A train starting from rest attains a velocity of `72 km h^(-1)` in 5 minute. Assuming that the acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for attaining this velocity.

Explanation

`u =0, v = 72 km h ^(-1) = 20 ms ^(-1) and t = 6 ` minute `= 300 s `
`(i) a = ((v-u))/( t) = (20 ms ^(-1) -0 ms ^(-1))/( 300s )= (1)/(15) ms ^(-2)`
(ii) `2 as = v ^(2) - u ^(2) -0 `
`therefore s = (v ^(2))/(2a)`
`= ((20 ms ^(-1)) ^(2))/(2 xx ((1)/(15) ms ^(-2)))`
`= (20 xx 20 xx 15)/(2) m = 3000 m = 3 km`
The acceleration of the train is `(1)/(15) ms ^(-2) and ` the distance travelled is 3 km.