A) \[\frac{2\upsilon (n+1)}{n}\]
B) \[\frac{\upsilon (n+1)}{n}\]
C) \[\frac{\upsilon (n-1)}{n}\]
D) \[\frac{2\upsilon (n-1)}{n}\]
Correct Answer: D
Solution :
| [d] \[\because \,\,v=0+na\Rightarrow \,\,a=v/n\] |
| Now, distance travelled in n sec. |
| \[\Rightarrow \] \[{{S}_{n}}=\frac{1}{2}\,a{{n}^{2}}\] |
| 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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