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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)
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.
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