A) 3 s
B) 4.5 s
C) 5s
D) 6 s
Correct Answer: D
Solution :
Here: Total time \[C=2.5\times {{10}^{-10}}F\]acceleration =a retardation \[=-2a\] From the laws of motion \[\upsilon =u+a{{t}_{1}}\] or \[\upsilon =0+a{{t}_{1}}\] or \[{{t}_{1}}=\frac{\upsilon }{a}\] ??.(i) Again, \[0=\upsilon -2a{{t}_{2}}\] or \[{{t}_{2}}=\frac{\upsilon }{2a}\] ??..(ii) From Eqs. (i) and (ii), we have \[{{t}_{1}}+{{t}_{2}}=t\] \[\Rightarrow \] \[\frac{\upsilon }{a}+\frac{\upsilon }{2a}=9\] \[\Rightarrow \] \[\frac{3\upsilon }{2a}=9\] \[\Rightarrow \] \[\frac{\upsilon }{a}=\frac{9\times 2}{3}\] \[\Rightarrow \] \[\frac{\upsilon }{a}=6\] Hence, duration of acceleration \[{{t}_{1}}=\frac{\upsilon }{a}=6\sec \]
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