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A boy starts from rest is accelerated un...

A boy starts from rest is accelerated uniformly for 30s. If `x_(1),x_(2),x_(3)` are the distance travelled in first 10s, next 10 s and last 10s respectively, then `x_(1) : x_(2) : x_(3) :` is

A

`1 : 2 : 3`

B

`1 : 1 : 1`

C

`1 : 3 : 5`

D

`1 : 3 : 9`

Text Solution

Verified by Experts

The correct Answer is:
C
According to Newton.s `2^(nd)` eqn of motion,
`x = ut + (1)/(2) at^(2)`
Therefore distance travelled in 10 s
`x = 0 + (1)/(2) a(10)^(2)`
x = 50 a
Distance travelled in 20 seconds,
`x + x_(2) = ut + (1)/(2) a(20)^(2)`
`x + x_(2) = 0 + (1)/(2) a xx 400`
`x + x_(2) = 200 a`
`x_(2) = 200 a - 50 a`
Distance travelled in 30 seconds
`x_(1) + x_(2) + x_(3) = 0 + (1)/(2) a(30)^(2)`
`x_(1) + x_(2) + x_(3) = 450 a`
`x_(3) = 250 a`
From eqns (i),(ii) and (iii), we get
`x_(1): x_(2) : x_(3) = 1 : 3 : 5`
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