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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