Time–Speed–Distance (TSD) – Study Notes
(Quantitative Aptitude – Competitive Exams – IIT‑JEE, JEE‑Main, NTSE, etc.)
1. Core Concepts
| Variable | Symbol | Unit | Meaning |
|---|---|---|---|
| Distance | (d) | km, m, mi, ft | Space travelled in a straight line |
| Speed | (v) | km h⁻¹, m s⁻¹ | Instantaneous rate of motion |
| Time | (t) | h, s, min | Duration of travel |
Key Relationship
[ v = \frac{d}{t}\quad\Longleftrightarrow\quad d = v,t\quad\Longleftrightarrow\quad t = \frac{d}{v} ]
2. Derivation from Instantaneous Velocity
For a continuous motion, instantaneous speed is (v(t) = \frac{ds}{dt}).
Integrating over a time interval ([0,T]):
[ \int_0^T v(t),dt = \int_0^T \frac{ds}{dt},dt \Longrightarrow \int_0^T v(t),dt = s(T)-s(0) ]
If (v(t)=) constant, the integral becomes (vT) giving the familiar (d = v,t).
Takeaway: TSD is a special case of the integral of speed over time.
3. Average Speed vs. Average Rate
| Situation | Average Speed | Average Rate |
|---|---|---|
| Uniform Motion | (v_{\text{avg}} = v_{\text{const}}) | Same as speed |
| Non‑Uniform Motion | (v_{\text{avg}} = \frac{d_{\text{total}}}{t_{\text{total}}}) | Same as average speed (for scalar motion) |
| Round Trip | (\displaystyle v_{\text{avg}} = \frac{2d}{t_{\text{out}} + t_{\text{back}}}) | Same |
Tricky Point – If the distance travelled is not a straight line (e.g., round trip), average speed uses total distance, not net displacement.
4. Relative Motion (Two Objects)
Let two objects (A) and (B) move along a straight line.
- Speeds: (v_A, v_B)
- Direction: + for same direction, − for opposite
4.1 Relative Speed
- If moving towards each other: (v_{\text{rel}} = v_A + v_B)
- If moving away: (v_{\text{rel}} = |v_A - v_B|)
4.2 Time to Meet
If initial separation is (d):
[ t_{\text{meet}} = \frac{d}{v_{\text{rel}}} ]
4.3 Application: Chase Problems
When one object starts later, add the delay:
(t = \frac{d}{v_{\text{rel}}} + \text{delay}).
5. Mixed‑Unit Conversion
| Unit | Conversion |
|---|---|
| 1 mi | 1.609 km |
| 1 ft | 0.3048 m |
| 1 hr | 3600 s |
| 1 min | 60 s |
Practice Tip: Convert all speeds into consistent units before applying TSD formulae.
6. Practical Formulae
| Scenario | Formula | Explanation |
|---|---|---|
| Constant speed | (d = vt) | Direct |
| Average speed | (v_{\text{avg}} = \frac{d_{\text{total}}}{t_{\text{total}}}) | Use for non‑uniform motion |
| Time for a given distance | (t = \frac{d}{v}) | |
| Speed for a given distance/time | (v = \frac{d}{t}) | |
| Relative speed | (v_{\text{rel}} = v_1 \pm v_2) | + for opposite directions |
| Speed of a moving object on a moving platform | (v_{\text{total}} = v_{\text{platform}} \pm v_{\text{object}}) |
7. Tricky Points & Common Pitfalls
| # | Pitfall | Correct Approach |
|---|---|---|
| 1 | Using displacement instead of distance in average speed. | Always use total distance travelled. |
| 2 | Neglecting direction in relative speed. | Treat opposite directions as subtraction; same direction as addition. |
| 3 | Unit mismatch (mi/h vs km/h). | Convert to a single unit system first. |
| 4 | Assuming instantaneous speed equals average speed for non‑uniform motion. | Use integration or weighted average. |
| 5 | Not accounting for delays in chase problems. | Add the delay to the meet time. |
| 6 | Confusing “speed” with “velocity”. | For TSD we need speed (scalar). |
8. Sample Problem Walkthrough
Problem:
A train travels 120 km from City A to City B. It then reverses and goes back to City A. The outbound journey takes 2 h 30 min, while the return takes 3 h 10 min.
Questions:
- What was the average speed of the train for the whole trip?
- What was the average speed during the outbound leg?
Solution:
- Total distance (d_{\text{tot}} = 120 + 120 = 240) km.
Total time (t_{\text{tot}} = 2.5 + 3.1667 \approx 5.6667) h.
[ v_{\text{avg whole}} = \frac{240}{5.6667} \approx 42.35\ \text{