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

Answer

`1//15 m//s^(2)` ; `3 km`

Explanation

Here , `u = 0 , v = 72 km//h = ( 72 xx 1000 m )/( 60 xx 60 s) = 20 m//s, t = 5 min = 5 xx 60 = 300 s`
Thus , `a = ( v - u)/( t) = ( 20 - 0)/( 300) = (1//15)m//s^(2)`
From `v^(2) - u^(2) = 2 as , s = ( v^(2) - u^(2))/( 2a) = ( 20^(2) - 0^(2))/( 2 xx 1//15) = 3000 m = 3 km`