The ratio of the distance travelled in the fourth and the third second by a starting particle, moving over a straight path from rest with constant acceleration is......

Options

A. `7/5`
B. `7/3`
C. `5/7`
D. `3/7`

Answer

A

Explanation

Distance covered in `n^(th)` second is given by
`d_(n) = v_(0) + a/2 (2n-1)`
`therefore` Distance covered by a particle in `4^(th)` second,
`d _(4) = 0 + a/2 (7) [because v _(0) = 0]`
`d _(4) = (7a)/(2)` and distance covered by a particle in `3^(rd)` second,
`d _(3) = 0 + (a)/(2) (5)`
`thereforfe d _(3) = (5a)/(2)`
`therefore` Ratio `(d_(4))/(d _(3)) = (7a)/(2) xx (2)/(5a) =7/5`