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kriti kanwal Grade: 9
        how to derivea relation for the distance travelled in nth second by a uniformly accelerated body
6 years ago

Answers : (2)

asd wer
31 Points
										

distance travelled in n seconds:


S1= un + 1/2a(n^2)


distance travelled in (n-1) seconds:


S2 = u(n-1) + 1/2a((n-1)^2)


distance travelled in nth second:


S1-S2


= u + 1/2a(2n-1)

6 years ago
vijay kumar pandey
32 Points
										

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.

6 years ago
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