1.

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.

Answer»

Solution :`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.


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