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Two parallel rail tracks run north-south...

Two parallel rail tracks run north-south Train `A` moves north with a speed of `54 km h^(-1)` and train `B` moves south with a speed of `90 km h^(-1)`. What is the
a. relative velocity of `B` with respect to `A`?
b. relative of a monkey running on the roof of the train `A` against its motion (with its velocity of `18 km h^(1)` with respect to the train `A` ) as observed by a man standing on the ground?

A

`- 40 ms^(-1)`, i.e. the train B appears to A to move with a speed of `40 ms^(-1)` from south to north` `10 ms^(-1)`

B

`40 ms^(-1)`, i.e. the train B appears to A to move with a speed of `40 ms^(-1)` from north to south` `10 ms^(-1)`

C

`- 40 ms^(-1)`, i.e. the train B appears to A to move with a speed of `40 ms^(-1)` from north to south` `20 ms^(-1)`

D

`- 40 ms^(-1)`, i.e. the train B appears to A to move with a speed of `40 ms^(-1)` from north to south` `10 ms^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D

Choose the positive direction of x-axis to be from south to north. Then.
`v_(A) = + 54 km h^(-1) = 15 ms^(-1)`
`v_(B) = - 90 km h^(-1) = - 25 ms^(-1)`
Relative velocity of B with respect to `A = v_(B) - v_(A) = - 40 ms^(-1)`, i.e. the train B appears to A to move with a speed of `40 ms^(-1)` from north to south.
Relative velocity of ground with respect to `B = 0 -v_(B) = 25 ms^(-1)`
In (c), let the velocity of the monkey with respect to ground be `v_(M)`. Relative velocity of the monkey with respect to A,
`v_(MA) = v_(M) - v_(A) = - 18 km h^(-1) = - 5 ms^(-1)`. Therefore, `v_(M) = (15 - 5) ms^(-1) = 10 ms^(-1)`
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