A) \[\frac{2v(n+1)}{n}\]
B) \[\frac{v(n+1)}{n}\]
C) \[\frac{v(n-1)}{n}\]
D) \[\frac{2v(n-1)}{n}\]
Correct Answer: D
Solution :
[d] \[\therefore v=0+na\Rightarrow a=v/n\] Now, distance travelled I n sec \[\alpha \Rightarrow {{S}_{n}}=\frac{1}{2}a{{n}^{2}}\]and distance travelled in \[(n-2)\]\[\sec \Rightarrow {{S}_{n-2}}=\frac{1}{2}a{{(n-2)}^{2}}\] \[\therefore \] distance travelled in last two seconds, \[={{S}_{n}}-{{S}_{n-2}}=\frac{1}{2}a{{n}^{2}}-\frac{1}{2}a{{(n-2)}^{2}}\] \[=\frac{a}{2}[{{n}^{2}}-{{(n-2)}^{2}}]=\frac{a}{2}[n+(n-2)][n-(n-2)]\] \[=a(2n-2)=\frac{v}{n}(2n-2)=\frac{2v(n-1)}{n}\]
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