A) 40 cm/sec
B) 20 cm/sec
C) 10 cm/sec
D) 5 cm/sec
Correct Answer: C
Solution :
Here, \[{{s}_{1}}=200\,\,cm,\,\,{{t}_{1}}=2\sec \] and \[{{s}_{2}}=220\,\,cm,\,\,{{t}_{2}}=4\,\,\sec \] We know, \[s=ut+\frac{1}{2}a{{t}^{2}}\] \[\therefore \] \[200=u\times 2+\frac{1}{2}\times a\times 4\] \[\Rightarrow \] \[a+u=100\] ... (i) and \[420=u(\times )6+\frac{1}{2}a\times {{(6)}^{2}}\] \[u+3a=70\] ... (ii) Solving equations (i) and (ii), we get \[u=115\,\,cm/\sec \], and\[a=-15\,\,cm/{{s}^{2}}\] Again, \[v=u+at\] \[=115-15\times 7\] \[\mathbf{=10}\,\,\mathbf{cm/sec}\mathbf{.}\]
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