Initially a body is at rest. If its acceleration is `5ms^(-2)` then the distance travelled in the `18^(th)` second is :-

Answer

B, C, D

Explanation

Motion A to C `rArr 17^(2)=7^(2)+2as`

Motion A to B `rArr v_(B)^(2)=7^(2)+2a(s/2)=(17^(2)+7^(2))/2`
`(A) v_(B)=sqrt((289+49)/2)=13 m//s`
(B) `lt v_(AB) gt =(7+13)/2=10 m//s`
(C) `t_(1)=(13-7)/a, t_(2)=(17-13)/a, t_(1)/t_(2)=6/4=3/2`
(D) `lt v_(BC) gt =(13+17)/2=15 m//s`