A) - 10 cm/s
B) 10 cm/s
C) 20 cm/s
D) 30 cm/s
Correct Answer: B
Solution :
[b] Let u be initial velocity & a be its acceleration Distance in first |
\[2\text{ }sec={{S}_{1}}=200\text{ }cm\] |
\[\Rightarrow \] \[u(2)+\frac{1}{2}a{{(2)}^{2}}=200cm\] |
\[\Rightarrow \] \[u+a=100\] ...(i) |
Distance in next 4 sec. \[={{S}_{2}}=220\,cm\] |
Distance in first 6 sec. |
\[={{S}_{1}}+{{S}_{2}}=200+220\,cm\] |
\[\Rightarrow \] \[u(6)+\frac{1}{2}a{{(6)}^{2}}=420\] |
\[\Rightarrow \] \[u+3a=70\] ...(ii) |
From equations (i) & (ii), we get |
\[a=-15\,cm/{{s}^{2}},\,\,u=115\,cm/s\] |
Hence, velocity at the end of 7 sec. from start |
\[=\] \[u+7a\] |
\[=\] \[115+7(-15)\] |
\[=\] \[10\,cm/s.\] |
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