Time‑Speed‑Distance (T‑S‑D) – SSC CPO Study Notes
(Quantitative Aptitude – Time‑Speed‑Distance)
1. Simple Explanation
The Time‑Speed‑Distance relationship is a basic physics formula that lets you find any one of the three quantities if the other two are known.
| Symbol | Quantity | Unit (common) | Short description |
|---|---|---|---|
| S | Distance | km, m, km/h, etc. | How far something travels |
| T | Time | hr, min, s | How long it takes |
| V | Speed (or velocity) | km/hr, m/s | Rate of motion (distance per unit time) |
Fundamental formula
[ \boxed{S = V \times T} ]
Rearranging gives the other two:
[ V = \frac{S}{T} \qquad\text{and}\qquad T = \frac{S}{V} ]
2. Key Points to Remember
| # | Concept | Why it matters for SSC CPO |
|---|---|---|
| 1 | Units must match | If distance is in km, speed must be km/hr and time in hours. |
| 2 | Conversion | 1 hr = 60 min, 1 km = 1000 m, 1 m/s = 3.6 km/hr. |
| 3 | Average speed | Use the total distance & total time, not instantaneous speeds. |
| 4 | Relative speed | When two objects move towards or away from each other: (V_{rel}=V_1\pm V_2). |
| 5 | Multiple legs | Sum distances, sum times, then find average speed if required. |
| 6 | Speed change | If speed changes in sections, handle each section separately. |
| 7 | Common pitfalls | Mixing up “speed” with “velocity”, forgetting to convert units, rounding mistakes. |
3. Worked Example
Question
A train leaves City‑A for City‑B at 08:00 am.
- It travels 120 km at 60 km/hr for the first 2 hours.
- Then it takes a 30‑minute break.
- After the break it continues at 80 km/hr for the remaining distance to City‑B, which is 160 km from the start.
How many hours will the train take to reach City‑B?
Step‑by‑step Solution
-
Find distance covered in the first segment
[ S_1 = V_1 \times T_1 = 60 \text{ km/hr} \times 2 \text{ hr} = 120 \text{ km} ] -
Check remaining distance
Total distance (S_{total}=160) km.
Remaining distance (S_2 = S_{total} - S_1 = 160 - 120 = 40) km. -
Time for second segment
[ T_2 = \frac{S_2}{V_2} = \frac{40 \text{ km}}{80 \text{ km/hr}} = 0.5 \text{ hr} = 30 \text{ min} ] -
Add break time
Break = 30 min = 0.5 hr. -
Total travel time
[ T_{total} = T_1 + \text{break} + T_2 = 2 + 0.5 + 0.5 = 3 \text{ hr} ] -
Arrival time
08:00 am + 3 hr = 11:00 am.
Quick Verification
- Distance travelled = (120 + 40 = 160) km ✔
- Total time = 3 hr ✔
Answer: The train reaches City‑B at 11:00 am and the travel duration is 3 hours.
4. Quick Revision Summary
| Topic | Formula | Typical Question Type |
|---|---|---|
| Basic T‑S‑D | (S = V \times T) | “How long will it take to travel X km at Y km/hr?” |
| Speed from distance & time | (V = \dfrac{S}{T}) | “What speed covers X km in Y hrs?” |
| Time from distance & speed | (T = \dfrac{S}{V}) | “How long does it take to cover X km at Y km/hr?” |
| Relative speed | (V_{rel} = V_1 \pm V_2) | “Two trains meet after Z hrs; find speeds.” |
| Multiple segments | Sum of distances = total; sum of times = total | “A journey consists of two legs with different speeds.” |
| Unit conversion | 1 hr = 60 min; 1 km = 1000 m; 1 m/s = 3.6 km/hr | “Convert 20 m/s to km/hr.” |
| Average speed | (V_{avg} = \dfrac{S_{total}}{T_{total}}) | “Find average speed for a round trip.” |
Study Tips
- Always write the units when you compute each quantity.
- Check consistency before solving (e.g., if distance is in km and time in minutes, convert time to hours).
- Practice sketching a simple diagram of the journey – it helps visualise multiple legs.
- Use a quick one‑liner:
S = V × T→T = S ÷ V→V = S ÷ T
- Solve past SSC CPO papers focusing on T‑S