Correct Answer - D
`u`and `4u` are speeds of two cars at a specific instant. Both cars are stopped form the spectific instant therefore,
For 1st acr:
`{:(v=0,,a=a_(1)t=1,,):}`
From equaiton of motion,
`v = u - at`
`0 = u - a_(1) t`
`a_(1)t = u ...(i)`
From equation of motion,
`v^(2) = u^(2) - 2as`
Here, `s = s_(1), v_(1) = 0`
`(s_(1)` is the distance convered by 1st car before coming to rest.)
`:. 0 = u^(2) - 2a_(1)s_(1)`
`2a_(1)s_(1) = u^(2) ....(ii)`
Putting the value of `a_(1)` in equaiton (iii), we get
`2.(u)/(t).s_(1) = u^(2)`
`s_(1) = (ut)/(2)`
For `2nd` car:
similarly, `a_(2)t = 4u`
`rArr a_(2) = (4u)/(t) ....(iii)`
and `0 = (4u^(2)) - 2a_(2)s_(2) ...(iv)`
From equaion (iii) and (iv), we get
`2.(4u)/(t).s_(2) = 16u^(2)` or `s_(2) = 2ut`
so, `(s_(1))/(s_(2)) = (ut)/(2.2ut)`
`rArr (s_(1))/(s_(2)) = (1)/(4) = 1:4`
Hence, the ratio of respective distance is `1: 4`
