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 :
As \[v=0+na\] \[\Rightarrow \] \[a=\frac{v}{n}\] Now, distance travelled in \[n\]sec \[\Rightarrow \] \[{{S}_{n}}=\frac{1}{2}a{{n}^{2}}\]and Distance travelled in \[(n-2)\sec \] \[\Rightarrow \] \[{{S}_{n-2}}=\frac{1}{2}a{{(n-2)}^{2}}\] Distance travelled in last 2 s, \[{{S}_{n}}-{{S}_{n-2}}=\frac{1}{2}a{{n}^{2}}-\frac{1}{2}a{{(n-2)}^{2}}\] \[=\frac{a}{2}[{{n}^{2}}-{{(n-2)}^{2}}]\] \[=\frac{a}{2}[n+(n-2)][n-(n-2)]\] \[=a(2n-2)=\frac{v}{n}(2n-2)=\frac{2v(n-1)}{n}\]
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