Time‑Speed‑Distance: Study Notes
(For Ladakh Police Sub‑Inspector (SI) – Quantitative Aptitude)
1. Simple Explanation
-
Speed / Velocity – How fast an object moves.
- Speed is a scalar (only magnitude).
- Velocity is a vector (magnitude + direction).
- In most police aptitude questions we use speed.
-
Distance – Total ground covered (path length).
-
Time – Duration of the motion.
The basic relationship:
[ \textbf{Speed} = \frac{\textbf{Distance}}{\textbf{Time}} ]
Rearranged for other variables:
[ \textbf{Distance} = \textbf{Speed} \times \textbf{Time} ] [ \textbf{Time} = \frac{\textbf{Distance}}{\textbf{Speed}} ]
Unit Consistency
Keep units the same while solving (e.g., km & hours, or meters & seconds).
1 km = 1000 m, 1 hr = 3600 s, 1 m/s = 3.6 km/h.
2. Key Points & Formulas
| Symbol | Meaning | Units | Typical Value in Police Context |
|---|---|---|---|
| (v) | Speed | km/h or m/s | 40–80 km/h for patrol cars |
| (d) | Distance | km or m | 5–30 km in an inspection route |
| (t) | Time | hr or s | 0.1–3 hr |
| (a) | Acceleration | km/h² or m/s² | Usually 0 in “constant‑speed” problems |
2.1 Conversion Factors
| from → to | Factor |
|---|---|
| km → m | ×1000 |
| km/h → m/s | ×(1000/3600) = ×0.2778 |
| m/s → km/h | ×3.6 |
| hr → min | ×60 |
| min → sec | ×60 |
2.2 Average Speed
When a vehicle moves at different speeds over equal distances:
[ v_{\text{avg}} = \frac{2v_1v_2}{v_1+v_2} ]
When it travels equal times at different speeds, use a weighted average.
2.3 Relative Speed
If two vehicles move toward each other:
[ v_{\text{rel}} = v_1 + v_2 ]
If they move in the same direction:
[ v_{\text{rel}} = |v_1 - v_2| ]
2.4 Distance Traveled While Accelerating/Decelerating
If acceleration is uniform, distance ( d = \frac{1}{2}(v_i + v_f)t ).
For exam‑level problems, acceleration is seldom used unless explicitly stated.
2.5 “Time for One Way / Return”
If a round trip is involved:
[ t_{\text{total}} = t_{\text{out}} + t_{\text{back}} ]
3. Worked Example (Police Scenario)
Problem
A patrol car leaves a police outpost at 10:00 AM.
- It travels 40 km to a checkpoint at a constant speed of 80 km/h.
- It then returns to the outpost at a constant speed of 60 km/h.
- How much time will it take to complete the round trip?
- At what time will it finish?
Solution
-
Outward trip
[ t_{\text{out}} = \frac{d}{v_{\text{out}}} = \frac{40\ \text{km}}{80\ \text{km/h}} = 0.5\ \text{h} = 30\ \text{min} ]
-
Return trip
[ t_{\text{back}} = \frac{d}{v_{\text{back}}} = \frac{40\ \text{km}}{60\ \text{km/h}} \approx 0.6667\ \text{h} = 40\ \text{min} ]
-
Total time
[ t_{\text{total}} = 30\ \text{min} + 40\ \text{min} = 70\ \text{min} = 1\ \text{h}\ 10\ \text{min} ]
-
Finish time
[ 10:00\ \text{AM} + 1\ \text{h}\ 10\ \text{min} = 11:10\ \text{AM} ]
Answer: The patrol car will finish the round trip at 11:10 AM, taking 1 h 10 min.
4. Quick Revision Summary
| Concept | Formula | Typical Use |
|---|---|---|
| Speed | (v = \frac{d}{t}) | Compute speed from distance/time |
| Distance | (d = v \times t) | Find distance covered |
| Time | (t = \frac{d}{v}) | Determine travel duration |
| Average speed (equal distances) | (v_{\text{avg}} = \frac{2v_1v_2}{v_1+v_2}) | Multi‑segment trips |
| Relative speed (opposite directions) | (v_{\text{rel}} = v_1 + v_2) | Pursuit or collision scenarios |
| Relative speed (same direction) | (v_{\text{rel}} = | v_1 - v_2 |
| Unit conversion | km↔m, hr↔min, km/h↔m/s | Keep units consistent |
| Round‑trip time | (t_{\text{total}} = t_{\text{out}} + t_{\text{back}}) | Patrol schedules |
Tips for the Exam
- Read carefully – identify if distances are equal or times are