A particle starts from rest with uniform acceleration a. Its velocity after n second is v. The displacement of the body in the last two second is
A
2v(n−1)n
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B
v(n−1)n
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C
v(n+1)n
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D
2v(2n+1)n
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Solution
The correct option is A2v(n−1)n We know, At t=nsec sn=12an2 v2=2asn =2a⋅12an2 =a2n2 ∴v=an ∴a=vn ...(i) Now, sn−sn−2=12a[n2−(n−2)2] =12a[4n−4] =2a(n−1) from (i) sn−s(n−2)=2v(n−1)n