Time–Speed–Distance (TSD) – Quick Study Notes
(For Chandigarh Police Constable – Quantitative Aptitude)
1. Simple Explanation
The TSD relationship connects three fundamental quantities:
- Speed (S) – distance covered per unit time (e.g., km/h, m/s).
- Time (T) – duration taken to cover a distance.
- Distance (D) – total ground covered.
The basic equation is:
[ \boxed{D = S \times T} ]
From this you can solve for any one variable if the other two are known.
2. Key Points
| Concept | What to remember |
|---|---|
| Units must match | If speed is in km/h, time must be in hours; if speed is in m/s, time in seconds. |
| Speed = Distance / Time | (S = \frac{D}{T}) |
| Time = Distance / Speed | (T = \frac{D}{S}) |
| Distance = Speed × Time | (D = S \times T) |
| Relative motion | When two objects move towards/away from each other: use relative speed = speed₁ ± speed₂. |
| Average speed | Only if the speed is constant; otherwise, use total distance / total time. |
| Conversions | 1 km = 1000 m; 1 hour = 3600 s. |
| Common traps | Forgetting to convert units; mixing up “relative speed” direction (plus for same direction, minus for opposite). |
| Useful shortcuts | For “half the distance in half the time” problems, the ratio of distance to time remains the same. |
3. Common Formulas
| Formula | Variables | Example |
|---|---|---|
| Distance | (D = S \times T) | A car travels at 60 km/hr for 2 hrs → (D = 60 \times 2 = 120) km |
| Speed | (S = \frac{D}{T}) | 150 km in 3 hrs → (S = \frac{150}{3} = 50) km/hr |
| Time | (T = \frac{D}{S}) | 300 km at 75 km/hr → (T = \frac{300}{75} = 4) hrs |
| Relative Speed | (S_{\text{rel}} = S_1 \pm S_2) | Two trains moving towards each other: 80 km/hr + 60 km/hr = 140 km/hr |
| Average Speed | (S_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}}) | 100 km + 200 km in 3 hrs + 5 hrs → (S_{\text{avg}} = \frac{300}{8} = 37.5) km/hr |
4. Worked Example
Problem
A cyclist travels 30 km north in 2 hours. A motorbike starts from the same point 30 minutes later, heading east at 60 km/h. How long will it take the motorbike to catch up to the cyclist if the cyclist continues at the same speed?
Step‑by‑step solution
-
Find the cyclist’s speed
[ S_{\text{cyclist}} = \frac{30,\text{km}}{2,\text{h}} = 15,\text{km/h} ] -
Find the cyclist’s lead distance when the motorbike starts
The cyclist rides for 0.5 h before the motorbike starts: [ D_{\text{lead}} = 15,\text{km/h} \times 0.5,\text{h} = 7.5,\text{km} ] -
Set up relative motion
The motorbike is moving east, the cyclist north; they are moving perpendicular to each other.
The distance between them at any time (t) (in hours after the motorbike starts) is: [ D(t) = \sqrt{(7.5)^2 + (60t)^2} ] We want the time when (D(t) = 0) – i.e., when the motorbike catches the cyclist, so we need a point where their positions coincide.
However, because directions are perpendicular, the motorbike can never catch the cyclist unless the cyclist changes direction.
This problem is actually a relative speed problem only if they move in the same direction.
Thus, the motorbike will never catch the cyclist.(If the cyclist turned east at the same speed, we would compute relative speed: 60 – 15 = 45 km/h. Then time to cover the 7.5 km lead: (t = \frac{7.5}{45} = 0.1667) h ≈ 10 min.)
Takeaway
- Always confirm the direction of motion before applying relative speed.
- For non‑parallel paths, use distance formula with Pythagoras.
5. Quick Revision Summary
| Topic | Quick Check |
|---|---|
| Basic TSD | (D = S \times T) |
| Speed | (S = \frac{D}{T}) |
| Time | (T = \frac{D}{S}) |
| Relative Speed | Same direction → +; opposite → – |
| Average Speed | Total distance ÷ Total time |
| Unit Consistency | Convert all to same units before plugging in |
| Common Pitfalls | Forgetting unit conversion; mis‑applying relative speed for perpendicular motion |
Quick Practice (to test yourself)
- A bus travels 120 km in 2.5 hrs. What