A) \[5\,\,cm\,\,{{s}^{-1}}\]
B) \[10\,\,cm\,\,{{s}^{-1}}\]
C) \[15\,\,cm\,\,{{s}^{-1}}\]
D) \[20\,\,cm\,\,{{s}^{-1}}\]
Correct Answer: B
Solution :
We have,\[{{S}_{1}}=200\,\,cm\]; \[{{t}_{1}}=2\,\,s\] \[{{S}_{2}}=420\,\,cm\]; \[{{t}_{2}}=6\,\,s\] \[{{S}_{1}}=u{{t}_{1}}+\frac{1}{2}at_{1}^{2}\Rightarrow u+a=100\] ? (1) \[{{S}_{2}}=u{{t}_{2}}+\frac{1}{2}at_{2}^{2}\Rightarrow u+3a=70\] ? (2) Solving (1) and (2), we get \[a=-15\,\,cm\,\,{{s}^{-2}}\]and\[u=115\,\,cm\,\,{{s}^{-1}}\] Since, \[v=u+at\] \[v=115-15\times 7=10\,\,cm\,\,{{s}^{-1}}\]
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