For converting Km/Hour to Meter/Second

x Km/hr=(x*5/18)m/s

For converting meter/second to Km/Hour

y m/s=(y*18/5)Km/hr

How to remember this formula

First converting kilometer to meter

it is way of converting High to low

So multiply by the low number 5/18

Second converting meter to Kilometer

It is way of converting low to high

So multiply by the high number 18/5

Time taken by train of length l meters to pass a stationary object of length b meters I s the time taken by the train to cover (l+b) meters

Suppose two train are moving in same direction with U m/s and V m/s where u>v then their relative speed is (u-v)m/s (u>v)

If the train going in opposite direction then there relative speed is =(u+v) m/s

Two train of length a meters and b meters are moving in opposite direction at um/s and v m/s then time taken to cross each other =(a+b)/(u+v)

If it is going same direction=(a+b)/(u-v) sec

x Km/hr=(x*5/18)m/s

For converting meter/second to Km/Hour

y m/s=(y*18/5)Km/hr

How to remember this formula

First converting kilometer to meter

it is way of converting High to low

So multiply by the low number 5/18

Second converting meter to Kilometer

It is way of converting low to high

So multiply by the high number 18/5

Time taken by train of length l meters to pass a stationary object of length b meters I s the time taken by the train to cover (l+b) meters

Suppose two train are moving in same direction with U m/s and V m/s where u>v then their relative speed is (u-v)m/s (u>v)

If the train going in opposite direction then there relative speed is =(u+v) m/s

Two train of length a meters and b meters are moving in opposite direction at um/s and v m/s then time taken to cross each other =(a+b)/(u+v)

If it is going same direction=(a+b)/(u-v) sec

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