this topic gives you an edge in sections like “Reasoning” & “Numerical Ability”.
1. Simple Explanation
| Quantity | Symbol | Unit | Relationship |
|---|---|---|---|
| Time | (t) | hours, minutes, seconds | Duration of travel |
| Speed | (s) | km/h, m/s, mph | Distance traversed per unit time |
| Distance | (d) | km, meters, miles | Total length covered |
The fundamental equation is:
[ \boxed{d = s \times t} ]
If any one quantity is known, the other two can be found.
Tip – Always keep the units consistent before plugging numbers into the formula.
2. Key Points
| # | Point | Why it matters | Quick Check |
|---|---|---|---|
| 1 | Unit consistency | Mixing km and m or hrs and mins leads to wrong answers. | Convert all to common base (e.g., km & hrs). |
| 2 | Average speed vs. instantaneous speed | Most exam problems use average speed. | If no other info, assume constant speed. |
| 3 | Speed = Distance ÷ Time | Useful when time is given. | (s = \frac{d}{t}) |
| 4 | Time = Distance ÷ Speed | Useful when speed is given. | (t = \frac{d}{s}) |
| 5 | Speed = 1 ÷ Time for unit distance | Handy for “how long for 1 km”. | (s_{\text{1 km}} = \frac{1}{t_{\text{1 km}}}) |
| 6 | Convert minutes to hours | 1 hr = 60 min. | (t_{\text{hrs}} = \frac{t_{\text{min}}}{60}) |
| 7 | Convert km to meters | 1 km = 1000 m. | (d_{\text{m}} = d_{\text{km}}\times 1000) |
| 8 | Relative speed in “meet” problems | If two objects move towards each other, relative speed = sum. | (s_{\text{rel}} = s_1 + s_2) |
| 9 | Relative speed in “one follows another” | When one follows the other, relative speed = difference. | (s_{\text{rel}} = |
| 10 | Time to cover a part of a journey | Use proportion or direct TSD. | (t_{\text{part}} = \frac{d_{\text{part}}}{s}) |
3. Worked Example
Problem
A bus travels from City A to City B, a distance of 240 km. It covers the first 120 km in 2 hrs and the next 120 km in 1 hr 30 min. What is the average speed of the bus for the whole journey?
Step‑by‑Step
-
Convert all times to hours
- First part: (2,\text{hrs})
- Second part: (1,\text{hr} , 30,\text{min} = 1 + \frac{30}{60} = 1.5,\text{hrs})
-
Total time
[ t_{\text{total}} = 2 + 1.5 = 3.5,\text{hrs} ] -
Total distance
[ d_{\text{total}} = 120 + 120 = 240,\text{km} ] -
Average speed
[ s_{\text{avg}} = \frac{d_{\text{total}}}{t_{\text{total}}} = \frac{240}{3.5} \approx 68.57,\text{km/h} ]
Answer: The bus’s average speed ≈ 68.6 km/h.
Check – The first half took 2 hrs → speed = 120/2 = 60 km/h.
The second half took 1.5 hrs → speed = 120/1.5 = 80 km/h.
Weighted average falls between 60 & 80 → 68.6 km/h is plausible.
4. Quick Revision Summary
| Concept | Formula | Example |
|---|---|---|
| Distance | (d = s \times t) | 50 km = 100 km/h × 0.5 h |
| Speed | (s = \dfrac{d}{t}) | 120 km ÷ 2 h = 60 km/h |
| Time | (t = \dfrac{d}{s}) | 180 km ÷ 60 km/h = 3 h |
| Unit convert | 1 hr = 60 min, 1 km = 1000 m | 45 min = 0.75 h |
| Relative speed (towards) | (s_{\text{rel}} = s_1 + s_2) | Two trains 80 km/h & 60 km/h → 140 km/h |
| Relative speed (following) | (s_{\text{rel}} = | s_1 - s_2 |
| Average speed (mixed times) | (s_{\text{avg}} = \dfrac{d_{\text{total}}}{t_{\text{total}}}) | 240 km in 3.5 h → 68.6 km/h |
Common Mistakes to Avoid
- Mixing km with m or hrs with min without conversion.
- Forgetting to add times when parts of a trip are given separately.
- Assuming “average speed” is just the mean of speeds when times differ.
- Misinterpreting relative speed: use sum