A man walking at 5 km/hr noticed that a 225 m long train coming in the opposite direction crossed him in 9 seconds. The speed of the train is
85 km/hour
Understanding Relative Speed in Opposite Direction
This problem involves calculating the speed of a train when it crosses a moving man. Since the train and the man are moving in opposite directions, we need to use the concept of relative speed.
When two objects move towards each other (or in opposite directions), their relative speed is the sum of their individual speeds. The distance covered by the train in crossing the man is equal to the length of the train.
Given Information
- Speed of the man (\(V_M\)): 5 km/hr
- Length of the train (\(L_T\)): 225 m
- Time taken to cross the man (\(t\)): 9 seconds
- Direction of motion: Opposite to each other
Unit Conversion
The given speeds are in km/hr, the length is in meters, and time is in seconds. To perform calculations, we need to convert all quantities to a consistent system of units, such as meters and seconds. The standard conversion factor from km/hr to m/s is \(\frac{5}{18}\).
- Convert man's speed from km/hr to m/s: \[V_M = 5 \text{ km/hr} \times \frac{5}{18} \text{ m/s per km/hr} = \frac{25}{18} \text{ m/s}\]
Let the speed of the train be \(V_T\) km/hr. We need to find \(V_T\). Let's also express \(V_T\) in m/s:
- Convert train's speed from km/hr to m/s: \[V_T \text{ (in m/s)} = V_T \text{ (in km/hr)} \times \frac{5}{18} \text{ m/s per km/hr} = \frac{5V_T}{18} \text{ m/s}\]
Calculating Relative Speed
When the train crosses the man moving in the opposite direction, the distance covered is the length of the train (225 m). The time taken is 9 seconds. Using the formula: Distance = Relative Speed \(\times\) Time.
Let the relative speed be \(V_{rel}\) m/s.
\[L_T = V_{rel} \times t\] \[225 \text{ m} = V_{rel} \text{ m/s} \times 9 \text{ s}\]Now, we can calculate the relative speed:
\[V_{rel} = \frac{225}{9} \text{ m/s}\] \[V_{rel} = 25 \text{ m/s}\]Relating Relative Speed to Individual Speeds
Since the train and the man are moving in opposite directions, the relative speed is the sum of their individual speeds:
\[V_{rel} = V_T \text{ (in m/s)} + V_M \text{ (in m/s)}\]We know \(V_{rel} = 25\) m/s and \(V_M = \frac{25}{18}\) m/s. Substituting these values:
\[25 = \frac{5V_T}{18} + \frac{25}{18}\]To solve for \(\frac{5V_T}{18}\), subtract \(\frac{25}{18}\) from both sides:
\[25 - \frac{25}{18} = \frac{5V_T}{18}\]Find a common denominator:
\[\frac{25 \times 18}{18} - \frac{25}{18} = \frac{5V_T}{18}\] \[\frac{450 - 25}{18} = \frac{5V_T}{18}\] \[\frac{425}{18} = \frac{5V_T}{18}\]Multiply both sides by 18:
\[425 = 5V_T\]Now, solve for \(V_T\) (which is currently in km/hr based on our equation setup, though we used the \(\frac{5}{18}\) factor):
\[V_T = \frac{425}{5}\] \[V_T = 85\]So, the speed of the train is 85 km/hr.
Alternative Approach (Working directly with relative speed in m/s and converting at the end)
Let the speed of the train in m/s be \(v_t\). The speed of the man in m/s is \(v_m = 5 \times \frac{5}{18} = \frac{25}{18}\) m/s.
When moving in opposite directions, the relative speed is \(v_{rel} = v_t + v_m\).
The distance is the length of the train, \(d = 225\) m. The time is \(t = 9\) s.
\[d = v_{rel} \times t\] \[225 = (v_t + v_m) \times 9\] \[v_t + v_m = \frac{225}{9} = 25 \text{ m/s}\]Substitute the value of \(v_m\):
\[v_t + \frac{25}{18} = 25\] \[v_t = 25 - \frac{25}{18}\] \[v_t = \frac{25 \times 18 - 25}{18} = \frac{450 - 25}{18} = \frac{425}{18} \text{ m/s}\]Now, convert the train's speed from m/s back to km/hr. The conversion factor from m/s to km/hr is \(\frac{18}{5}\).
\[V_T \text{ (in km/hr)} = v_t \text{ (in m/s)} \times \frac{18}{5} \text{ km/hr per m/s}\] \[V_T = \frac{425}{18} \times \frac{18}{5}\] \[V_T = \frac{425}{5}\] \[V_T = 85 \text{ km/hr}\]Conclusion
The speed of the train is 85 km/hr.
| Quantity | Value (Given) | Value (Converted to m/s) |
|---|---|---|
| Man's Speed (\(V_M\)) | 5 km/hr | \(\frac{25}{18}\) m/s |
| Train's Length (\(L_T\)) | 225 m | 225 m |
| Time Taken (\(t\)) | 9 s | 9 s |
| Relative Speed (\(V_{rel}\)) | - | 25 m/s |
| Train's Speed (\(V_T\)) | 85 km/hr | \(\frac{425}{18}\) m/s |
Revision Table: Relative Speed Concepts
| Scenario | Relative Speed (\(V_{rel}\)) | Distance Covered for Crossing |
|---|---|---|
| Objects moving in the same direction (faster overtakes slower) | \(|V_1 - V_2|\) | Sum of their lengths |
| Objects moving in opposite directions | \(V_1 + V_2\) | Sum of their lengths |
| Train crossing a stationary object (like a pole or a man) | Speed of the train | Length of the train |
| Train crossing a moving object (like a man) | Relative speed of train w.r.t man | Length of the train |
Additional Information: Speed, Distance, and Time
The fundamental relationship between speed, distance, and time is:
\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} \]This can be rearranged to find distance or time:
- Distance = Speed \(\times\) Time
- Time = \(\frac{\text{Distance}}{\text{Speed}}\)
It is crucial to ensure that the units are consistent when using these formulas. Common units include:
- Speed: m/s, km/hr
- Distance: meters (m), kilometers (km)
- Time: seconds (s), hours (hr)
The conversion factor between km/hr and m/s is derived from:
\[ 1 \text{ km/hr} = \frac{1 \text{ km}}{1 \text{ hr}} = \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{10}{36} \text{ m/s} = \frac{5}{18} \text{ m/s} \]And conversely:
\[ 1 \text{ m/s} = \frac{1 \text{ m}}{1 \text{ s}} = \frac{1/1000 \text{ km}}{1/3600 \text{ hr}} = \frac{3600}{1000} \text{ km/hr} = \frac{18}{5} \text{ km/hr} \]Similar Questions
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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?
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