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A ship A is moving Westwards with a speed of 10 km/h and a ship B 100 km South of A is moving northwards with a speed of 10 km/h . The time after which the distance between them becomes shortest is |
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Answer» 0h If we SUPPOSE the x-axis along eastward direction and the y-axis along northward direction , then the positionof ship A after time t, `vecr_A=(-10hati)t` The position of ship B after timet, `vec_B=-100hatj+(10hatj)t` Therefore, position of B with respect to A, `vec_B-vec_A =(10t)hati+(10 t-100)hatj` so, `|vec_B-vec_A| =sqrt((10t)^2+(10t-100)^2)` `=sqrt(100t^2+100t^2-2000t+10000)` `=10 sqrt(2) sqrt(t^2-10t+50)` This DISTANCE becomes the shortest when `(t^2-10t+50)` becomesminimum, i,e., `d/(dt) (t^2-10t+50) =0 or, 2t -10 =0` `therefore t=5 hours` |
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