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Two persons A and B are located in X-Y plane at the points (0,0) and (0,10) respectively, The distances are measured in MKS unit. At a time t = 0, they start moving simultaneously with velocities vec v_(A) = 2hatj ms^(-1) and vec v_(B) = 2hat i ms^(-1) respectively. The time after which A and B are at their closest distance is |
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Answer» (a) 2.5 s Distance `d =sqrt((0 - V_(B)t)^(2) + (V_(A)t - 10)^(2))` `RARR d =sqrt((2t^(2) + (2t - 10)^(2))` ` rArrd^(2) = 4T^(2) + 4t^(2) - 40t + 100` `rArr d^(2) = l = 8t^(2) - 40t + 100` `rArr dl/(dt) = 16t - 40` `dl/(dt) = 0` `rArr 16t - 40 = 0` `:. t = 40/(16) = 2.5 s` 
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