Time‑Speed‑Distance: Quick Study Notes
(Coast Guard – Quantitative Aptitude)
1. Simple Explanation
The Time‑Speed‑Distance (T‑S‑D) relationship is one of the most frequently used concepts in quantitative aptitude. It links the three quantities:
| Quantity | Symbol | Unit |
|---|---|---|
| Distance | (D) | km, m, nautical miles, etc. |
| Speed (or Velocity) | (S) | km/h, m/s, knots, etc. |
| Time | (T) | hours, minutes, seconds, etc. |
The core formula is:
[ \boxed{D ;=; S \times T} ]
This means Distance = Speed × Time.
If any two of the three values are known, the third can be found by rearranging the formula:
- (S = \dfrac{D}{T})
- (T = \dfrac{D}{S})
2. Key Points to Remember
| # | Point | Practical Tip |
|---|---|---|
| 1 | Units must match. If speed is km/h, time must be in hours; if speed is knots, use nautical miles. | Convert all values to the same unit before calculation. |
| 2 | Speed can be constant or variable. For most exam questions, speed is assumed constant unless stated otherwise. | If speed changes, split the journey into segments. |
| 3 | Time can be in fractional hours (e.g., 1 h 30 min = 1.5 h). | Convert minutes to hours: divide by 60. |
| 4 | Distance is cumulative. If you travel 100 km in 2 h and then another 50 km in 1 h, total distance = 150 km. | Use addition of distances for multi‑segment problems. |
| 5 | Useful derived formulas: | |
| • Average Speed (S_{\text{avg}} = \dfrac{\text{Total Distance}}{\text{Total Time}}). | ||
| • If two speeds and a ratio of distances are given, set up proportions. | ||
| 6 | Conversion between units: | |
| • 1 knot = 1.852 km/h. | ||
| • 1 nautical mile = 1.852 km. | ||
| • 1 hour = 60 minutes. 1 minute = 60 seconds. | ||
| 7 | Common traps: | |
| • Forgetting to convert minutes to hours. | ||
| • Mixing km/h with knots or nautical miles without converting. | ||
| • Using “speed” when “velocity” is actually required (negative direction). |
3. Worked Example
Problem
A Coast Guard cutter travels from Port A to Port B at a constant speed of 30 knots. The distance between the two ports is 120 nautical miles.
- How long will the journey take?
- If the cutter returns at a speed of 25 knots, how long will the return trip take?
- What is the total time for the round trip?
Solution
-
Outbound Trip
- Speed (S_1 = 30) knots
- Distance (D = 120) nautical miles
- Time (T_1 = \dfrac{D}{S_1} = \dfrac{120}{30} = 4) hours
-
Return Trip
- Speed (S_2 = 25) knots
- Distance (D = 120) nautical miles
- Time (T_2 = \dfrac{120}{25} = 4.8) hours
- Convert (0.8) hr → (0.8 \times 60 = 48) minutes
- So, (T_2 = 4) hr (48) min
-
Total Time
[ T_{\text{total}} = T_1 + T_2 = 4,\text{h} + 4,\text{h}48,\text{min} = 8,\text{h}48,\text{min} ]
Answer
- Outbound: 4 hours
- Return: 4 hours 48 minutes
- Round trip: 8 hours 48 minutes
Quick Calculation Tips
| Scenario | Formula | Example |
|---|---|---|
| Find Time | (T = \dfrac{D}{S}) | (T = \dfrac{150\text{ km}}{50\text{ km/h}} = 3\text{ h}) |
| Find Speed | (S = \dfrac{D}{T}) | (S = \dfrac{200\text{ km}}{4\text{ h}} = 50\text{ km/h}) |
| Find Distance | (D = S \times T) | (D = 55\text{ km/h} \times 2.5\text{ h} = 137.5\text{ km}) |
4. Quick Revision Summary
- Core Formula: (D = S \times T)
- Rearrange to find any missing quantity.
- Units: always convert to compatible units.
- Time in hours: minutes ÷ 60; seconds ÷ 3600.
- Average Speed: total distance ÷ total time.
- Segmented journeys: split into parts, compute each, then sum.
- Common pitfalls: unit mismatch, forgetting fractional hour conversion, misusing “speed” vs. “velocity.”
Practice Question
A buoy‑search operation covers a 250‑km stretch at 20 km/h. After a 30‑minute break, the team continues at 25 km/h.
- How long did the entire operation take?
(Use the same steps as above to solve.)
End of notes. Good luck with your preparation!