A small block slides without friction down an inclined starting form rest. Let S n be the distance travelled from time t = n-1 to t = n. Then S n /S n+1 is - (A) 2n-1/2n (B) 2n+1/2n+1 (C) 2n-1/2n+1 (D) 2n/2n+1

							
Dear Shivam
S=ut+0.5at2= 0.5at2   (since u=0)
At t=n-1
S= 0.5a(n-1)2
At t=n
S'=0.5a(n)2
Sn= S'-S =0.5 [n2 - (n-1)2] = 0.5a (n-n+1)*(n+n-1)
Sn= 0.5a*(2n-1)
put n+1 in place of n in expression of Sn
Sn+1= 0.5a (2(n+1)-1)= 0.5a (2n+1)
Sn/Sn+1 = 2n-1/2n+1
option (C)
 
All the best.                                                           
AKASH GOYAL
AskiitiansExpert-IITD
 
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a is the accleration of block then
 S = 1/2 a t2                         (using this eq)
Sn = 1/2 a [n2-(n-1)2]
     = 1/2 a [2n-1]             ...........1
Sn+1 = 1/2a[(n+1)2-n2]
         =1/2 a [2n+1]          .........2
dividing eq 1 & 2
Sn/Sn+1 = 2n-1/2n+1
						

							Sir it is inclined plane so how do u took a=g as it is not horizontal plane So akash sir plz correct it