Time‑Speed‑Distance – Study Notes
For Jharkhand Teacher Eligibility Test (TET) – Quantitative Aptitude
1. Simple Explanation
The time‑speed‑distance relationship is the foundation of many quantitative problems.
It tells us how far an object travels, how fast it moves, and how long it takes.
-
Speed (or Velocity) – How fast an object moves.
Units: km/h, m/s, mph, etc. -
Distance – Total ground covered, irrespective of direction.
Units: km, m, miles, etc. -
Time – How long the motion lasts.
Units: hours, minutes, seconds, etc.
The basic relationship is:
Distance = Speed × Time
Speed = Distance ÷ Time
Time = Distance ÷ Speed
2. Key Formulae
| Quantity | Formula | Example |
|---|---|---|
| Distance | ( d = v \times t ) | 60 km/h × 3 h = 180 km |
| Speed | ( v = \dfrac{d}{t} ) | 120 km ÷ 2 h = 60 km/h |
| Time | ( t = \dfrac{d}{v} ) | 200 km ÷ 80 km/h = 2.5 h |
NOTE
- Always convert units to a common system before using the formulas.
- For relative speed problems (two objects moving towards/away from each other), add or subtract speeds accordingly.
3. Important Points to Remember
| Point | Details |
|---|---|
| Unit Consistency | If distance in km and time in hours → speed in km/h. For m/s, use meters and seconds. |
| Relative Motion | - Same direction: relative speed = |
| Changing Units | 1 km = 1000 m; 1 hour = 60 min = 3600 s; 1 mph = 1.609 km/h. |
| Speed vs. Velocity | Speed is scalar (no sign). Velocity is vector (has direction). In most TET problems, we deal with speed (scalar). |
| Work‑Time Relationship | For problems involving work (e.g., “A can do a job in 5 days, B in 8 days, how long together?”) – treat “work” as a type of distance. |
| Average Speed | If speed varies, average speed = Total distance ÷ Total time. |
| Instantaneous Speed | Not required for TET; only average speeds used. |
| Typical Question Types | 1. Single object: Find missing quantity. 2. Two objects moving toward/away. 3. Relative speed (e.g., a train overtaking another). 4. Work–time problems. 5. Unit conversion problems. |
4. Worked Example
Problem
A train travels from City A to City B (200 km) at a constant speed of 80 km/h. Another train starts from City B to City A at 10 km/h less speed, i.e., 70 km/h. If both trains start at the same time, when and where will they meet?
Step‑by‑Step Solution
| Step | Calculation | Explanation |
|---|---|---|
| 1. Relative speed | ( v_{\text{rel}} = 80 + 70 = 150 ) km/h | Since they move towards each other, add speeds. |
| 2. Time to meet | ( t = \dfrac{200}{150} = \dfrac{4}{3} ) h ≈ 1 h 20 min | Use ( t = d / v_{\text{rel}} ). |
| 3. Distance from City A | ( d_A = 80 \times \dfrac{4}{3} = \dfrac{320}{3} ≈ 106.67 ) km | Distance = speed × time. |
| 4. Distance from City B | ( d_B = 70 \times \dfrac{4}{3} = \dfrac{280}{3} ≈ 93.33 ) km | Same formula. |
| 5. Verification | ( d_A + d_B = 200 ) km | Checks out. |
Answer:
They will meet after 1 hour 20 minutes from the start, 106.67 km from City A (or 93.33 km from City B).
5. Quick Revision Summary
| Topic | Key Takeaway |
|---|---|
| Basic Relationship | (d = v \times t) – use to solve any missing variable. |
| Unit Conversion | Convert all distances to km or m and all times to hours or seconds before using formulas. |
| Relative Speed | Add speeds when objects move towards each other; subtract when in the same direction. |
| Work‑Time | Treat “work” like distance; use ( \text{Rate} = \frac{1}{\text{Time}}). |
| Average Speed | ( \text{Avg. Speed} = \frac{\text{Total Distance}}{\text{Total Time}}). |
| Common Mistake | Forgetting to convert minutes to hours or vice‑versa. |
| Practice Tip | Solve at least 5 problems of each type (single, relative, work) before the exam. |
Quick Self‑Check Questions (Optional)
- A cyclist travels 30 km in 1 h 15 min. What is his average speed in km/h?
- Two cars start from opposite ends of a 300 km road at the same time. Car A moves at 90 km/h, Car B at 60 km/h. How long until they meet