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Question: A body is moving from rest under constant acceleration and let \(S_{1}\) be the displacement in the ...

A body is moving from rest under constant acceleration and let S1S_{1} be the displacement in the first (p1)(p - 1) sec and S2S_{2} be the displacement in the first psec.psec. The displacement in (p2p+1)th(p^{2} - p + 1)^{th} sec. will be

A

S1+S2S_{1} + S_{2}

B

S1S2S_{1}S_{2}

C

S1S2S_{1} - S_{2}

D

S1/S2S_{1}/S_{2}

Answer

S1+S2S_{1} + S_{2}

Explanation

Solution

From S=ut+12a6mut2S = ut + \frac{1}{2}a\mspace{6mu} t^{2}

S1=12a(P1)2S_{1} = \frac{1}{2}a(P - 1)^{2} and S2=12a6muP2S_{2} = \frac{1}{2}a\mspace{6mu} P^{2} [As6muu=0\lbrack As\mspace{6mu} u = 0]

From Sn=u+a2(2n1)S_{n} = u + \frac{a}{2}(2n - 1)

S(P2P+1)th=a2[2(P2P+1)1]S_{(P^{2} - P + 1)^{th}} = \frac{a}{2}\left\lbrack 2(P^{2} - P + 1) - 1 \right\rbrack =a2[2P22P+1]= \frac{a}{2}\left\lbrack 2P^{2} - 2P + 1 \right\rbrack

It is clear that S(P2P+1)th=S1+S2S_{(P^{2} - P + 1)^{th}} = S_{1} + S_{2}