A body starts from rest and moves with constant acceleration the ratio of the distance covered in the nth second to that covered in ‘n’ seconds is
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a
2n2+1n
b
2n2−1n
c
2n+1n2
d
2n−1n2
answer is D.
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Detailed Solution
Given that u=0∴ The distance travelled in the nth second is Sn=u+an−12 Sn=0+a2n−1.........(i) The distance travelled in seconds isS=u(n)+12a(n)2S=12a(n)2……….(ii)∴SnS=(2n−1)a2an22=(2n−1)n2=2n−1n2