Time–Speed–Distance: Study Notes
Program: Sikkim Forest Guard
Subject: Quantitative Aptitude
Topic: Time–Speed–Distance
1. Simple Explanation
The Time–Speed–Distance (TSD) relationship links three fundamental variables of motion:
| Symbol | Meaning | Unit (SI) |
|---|---|---|
| (t) | Time taken | hours, minutes, seconds |
| (s) | Speed (average) | km/h, m/s |
| (d) | Distance traveled | km, m |
The core equation is:
[ \boxed{d = s \times t} ]
- Distance = Speed × Time
- Rearranged versions allow you to find any one variable when the other two are known:
[ s = \frac{d}{t}\quad,\quad t = \frac{d}{s} ]
2. Key Points to Remember
| Point | Why it matters |
|---|---|
| Units must match | Mixing km with miles or hours with minutes gives wrong answers. Convert if necessary. |
| Average speed | In most aptitude questions, “speed” refers to average speed over the whole journey. |
| Multiple legs | If a trip has several segments with different speeds, compute distance for each segment separately and sum them. |
| Time in fractions | 1 h 30 min = 1.5 h; 45 min = 0.75 h. |
| Speed change | “Speed increased by 20 km/h” → add 20 to the original speed, not multiply. |
| Distance change | “Distance increased by 30 km” → add 30 to the original distance, not multiply. |
| Round‑off | Keep at least one decimal place when intermediate steps involve fractions. |
| Check plausibility | Result should be realistic (e.g., a forest guard cannot cover 500 km in 1 h at 50 km/h). |
3. Worked Example
Problem (Forest‑Guard Context)
A forest guard starts from Camp A and walks to Camp B (120 km away) at a speed of 6 km/h.
After reaching Camp B, he immediately turns around and walks back to Camp A, but this time at 8 km/h.a) How long does the entire round trip take?
b) What is his average speed for the whole journey?
Step‑by‑Step Solution
-
Compute time for each leg
-
A → B
[ t_1 = \frac{d}{s_1} = \frac{120\ \text{km}}{6\ \text{km/h}} = 20\ \text{h} ] -
B → A
[ t_2 = \frac{d}{s_2} = \frac{120\ \text{km}}{8\ \text{km/h}} = 15\ \text{h} ]
-
-
Total time
[ t_{\text{total}} = t_1 + t_2 = 20\ \text{h} + 15\ \text{h} = 35\ \text{h} ]
-
Total distance
[ d_{\text{total}} = 120\ \text{km} + 120\ \text{km} = 240\ \text{km} ]
-
Average speed for round trip
[ s_{\text{avg}} = \frac{d_{\text{total}}}{t_{\text{total}}} = \frac{240\ \text{km}}{35\ \text{h}} \approx 6.86\ \text{km/h} ]
Answers
| Part | Result |
|---|---|
| a) Time for round trip | 35 hours |
| b) Average speed | ≈ 6.86 km/h |
4. Quick Revision Summary
| What | How to Apply |
|---|---|
| TSD formula | (d = s \times t) (or its inverses) |
| Unit consistency | Convert all units to the same system before calculation |
| Multiple segments | Sum distances and times separately, then apply formulas |
| Average speed | Total distance ÷ total time |
| Common traps | Mixing up “speed increased by” vs “speed multiplied by”; forgetting to convert minutes to hours |
Quick Practice Question (for you to try)
A forest guard patrols a 75 km trail at 5 km/h. After a break, he patrols the same trail again, but now at 7 km/h.
Find:
- Total time taken for both patrols.
- Average speed for the entire patrol.
Answer:
- (t_1 = 75/5 = 15) h, (t_2 = 75/7 \approx 10.71) h → total ≈ 25.71 h.
- (s_{\text{avg}} = 150/25.71 \approx 5.83) km/h.
Keep revisiting these steps, and you'll master Time–Speed–Distance problems with confidence!