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,

Options

A. `4 h, 80 sqrt(5) km`
B. `50 sqrt(2) h, sqrt(10) km `
C. `100 sqrt(2) h, 2 sqrt(10) km`
D. `80 sqrt(5) h, 4 km`

Answer

A

Explanation

According to the question,
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) ` for finding value of `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` .