A train moving with a speed of 60 km per hour crosses an electric pole in 30 seconds. What is the length of train in meters?
500
Calculating Train Length from Speed and Time
This question asks us to find the length of a train given its speed and the time it takes to cross a fixed point, specifically an electric pole. When a train crosses a pole, the distance the train covers is equal to its own length.
We are given the following information:
- Train Speed = 60 km/hour
- Time taken to cross the pole = 30 seconds
To find the length of the train, we can use the basic formula relating distance, speed, and time:
\( \text{Distance} = \text{Speed} \times \text{Time} \)
Before we can use this formula, we need to ensure that the units are consistent. The speed is given in kilometers per hour (km/h), and the time is in seconds (s). We need to convert the speed into meters per second (m/s) to get the distance (length of the train) in meters.
Speed Unit Conversion (km/h to m/s)
To convert speed from km/h to m/s, we use the conversion factors:
- 1 kilometer = 1000 meters
- 1 hour = 3600 seconds
So, to convert km/h to m/s, we multiply by \( \frac{1000}{3600} \), which simplifies to \( \frac{5}{18} \).
Let's convert the train's speed:
\( \text{Speed in m/s} = 60 \, \text{km/h} \times \frac{5}{18} \, \frac{\text{m/s}}{\text{km/h}} \)
\( \text{Speed in m/s} = \frac{60 \times 5}{18} \, \text{m/s} \)
\( \text{Speed in m/s} = \frac{300}{18} \, \text{m/s} \)
We can simplify this fraction:
\( \frac{300}{18} = \frac{150}{9} = \frac{50}{3} \, \text{m/s} \)
So, the speed of the train is \( \frac{50}{3} \) m/s.
Calculating the Length of the Train
Now that we have the speed in m/s and the time in seconds, we can calculate the distance covered, which is the length of the train.
\( \text{Length of train} = \text{Speed} \times \text{Time} \)
\( \text{Length of train} = \frac{50}{3} \, \text{m/s} \times 30 \, \text{s} \)
\( \text{Length of train} = \frac{50}{3} \times 30 \, \text{meters} \)
\( \text{Length of train} = 50 \times \frac{30}{3} \, \text{meters} \)
\( \text{Length of train} = 50 \times 10 \, \text{meters} \)
\( \text{Length of train} = 500 \, \text{meters} \)
Therefore, the length of the train is 500 meters.
Let's verify this against the given options:
| Option | Length (meters) |
|---|---|
| 1 | 300 |
| 2 | 400 |
| 3 | 500 |
| 4 | 600 |
Our calculated length of 500 meters matches Option 3.
Revision Table: Key Concepts for Train Problems
| Concept | Explanation | Formula/Rule |
|---|---|---|
| Train crossing a pole/point object | The distance covered by the train is equal to its own length. Time taken is the time from when the engine reaches the point until the last coach leaves the point. | Distance = Length of train |
| Train crossing a platform/bridge/tunnel | The distance covered by the train is the sum of the train's length and the length of the platform/bridge/tunnel. | Distance = Length of train + Length of object |
| Train crossing another train (moving in opposite direction) | The distance covered is the sum of the lengths of both trains. Relative speed is the sum of their speeds. | Distance = L1 + L2 Relative Speed = S1 + S2 |
| Train crossing another train (moving in same direction) | The distance covered is the sum of the lengths of both trains. Relative speed is the difference of their speeds (faster train minus slower train). | Distance = L1 + L2 Relative Speed = |S1 - S2| |
| Speed Conversion (km/h to m/s) | Multiply the speed in km/h by \( \frac{5}{18} \). | \( S_{\text{m/s}} = S_{\text{km/h}} \times \frac{5}{18} \) |
| Speed Conversion (m/s to km/h) | Multiply the speed in m/s by \( \frac{18}{5} \). | \( S_{\text{km/h}} = S_{\text{m/s}} \times \frac{18}{5} \) |
Additional Information: Understanding Speed, Distance, and Time
Speed, distance, and time are fundamental concepts in physics and are often encountered in quantitative aptitude problems involving motion, such as those with trains, cars, or boats.
- Speed: It is the rate at which an object moves. It is defined as the distance covered per unit of time. Common units are meters per second (m/s) or kilometers per hour (km/h).
- Distance: It is the total path length covered by a moving object. Units can be meters (m), kilometers (km), etc.
- Time: It is the duration for which the motion occurs. Units can be seconds (s), minutes (min), hours (h), etc.
The relationship \( \text{Distance} = \text{Speed} \times \text{Time} \) is a cornerstone for solving motion problems. It can be rearranged to find speed or time if the other two quantities are known:
- \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \)
- \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \)
It is crucial to always use consistent units for speed, distance, and time when using these formulas. If distance is in meters and speed is in m/s, time will be in seconds. If distance is in kilometers and speed is in km/h, time will be in hours.
In problems involving trains or other objects with length, the concept of distance changes depending on what the train is crossing (a point, a platform, another train). Understanding these scenarios is key to correctly applying the distance-speed-time formula.
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Important Questions from Problem on Trains
Eight railway stations A, B, C, D, E, F, G and H are connected either by two-way passages or one-way passages. One-way passages are from C to A, E to G, B to F, D to H, G to C, E to C and H to G. Two-way passages are between A and E, G and B, F and D, and E and D.
If the route between G and C is closed, which one of the following stations need not be passed through while travelling from H to C?
A daily train is to be introduced between station A and station B starting from each end at 6 AM and the journey is to be completed in 42 hours. What is the number of trains needed in order to maintain the Shuttle Service?
A train with a uniform speed passes a 122 meters long platform in 17 seconds and a 210 meters long bridge in 25 seconds. The speed of the train is:
How long does a train 153 meters long running at the rate of 90 kmph take to cross a bridge 622 meters in length?
A train passes a 360 metre long platform in 40 seconds and a man standing on the platform in 16 seconds. The speed of the train is:
