Question

the ratio of displacement in n second to Nth second for a particle starting from rest and uniform acceleration is​

Answer 1

Answer:

Hope this is useful

Explanation:

Answer 2

Answer:

$\frac{s}{s_{n}} = \frac{n^2}{2n-1}$

Explanation:

Distance travelled by a body in 'n'th second is

$s_{n} = u +\frac{1}{2}a(2n-1)$....(1)

Distance travelled in 'n' seconds is

$s = un + \frac{1}{2}an^{2}$...(2)   (as $s = ut + \frac{at^{2}}{2}$ and $t = n$)

here $u = 0$ since the body starts from rest (given)

Now, dividing equation (1) with (2) we get,

  $\frac{s_{n}}{s} = \frac{\frac{a}{2}(2n-1) }{\frac{a}{2}n^{2} }$

therefore, by simplifying we get

           $\frac{s_{n}}{s} = \frac{2n-1}{n^2}$

i.e    $\frac{s}{s_{n}} = \frac{n^2}{2n-1}$

#SPJ2