Relative Speed Complex Train Problems

In this class, We discuss Relative Speed Complex Train Problems.

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The reader should have prior knowledge of the relative speed formula. Click Here.

Example 1:

A train running at 54 kmph takes 20 seconds to cross a platform.

The train takes 12 seconds to cross a man walking at six kmph in the same direction the train is moving.

Find the length of the train and the length of the platform?

Solution:

The below diagram shows the graphical intuition of the example.

Important: We know an important point from the formula and previous examples.

Train passing a man takes the length of the train distance.

From this point, we find the length of the train. Once train length is obtained, we can find platform length.

Time = Distance / speed

D = S * T

D = length of the train = (40/3) * 12

D = 160 meters.

Time to pass platform = Distance / Speed

Distance = length of the train + length of the platform

Speed = train speed

speed = 54 * 5/18 = 15 m/s

t = (x + y)/15

20 = (x + y)/15

x + y = 300 where x = 160

y = 140

Example 2:

A man is sitting on a train that is moving at 50 kmph.

The man observes that the goods train traveling in the opposite direction takes 9 seconds to pass him.

If the goods train is 280 meters long, find its speed?

Solution:

Important: Man sitting on a moving train observed the goods. So man is also moving at the speed of a train.

Both are moving, so we go with relative speed.

Relative speed = distance / time

Distance = length of the goods train

Relative speed = 280/9

Relative speed = (280/9) * (18/5)

Relative speed = 112 kmph.

speed of goods train = 112 – 50 = 62 kmph where relative speed = (u +v)