A body starts from rest and is uniformly accelerated for `30` 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. `1:1:1`
C. `1:3:5`
D. `1:3:9`

Answer

C

Explanation

Using , `s=ut +1/2 "at"^2`
Here , ` u = 0 , :. X_(1) = 0 +1/2 a xx 10^(2) = 50 a " "...(i)`
`x_(1) +x_(2) = 0 +1/2 a (10 + 10)^(2) = 200a `
`:. X_(2) = 200a - x_(1) = 200a - 50 a ` (Using (i))
or `x_(2) = 150 a " "...(ii))`
`x_(1) +x_(2) +x_(3)=0 +1/2 a (10+10 +10)^(2) = 450 a `
`:. x_(3) = 450 a - x_(1) -x_(2)`
`= 450 a - 50 a - 150 a " "` (Using (i) and (ii))
or `x_(3) = 250 a `
Hence, `x_(1) : x_(2) : x_(3) = 50 a : 150 a : 250 a = 1 : 3 : 5 `