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Two trains cross each other in 14 seconds when running in opposite directions along parallel tracks. The faster train is 160 m long and crosses a lamp post in 12 seconds. IfΒ the speed of the other train is 6 km/hr less than the faster one, its length, in m, is
Speed of the faster train =Β $$\frac{160}{12}=\frac{40}{3}\ $$ m/s
Speed of the slower train =Β $$\frac{40}{3}-\left(6\times\ \frac{5}{18}\right)=\frac{35}{3}$$ m/s
Sum of speeds (when the trainsΒ travel towards each other) =Β $$\frac{40}{3}+\frac{35}{3}=25$$ m/s
Let the slower train be $$x$$ metres long; then:Β $$\frac{160+x}{25}=14$$
On solving,Β $$x=190\ m$$
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