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# A small block slides without friction down an inclined starting form rest. Let Sn be the distance travelled from timet = n-1 to t = n. Then Sn/Sn+1 is -(A)   2n-1/2n(B)   2n+1/2n+1(C)   2n-1/2n+1(D)   2n/2n+1 AKASH GOYAL AskiitiansExpert-IITD
419 Points
10 years ago

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

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10 years ago

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 Kushagra Madhukar
one year ago
Dear student,

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
Hence option (C) is correct

Thanks and regards,
Kushagra