InterviewSolution
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InterviewSolution
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A passenger reaches the platform and finds that the second least boggy of the train is passing him. The second last boggy takes `3 s` to pass the passenger, and the last boggy takes `2 s` to pass him. Find the time by which the passnger late for the departure of the trains? Assume that the train accelerates at constant rate and all the boggies are of equal length. |
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Answer» Correct Answer - `3.5 s` `(n-2)l=(1)/(2) at^(2)` `(n-1)l=(1)/(2)a(t+3)^(2)` `nl=(1)/(2)a (t+5)^(2)` `l=(1)/(2)a [(1+3)^(2)-t^(2)]=(1)/(2) a[(t+5)^(2)-(t+3)^(2)]` `t^(2)+9+6t-t^(2)=t^(2)+25+10 t-t^(2)-9-6t` `9+6t =16 +4t` `2t=7` `t=3.5 sec`. |
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