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 m/s Final velocity \[(\upsilon )\] = 100 m/s distance s = 10 cm = 0.1 m Using the relation of equation of motion \[{{\upsilon }^{2}}={{u}^{2}}+2as\] \[a=\frac{{{\upsilon }^{2}}-{{u}^{2}}}{2s}=\frac{{{(100)}^{2}}-{{(200)}^{2}}}{2\times 0.1}\] \[=\frac{10000-40000}{2\times 0.1}=\frac{-30000}{2}\] \[\begin{align} & =-150000\,m/{{s}^{2}} \\ & =-15\times {{10}^{4}}\,m/{{s}^{2}} \\ \end{align}\] (-) minus sign denotes retardation
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