A train, 250 m long, passes a railway platform 200 m long, in 45 s with a uniform speed. What is the time (in seconds) taken by the train to pass a man cycling in the direction of the train at a speed of 6 km/h?
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Understanding the Problem: Train Passing Platform and a Man
This problem involves calculating the speed of a train using the information about it passing a railway platform and then using that speed to find the time it takes for the train to pass a man cycling in the same direction.
We need to apply the concepts of speed, distance, time, and relative speed to solve this problem. The units need to be consistent throughout the calculation, so we'll convert speeds to metres per second (m/s) as the lengths and time are given in metres and seconds.
Step 1: Calculate the Train's Speed
When a train passes a platform, the total distance covered by the train is equal to the sum of its own length and the length of the platform.
- Length of the train = 250 m
- Length of the platform = 200 m
- Total distance covered = Length of train + Length of platform = 250 m + 200 m = 450 m
- Time taken to pass the platform = 45 s
The speed of the train can be calculated using the formula: Speed = Distance / Time.
Speed of train = $\frac{\text{Total distance}}{\text{Time taken}}$
Speed of train = $\frac{450 \text{ m}}{45 \text{ s}}$ = 10 m/s
Step 2: Convert the Man's Speed to m/s
The speed of the man is given in kilometers per hour (km/h). We need to convert this speed to meters per second (m/s) to maintain consistent units.
- Man's speed = 6 km/h
To convert km/h to m/s, we multiply by the conversion factor $\frac{5}{18}$.
Man's speed in m/s = $6 \times \frac{5}{18}$ m/s
Man's speed in m/s = $\frac{30}{18}$ m/s = $\frac{5}{3}$ m/s
Step 3: Calculate the Relative Speed
The man is cycling in the same direction as the train. When two objects move in the same direction, their relative speed is the difference between their individual speeds.
- Speed of train = 10 m/s
- Speed of man = $\frac{5}{3}$ m/s
Relative speed = Speed of train - Speed of man
Relative speed = $10 - \frac{5}{3}$ m/s
To subtract, we find a common denominator:
Relative speed = $\frac{30}{3} - \frac{5}{3}$ m/s
Relative speed = $\frac{30 - 5}{3}$ m/s = $\frac{25}{3}$ m/s
This relative speed is the speed at which the train gains on the man.
Step 4: Calculate the Time to Pass the Man
When a train passes a man (or any point object), the distance covered by the train relative to the man is the length of the train itself.
- Distance to be covered (length of train) = 250 m
- Relative speed = $\frac{25}{3}$ m/s
The time taken to pass the man is calculated using the formula: Time = Distance / Relative Speed.
Time taken = $\frac{\text{Length of train}}{\text{Relative speed}}$
Time taken = $\frac{250 \text{ m}}{\frac{25}{3} \text{ m/s}}$
Time taken = $250 \times \frac{3}{25}$ s
Time taken = $(10 \times 25) \times \frac{3}{25}$ s
Time taken = $10 \times 3$ s = 30 s
So, the time taken by the train to pass the man cycling in the same direction is 30 seconds.
| Quantity | Value | Units |
|---|---|---|
| Train Length | 250 | m |
| Platform Length | 200 | m |
| Time to pass Platform | 45 | s |
| Man's Speed | 6 | km/h |
| Train Speed | 10 | m/s |
| Man's Speed | $\frac{5}{3} \approx 1.67$ | m/s |
| Relative Speed (Train vs Man) | $\frac{25}{3} \approx 8.33$ | m/s |
| Time to pass Man | 30 | s |
Revision Table: Key Concepts in Train Problems
| Scenario | Distance Covered | Speed Used |
|---|---|---|
| Train passing a stationary point object (man, pole, etc.) | Length of the train | Speed of the train |
| Train passing a stationary length (platform, bridge, tunnel, etc.) | Length of train + Length of stationary object | Speed of the train |
| Train passing a moving object (another train, man cycling, etc.) in the same direction | Length of the train (if passing a point object like a man) <br> Sum of lengths (if passing another train) |
Relative speed (Difference of speeds) |
| Train passing a moving object (another train, man walking, etc.) in the opposite direction | Length of the train (if passing a point object like a man) <br> Sum of lengths (if passing another train) |
Relative speed (Sum of speeds) |
Additional Information: Units and Conversions
It is crucial to use consistent units when solving physics or speed-distance-time problems. The standard units in the SI system are metres (m) for distance and seconds (s) for time, resulting in metres per second (m/s) for speed.
Common speed unit conversions:
- To convert km/h to m/s: Multiply by $\frac{5}{18}$. For example, 1 km/h = $1 \times \frac{5}{18}$ m/s.
- To convert m/s to km/h: Multiply by $\frac{18}{5}$. For example, 1 m/s = $1 \times \frac{18}{5}$ km/h.
Remember these conversion factors to easily switch between units during calculations.
Important Questions from Problem on Trains
A 253 m long train running at a speed of 60 km/h takes 42 seconds to cross a bridge. The length (in m) of the bridge is:
The distance between two stations A and B is 700km. A train covers the journey from A to B at a speed of 80 km/h and returns back to A with a uniform speed of 65 km/h. The average speed of train during the whole journey, is closes to:
A train X of length 345 m running at 50 km/h crosses another train Y running at 76 km/h in the opposite direction in 22 seconds. Train Y will cross a bridge of length 905 m in:
A train running at a speed of 60 km/h crossed a pole in 1.5 min.The length of the train (in. m) is:
A 600 m long train is running at the speed of 72 km/h. How much time will it take to cross a 200 m long bridge?
