Time – Speed – Distance (TSD)
Delhi TET – Quantitative Aptitude
(Simple, clear, and concise notes for quick revision)
1. Simple Explanation
The Time–Speed–Distance relationship is the foundation of all kinematics problems.
It tells us how long it takes, how fast a vehicle moves, and how far it travels.
The basic relation is
[ \boxed{\text{Distance} = \text{Speed} \times \text{Time}} ]
or, in other useful forms,
[ \text{Speed} = \frac{\text{Distance}}{\text{Time}}, \qquad \text{Time} = \frac{\text{Distance}}{\text{Speed}} ]
All quantities are expressed in compatible units (e.g., km, m, h, s).
2. Key Formulas
| Quantity | Formula | Example of Use |
|---|---|---|
| Distance (D) | (D = S \times T) | Find distance traveled at 60 km/h for 2 h |
| Speed (S) | (S = \frac{D}{T}) | Find speed when 120 km is covered in 3 h |
| Time (T) | (T = \frac{D}{S}) | Find time to cover 90 km at 30 km/h |
| Relative Speed (S₁+S₂) | (S_{\text{rel}} = S_1 + S_2) (for two objects moving towards each other) | Two trains moving towards each other |
| Relative Speed (S₁−S₂) | (S_{\text{rel}} = S_1 - S_2) (for two objects moving in same direction) | Faster train overtakes a slower one |
| Distance with Stop Time | (D = S \times (T - \text{stop})) | When a vehicle stops part of the journey |
Tip: Always check units before plugging numbers. Convert km to m or h to s if needed.
3. Key Points to Remember
- Unit Consistency – Convert all values to the same unit system before calculation.
- Ex: 60 km/h → 16.67 m/s (divide by 3.6).
- Direction Matters – For relative speed, add speeds when moving toward each other; subtract when moving in the same direction.
- Stop Time – If a vehicle stops for (t_{\text{stop}}) hours, subtract it from total travel time before multiplying by speed.
- Average Speed – If speed varies, use average speed (\displaystyle \bar{S} = \frac{D_{\text{total}}}{T_{\text{total}}}).
- Common Mistakes
- Mixing up units (km with m, h with min).
- Forgetting to subtract stop time.
- Using wrong sign for relative speed.
4. Worked Example
Problem
Two trains start from stations A and B, which are 200 km apart.
- Train‑1 departs from A heading east at 80 km/h.
- Train‑2 departs from B heading west at 60 km/h after a 30‑minute delay.
At what time will the trains meet, and how far from station A will they meet?
Solution
-
Convert delay to hours
[ 30 \text{ min} = 0.5 \text{ h} ] -
Relative speed
Both trains move toward each other:
[ S_{\text{rel}} = 80 + 60 = 140 \text{ km/h} ] -
Time to meet after Train‑2 starts
[ T_{\text{meet}} = \frac{\text{Distance}}{S_{\text{rel}}} = \frac{200}{140} = 1.\overline{4286} \text{ h} ] -
Total time from Train‑1’s departure
Train‑1 has already been traveling for (0.5) h when Train‑2 starts:
[ T_{\text{total}} = 0.5 + 1.\overline{4286} = 1.9286 \text{ h} ] -
Distance from station A
Train‑1’s distance: [ D_A = 80 \times 1.9286 \approx 154.29 \text{ km} ] -
Check
Distance covered by Train‑2:
[ D_B = 60 \times 1.\overline{4286} \approx 85.71 \text{ km} ]
Sum (154.29 + 85.71 = 240) km? Wait – we mis‑added.
Actually, the correct check:
[ D_A + D_B = 154.29 + 85.71 = 240 \text{ km} ]
But the stations are 200 km apart.
Fix: We double‑counted because we used relative speed incorrectly?
Oops! The proper method:
[ T_{\text{meet}} = \frac{200}{140} = 1.\overline{4286} ]
Distance from A: [ D_A = 80 \times 1.\overline{4286} \approx 114.29 \text{ km} ]
Distance from B: [ D_B = 60 \times 1.\overline{4286} \approx 85.71 \text{ km} ]
Sum (114.29 + 85.71 = 200) km ✔️
Answer
- They meet (1.43) h after Train‑2 starts, i.e. (1.93) h after Train‑1 departs.
- The meeting point is about 114 km from station A (or 86 km