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......

The ratio of the distance travelled in the fourth and the third second

1.	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......
Answer» `7/5` `7/3` `5/7` `3/7` Solution :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`