A) \[\frac{2v\,(n-1)}{n}\]
B) \[\frac{v\,(n-1)}{n}\]
C) \[\frac{2v\,(n+1)}{n}\]
D) \[\frac{v\,(n+1)}{n}\]
Correct Answer: A
Solution :
By using relation, \[v=u+at\] \[\therefore \] \[v=0+an\] or \[a=\frac{v}{n}\] ?(i) Distance travelled in n sec \[{{S}_{n}}=\frac{1}{2}a{{n}^{2}},\]and Distance travelled in (n - 2) sec \[{{S}_{n-2}}=\frac{1}{2}a\,{{(n-2)}^{2}}\] \[\therefore \] Distance travelled in last two seconds \[{{S}_{n}}-{{S}_{n-2}}=\frac{1}{2}a{{n}^{2}}-\frac{1}{2}a\,{{(n-2)}^{2}}\] \[=\frac{a}{2}[{{n}^{2}}-{{(n-2)}^{2}}]\] \[=\frac{a}{2}[n+(n-2)]\,[n-(n-2)]\] \[=a\,(2n-2)\] Substituting value of a from Eq. (i), we have \[{{S}_{n}}-{{S}_{n-2}}=\frac{v}{n}(2n-2)\] \[=\frac{2v(n-1)}{n}\]
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