The ratio of distances travelled by a body starting from rest with constant acceleration in g^(th) and 8^(th) second is .........

The ratio of distances travelled by a body starting from rest with con

1.	The ratio of distances travelled by a body starting from rest with constant acceleration in g^(th) and 8^(th) second is .........
Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/17-278001" style="font-weight:bold;" target="_blank" title="Click to know more about 17">17</a>/15` <br/>`8/9`<br/>`15/17`<br/>`9/8`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Distance <a href="https://interviewquestions.tuteehub.com/tag/covered-2552266" style="font-weight:bold;" target="_blank" title="Click to know more about COVERED">COVERED</a> by particle in `n^(th)` second, <br/> `<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a> _(n) = v _(0)+ a/2 (2n -1)` <br/> `therefore` Distance coverd in `9^(th)` second, <br/> `x _(9) =(a)/(2) (2 (9) -1) = (17a)/(2) (because v _(0) =0)` <br/> Distance covered in `8^(th)` second, <br/> `x _(8) = a/2(2 (8) -1) = (15a )/(2)` <br/> `therefore (x _(9))/(x _(8)) = (17a)/(2) xx (2)/(15 a)= (17)/(15)`</body></html>