The faster train, measuring 160 meters in length, traverses a lamppost in 12 seconds.
The speed of the faster train is calculated as: \[ \text{Speed of faster train} = \frac{160}{12} = \frac{40}{3} \, \text{m/s} \]
The speed of the slower train is determined to be: \[ \text{Speed of slower train} = \frac{40}{3} - \left(6 \times \frac{5}{18}\right) = \frac{35}{3} \, \text{m/s} \]
When the two trains move in opposing directions, their relative speed is the sum of their individual speeds: \[ \text{Relative speed} = \frac{40}{3} + \frac{35}{3} = 25 \, \text{m/s} \]
The trains pass each other in 14 seconds. The combined length of both trains is the sum of their individual lengths. Let \( x \) represent the length of the slower train. The relative speed is related to the total length and crossing time by the formula: \[ \text{Relative speed} = \frac{\text{Total length of two trains}}{\text{Time taken to cross each other}} \] Substituting the known values yields: \[ \frac{160 + x}{25} = 14 \] Solving for \( x \): \[ 160 + x = 14 \times 25 = 350 \] \[ x = 350 - 160 = 190 \, \text{meters} \]
The slower train measures \( \boxed{190} \, \text{meters} \) in length.