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

S_n= S'-S =0.5 [n² - (n-1)²] = 0.5a (n-n+1)*(n+n-1)

S_n= 0.5a*(2n-1)

put n+1 in place of n in expression of S_n

S_n+1= 0.5a (2(n+1)-1)= 0.5a (2n+1)

S_n/S_n+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)

S_n = 1/2 a [n²-(n-1)²]

     = 1/2 a [2n-1]             ...........1

S_n+1 = 1/2a[(n+1)²-n²]

         =1/2 a [2n+1]          .........2

dividing eq 1 & 2

S_n/S_n+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