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Ship A is moving Westwards with a speed of 20kmh^(-1)and another ship B which is at 200 km South of A is moving Northwards with a speed of 10km h^(-1). The time after which the distance between them is shortest and the shortest distance between them respectively, |
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Answer» `4 h, 80 SQRT(5) km` Let ship A travel `x_(A)`DISTANCE and ship B travel `x_(B)`distance, in TIME t . So, `u_(A) = (x_(A))/(t)rArrt = (x_(A))/(20)` and`u_(B) = (x_(B))/(t) rArr t = (x_(B))/(10) rArr (x_(A))/(20) = (x_(B))/(0) rArr x_(A) = 2x_(B)` So, `AB = sqrt(x_(4)^(2)+(200-x_(B))^(2))` `= sqrt(4x_(B)^(2)+ 40000+x_(B)^(2) - 400x_(B))` `= sqrt(5x_(B)^(2)-400x_(B) +40000)` Differentiate distance AB w.r.t. `x_(B) ` forfinding valueof `x_(B)` `(d(AB))/(dx_(B)) =(1)/(2sqrt(5x_(B)^(2)-400x_(B)+40000))(10x_(B) - 400)=0` or`x_(B) = 40 m ` Again differentiating, SO, `((d^(2)(AB))/(dx_(B)^(2)))_(x_(B)= 40 m)gt 0""[:. "Distance always greater than zero"]` Hence `x_(B) `at point, `x_(B) =40 `m distance, AB will be shortest. So, `AB = sqrt(5XX 40^(2) - 400 xx40 +4000)` `AB = sqrt(32000)` or `AB = 80sqrt(5) km` The time after which the distance AB is shortest `t = (x_(A))/(20) = (x_(B))/(10)=(40)/(10)= 4 hr` . 
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