A body starts from rest and is uniformly accelerated is `30m//s` . The distance travelled in the first `10` s is `x_(1)`, next `10` s is `x_(2)` and the last `10` s is `x_(3)`. Then `x_(1):x_(2):x_(3)` is the same as :-

Options

A. `1 : 2 : 3`
B. `1 : 2 : 5`
C. `1 : 3 : 5`
D. `1 : 4 : 9`

Answer

C

Explanation

`x_(1)=1/2 a(10)^(2)`
`x_(1)+x_(2)=1/2 a(20)^(2)`
`x_(1)+x_(2)+x_(3)=1/2 a(30)^(2)rArr x_(1):x_(2):x_(3)=1:3:5`