Question

$If\:initial\:velocity\:of\:an\\object\:is\:'u'\:and\\acceleration\:is\:'a'\:then\\find\:the\:distance\\travelled\:in\:{n}^{th}\:second.\\(A){S}_{n}=un+\frac{a{n}^{2}}{2}\\(B) {S}_{n}=u+\frac{a}{2}(2n+1)\\(C){S}_{n}=un+a{n}^{2}\\(D){S}_{n}=(n+\frac{a}{2}){n}^{2}$

Answer 1

Given,

Initial velocity  of an  object  is  u and  acceleration is  a.

Let the ditance traveled in n second  from initial position is  $S_n$ .

According  second  equation of  motion,

Distance traveled in  t second,

$S=ut+\frac{1}{2}at^{2}$

Distance traveled  in  n second,

$S_n=un+\frac{1}{2}an^{2}...........i)$

Distance traveled in (n-1) second,

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

Hence,

Distance traveled in nth second,

$S_{n^{th}}=S_n-S_{n-1}\\\\\;\;\;\;\;\;=un+\frac{1}{2}an^{2}-[u(n-1)+\frac{1}{2}a(n-1)^{2}]\\\\=un+\frac{1}{2}an^{2}-un+u-\frac{1}{2}a(n-1)^{2}\\\\=un+\frac{1}{2}an^{2}-un+u-\frac{1}{2}a(n^{2}+1-2n)\\\\=u+\frac{1}a(n^{2}-n^{2} -1+2n)\\ \\S_{n^{th} }=u+\frac{1}{2}a(2n-1)$

Answer 2

Distance traveled in nth second,

$\begin{lgathered} \sf \: S_{n^{th}}=S_n-S_{n-1}\\\\\;\;\;\;\;\; \sf=un+\frac{1}{2}an^{2}-[u(n-1)+\frac{1}{2}a(n-1)^{2}]\\\\ \sf \: =un+\frac{1}{2}an^{2}-un+u-\frac{1}{2}a(n-1)^{2}\\\\ \sf \: =un+\frac{1}{2}an^{2}-un+u-\frac{1}{2}a(n^{2}+1-2n)\\\\ \sf \: =u+\frac{1}a(n^{2}-n^{2} -1+2n)\\ \\ \sf \red{S_{n^{th} }=u+\frac{1}{2}a(2n-1)}\end{lgathered} </p><p></p><p>$