Karnataka Forest Guard – Quantitative Aptitude
Topic: Time‑Speed‑Distance (TSD)
1. Simple Explanation
The Time‑Speed‑Distance relationship is a basic kinematic formula used to solve everyday problems of motion.
Formula
[ \text{Distance (D)} = \text{Speed (S)} \times \text{Time (T)} ] or in other forms
[ \text{Speed} = \frac{D}{T}, \qquad \text{Time} = \frac{D}{S} ]
Units
- Speed: km/hr, m/s, ft/min, etc.
- Time: hr, min, sec.
- Distance: km, m, ft, etc.
Tip: Convert all quantities to the same unit before applying the formula.
2. Key Points to Remember
| Concept | How to Apply | Common Trap |
|---|---|---|
| Uniform motion | Use single speed for the whole distance. | Forgetting unit conversion (e.g., hr ↔ min). |
| Non‑uniform motion | Break the trip into segments, compute each segment’s distance/time, then add. | Assuming the same speed for all segments. |
| Average speed | (\displaystyle \text{Avg Speed} = \frac{\text{Total Distance}}{\text{Total Time}}) | Mixing up ‘average speed’ with ‘average of speeds’. |
| Speed conversion | 1 km/hr = 5/18 m/s. 1 m/s = 3.6 km/hr. | Using wrong conversion factor. |
| Distance from speed & time | (D = S \times T). | Using wrong units for speed or time. |
| Time from speed & distance | (T = \frac{D}{S}). | Dividing distance by wrong speed (e.g., converting to m/s but keeping distance in km). |
| Speed from distance & time | (S = \frac{D}{T}). | Not converting time from minutes to hours when speed is in km/hr. |
| Relative speed | When two objects move towards/away from each other, add/subtract speeds. | Forgetting direction (toward = +, away = –). |
| Maximum/minimum speed | Recognise constraints like “at constant speed” or “maximum speed under a limit”. | Ignoring given constraints. |
| Estimation | Round numbers to nearest 10/5 for quick mental math. | Over‑estimating or under‑estimating by a large margin. |
3. Worked Example
Problem (Typical Forest Guard Question)
A forest guard drives a jeep from Point A to Point B covering 120 km in 2 hr 30 min.
a) Find the average speed of the jeep.
b) If the guard returns at 80 km/hr, how long will the return journey take?
Step‑by‑Step Solution
a) Average Speed
-
Convert time to hours:
(2 \text{ hr } 30 \text{ min} = 2 + \frac{30}{60} = 2.5 \text{ hr}) -
Apply formula:
[ \text{Avg Speed} = \frac{D}{T} = \frac{120 \text{ km}}{2.5 \text{ hr}} = 48 \text{ km/hr} ]
Answer a) 48 km/hr.
b) Return Journey Time
-
Distance of return = 120 km (same path).
-
Speed on return = 80 km/hr.
-
Time = ( \frac{D}{S} = \frac{120}{80} = 1.5 \text{ hr}).
-
Convert to H M:
(1.5 \text{ hr} = 1 \text{ hr } 30 \text{ min}).
Answer b) 1 hr 30 min.
4. Quick Revision Summary
- Core Formula: (D = S \times T)
(and its inverses: (S = D/T), (T = D/S)) - Unit consistency: always convert to the same unit before calculation.
- Average speed: total distance ÷ total time.
- Break it down: for varying speeds, compute each leg separately.
- Relative motion: add speeds when moving towards each other; subtract when moving apart.
- Common conversions:
- km/hr ↔ m/s: multiply/divide by ( \frac{5}{18}).
- hr ↔ min: 1 hr = 60 min.
- km ↔ m: 1 km = 1000 m.
- Estimation: round numbers to nearest 10/5 for speed, 5/10 km for distance, 5/10 min for time.
Tip for exam:
- Read the problem carefully and note all given values.
- Write down the known quantities and the required unknown.
- Choose the right formula, convert units if needed, and solve.
- Double‑check the answer’s plausibility (e.g., speed should be reasonable for a jeep, time should be positive).
Good luck with your preparation!