Question

A body starts from rest and moves with constant acceleration the ratio of distance covered by the body in nth second to that covered in n second isn

Answer 1

Answer:

hope this helps

Explanation:

actually the other given answer is also correct.

displacement in the nth second,we r talking about.suppose accleration is a and the initial velocity is u.

s=ut+1/2at^2.

if i will be said to find out the displacement in the nth second.so i will find out the displacement in the n-1 th second,can be easily found by the above given equation.

s(n-1)=u(n-1)+1/2a(n-1)^2=un-u+1/2a(n^2-2n+1)=un-u+1/2an^2-an+1/2a

now this is the total displacement in n-1 second.in n second the displacent is :

s(n)=un+1/2an^2

so i f i want the displacement in that n second nly,so i need to subtract the total displacement in n second and in n-1 second.

s(n)-s(n-1)=un+1/2an^2-un+u-1/2an^2+an-1/2a=u+an-1/2a=u+1/2a(2n-1).

actually to make this answer correct u need to write the seconds (dimensionally correct)

so,u(1s)+1/2a(2n-1s)(1s) is the correct answer.if u will keep on writing the units from the first then u will get this equation only.