Question

The ratio of the distances travelled in the fourth and the third second by a particle, moving over a straight path with constant acceleration, is ................. .
(A) 7/5
(B) 5/7
(C) 7/3
(D) 3/7

Answer 1

Answer:

a ) one is answer

I hope it is helpful for you

Answer 2

Answer:

Option (A) is correct.

Explanation:

we need to find the the ratio of the distances traveled in the fourth and the third second by a particle, moving over a straight path with constant acceleration.

we know the formula for the distance traveled in nth second.

$S_{n}=u+\frac{a(2n-1)}{2}$................(1)

we assume object starts from rest so u=0.

apply equation 1 for forth second i.e. n=4

$S_{4}=0+\frac{a(7)}{2}$

$S_{4}=\frac{7a}{2}$.................(2)

apply equation 1 for third second i.e. n=3

$S_{3}=0+\frac{a(5)}{2}$

$S_{3}=\frac{5a}{2}$................(3)

divide equation (2) and (3) we get,

$\frac{S_{4} }{S_{3} }=\frac{7}{5}$ as acceleration is constant.

So the required is (A).