A) \[15\times {{10}^{4}}m/{{s}^{2}}\]
B) \[13.5\times {{10}^{4}}m/{{s}^{2}}\]
C) \[12\times {{10}^{4}}m/{{s}^{2}}\]
D) none of these
Correct Answer: A
Solution :
Here: initial velocity \[=200\text{ }m/s\] Final velocity \[\text{(}\upsilon \text{)}=100\text{ }m/s\] distance \[s=10cm=0.1\text{ }m\] Using the-relation of equation of motion \[{{\upsilon }^{2}}={{u}^{2}}+2as\] \[a=\frac{{{\upsilon }^{2}}-{{u}^{2}}}{2s}=\frac{{{(100)}^{2}}-{{(200)}^{2}}}{2\times 0.1}\] \[=\frac{10000-40000}{2\times 0.1}=\frac{-300000}{2}\] \[=-150000m/{{s}^{2}}\] \[=-15\times {{10}^{4}}m/{{s}^{2}}\] \[(-)\] minus sign denotes retardation
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