A train is travelling at a peed of `90 km h ^(-1).` Breakes are applied so as to produce a uniform acceleration of `- 0.5 ms ^(-2).` Find how far the train will go before it is brought to rest.

Explanation

Here, `u = 90 km h ^(-1) =(90 xx 1000)/(3600) = 25 ms ^(-1) ,`
`a =- 0.5 ms ^(-2),`
`v =0` (Because th train is brought to rest.)
`v ^(2) -u ^(2) = 2 as`
`therefore 0- (25) ^(2) =2 xx (-0.5) xx s`
`therefore s = (- (25 xx 25))/(- 2 xx 0.5) = 625 m`
The tranin will cover a distance of 625 m before it comes to rest.