A body `A` starts from rest with an acceleration `a_1`. After `2` seconds, another body `B` starts from rest with an acceleration `a_2`. If they travel equal distances in the `5^(th)` second, after the start of `A`, then the ratio `a_1 : a_2` is equal to :

Options

A. `9:5`
B. `5:7`
C. `5:9`
D. `7:9`

Answer

C

Explanation

The distance covered by the body in the `nth` second motion is
`S_(n)= u+a/2(2n-1)`
When `u` is initial velocity and `a` is acceleration Distance covered by the body `A` in `5th` second after its starts, with acceleration `a_(1)`, is
`(S_(5))_(A)= 0+(a_(1))/2(2xx5-1)= (9a_(1))/2`
Time taken by second body `=5 -2= 3s`
`(S_(3))_(B)= 0+(a_(2))/2(2xx3-1)= (5a_(2))/2` Given, `(S_(4))A= (S_(3))_(B)`
`:. (a_(1))/(a_(2))=5/9`