To find the speed of the train, we need to determine the relative speed at which the train crosses the man and then isolate the train's speed.
For calculations, it's essential to use consistent units. We'll convert the man's speed from km/h to m/s.
Conversion factor: \(1 \text{ km/h} = \frac{5}{18} \text{ m/s}\)
Man's Speed in m/s: \(S_m = 10.1 \times \frac{5}{18} = \frac{50.5}{18} \text{ m/s}\)
When objects move in opposite directions, their speeds add up to give the relative speed (\(S_{rel}\)). The distance covered when crossing a man is the length of the train.
Formula: \(L = S_{rel} \times t\)
Calculate \(S_{rel}\): \(S_{rel} = \frac{L}{t} = \frac{628 \text{ m}}{12 \text{ s}} = \frac{157}{3} \text{ m/s}\)
The relative speed is the sum of the train's speed (\(S_t\)) and the man's speed (\(S_m\)).
\(S_{rel} = S_t + S_m\)
Rearrange to find \(S_t\): \(S_t = S_{rel} - S_m\)
\(S_t = \frac{157}{3} \text{ m/s} - \frac{50.5}{18} \text{ m/s}\)
To subtract, find a common denominator (18):
\(S_t = (\frac{157 \times 6}{18}) - (\frac{50.5}{18}) = \frac{942 - 50.5}{18} = \frac{891.5}{18} \text{ m/s}\)
Convert the train's speed from m/s back to km/h.
Conversion factor: \(1 \text{ m/s} = \frac{18}{5} \text{ km/h}\)
\(S_t (\text{km/h}) = \frac{891.5}{18} \times \frac{18}{5}\)
\(S_t (\text{km/h}) = \frac{891.5}{5} = 178.3 \text{ km/h}\)
The speed of the train is 178.3 km/h.
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