A body starts from rest and moves with a uniform acceleration. The ratio of the distance covered in the nth second to the distance covered in n second

Question

A body starts from rest and moves with a uniform acceleration. The ratio of the distance covered in the nth second to the distance covered in n second (A)  (2/n) - (1/n2) (B)  (1/n2) - (1/n) (C)  (2/n2) - (1/n2) (D)  (2/n) + (1/n2)

Tardigrade Admin · Accepted Answer

Sn=u+(a/2)(2 n-1) (Distance travelled in nth second or Sn=0+(a/2)(2 n-1) dots(i) also. S =u t+(1/2) a t2 =0+(1/2) a n2 dots(ii) (Distance travelled in n th second) ∴ (Sn/S)=((a/2)(2 n-1)/a n2 / 2) =(2/n)-(1/n2)

Q. A body starts from rest and moves with a uniform acceleration. The ratio of the distance covered in the nth second to the distance covered in n second

$\frac{2}{n} - \frac{1}{n ^{2}}$

$\frac{1}{n ^{2}} - \frac{1}{n}$

$\frac{2}{n ^{2}} - \frac{1}{n ^{2}}$

$\frac{2}{n} + \frac{1}{n ^{2}}$

Q. A body starts from rest and moves with a uniform acceleration. The ratio of the distance covered in the nth second to the distance covered in n second

n2​−n21​

n21​−n1​

n22​−n21​

n2​+n21​

Sn​=u+2a​(2n−1) (Distance travelled in nth second or Sn​=0+2a​(2n−1)…(i) also. S=ut+21​at2 =0+21​an2…(ii) (Distance travelled in n th second) ∴SSn​​=an2/22a​(2n−1)​ =n2​−n21​

$\frac{2}{n} - \frac{1}{n ^{2}}$

$\frac{1}{n ^{2}} - \frac{1}{n}$

$\frac{2}{n ^{2}} - \frac{1}{n ^{2}}$

$\frac{2}{n} + \frac{1}{n ^{2}}$