Time‑Speed‑Distance – Quantitative Aptitude
(Sikkim PSC – Quantitative Aptitude – Time‑Speed‑Distance)
Objective:
- Understand the basic relationship between time, speed and distance.
- Apply formulas to solve simple and slightly‑modified problems.
- Memorise the most common variations that appear in the Sikkim PSC exam.
1. Simple Explanation
| Quantity | Symbol | Unit | How It’s Defined |
|---|---|---|---|
| Distance | (d) | km / m / km | The total length travelled. |
| Speed | (v) | km/h or m/s | The rate at which distance is covered. (Average speed if the rate is not constant.) |
| Time | (t) | h / s | The duration of the travel. |
The fundamental relationship is:
[ \boxed{d = v \times t} ]
- If you know any two of the three quantities, you can find the third by rearranging the formula:
[ \begin{aligned} t &= \frac{d}{v} \ v &= \frac{d}{t} \end{aligned} ]
Key idea:
Speed is a ratio of distance to time.
All other formulas are just algebraic rearrangements of the same relationship.
2. Key Points & Variations
| Variation | Formula | Typical Question |
|---|---|---|
| Average speed (constant speed) | (v_{\text{avg}} = \frac{d}{t}) | “A bus covers 120 km in 3 hrs. What is its average speed?” |
| Distance with time | (d = v \times t) | “A train travels at 80 km/h for 2.5 hrs. Find the distance.” |
| Time with speed | (t = \frac{d}{v}) | “A cyclist covers 50 km at 20 km/h. How long does it take?” |
| Speed with distance | (v = \frac{d}{t}) | “A runner completes 10 km in 30 min. What is the speed?” |
| Relative motion (two moving objects) | (d_{\text{rel}} = (v_1 \pm v_2) \times t) | “Two cars start from the same point, one moving east, the other west, at 60 km/h and 40 km/h respectively. After 1.5 hrs, how far apart are they?” |
| Distance after a break | (d_{\text{total}} = v \times t_{\text{first}} + v \times t_{\text{second}}) | “A bike travels 30 km in 1 hr, rests, then travels 20 km in 0.7 hr.” |
| Speed change (constant acceleration) | (d = v_{\text{avg}} \times t) where (v_{\text{avg}} = \frac{v_1 + v_2}{2}) | “A car accelerates uniformly from 0 to 40 km/h in 5 min. Find distance travelled.” |
| Unit conversion | 1 km/h = ( \frac{5}{18} ) m/s | “Convert 60 km/h to m/s.” |
| Problem with fractions | Use decimals or fractions carefully | “A journey of 90 km takes 1(\frac{1}{2}) hrs.” |
Common Pitfalls
- Mixing units – always convert to the same unit before applying the formula.
- Average speed vs. speed at a point – the problem statement will tell you if the speed is constant or average.
- Relative motion sign – use “+” for same direction, “−” for opposite directions.
- Time in hours – if time is given in minutes, divide by 60.
3. Worked Example
Problem
A train travels from City A to City B at a constant speed of 90 km/h.
The journey takes 2 hours 30 minutes.
- Find the distance between the two cities.
- If the train returns to City A at 60 km/h, how long will the return journey take?
Step‑by‑step Solution
1. Distance from A to B
Convert time to hours:
(2 \text{ hrs } 30 \text{ min} = 2 + \frac{30}{60} = 2.5) hrs
Use (d = v \times t):
[ d = 90 \text{ km/h} \times 2.5 \text{ h} = 225 \text{ km} ]
Answer: 225 km.
2. Time for return journey
Distance is the same: 225 km.
Speed on return = 60 km/h.
Use (t = \frac{d}{v}):
[ t = \frac{225 \text{ km}}{60 \text{ km/h}} = 3.75 \text{ h} ]
Convert 0.75 h to minutes:
(0.75 \times 60 = 45) min.
Answer: 3 hours 45 minutes.
Quick Check
- Distance = Speed × Time – both directions use the same distance.
- Return time is longer because speed is lower.
4. Quick Revision Summary
| Concept | Formula | When to Use |
|---|---|---|
| Distance | (d = v \times t) | Known speed & time |
| Time | (t = \frac{d}{v}) | Known distance & speed |
| Speed | (v = \frac{d}{t}) | Known distance & time |
| Relative motion | (d_{\text{rel}} = (v_1 \pm v_2) \times t) | Two objects moving |
| Acceleration (uniform) | (d = \frac{(v_1 + |