Two trains are moving in opposite directions @ 60 km/hr and 90 km/hr Their lengths are 1.10 km and 0.9 km respectively Find the time taken by the slower train to cross the faster train in seconds

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Solution

Relative speed $= (60 + 90) k m / h r$
$= (150 \times \frac{5}{18}) m / s e c = (\frac{125}{3}) m / s e c$
Distance covered $= (1.10 + 0.9) k m = 2 k m = 2000 m$
Required time = $(2000 \times \frac{3}{125}) s e c = 48 s e c$

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Two trains are moving in opposite directions @ 60 km/hr and 90 km/hr Their lengths are 1.10 km and 0.9 km respectively Find the time taken by the slower train to cross the faster train in seconds

Relative speed=(60+90)km/hr =(150×518)m/sec=(1253)m/secDistance covered =(1.10+0.9)km=2km=2000mRequired time = (2000×3125)sec=48sec

Relative speed $= (60 + 90) k m / h r$
$= (150 \times \frac{5}{18}) m / s e c = (\frac{125}{3}) m / s e c$
Distance covered $= (1.10 + 0.9) k m = 2 k m = 2000 m$
Required time = $(2000 \times \frac{3}{125}) s e c = 48 s e c$