An any instant of time, two ships A and B are 70 km apart along a line AB which is directed from north to south . A starts moving towards west at `25km*h^(-1)` and at the same time B starts moving towards north at `25km*h^(-1)` . Find the distance of closest approach between the two ships and the time required for this.

Explanation

Let us choose the origin at point A, x-axis along east,y-axis along north.
Initial position of ship A=0 and that of ship B=`-70hatjkm.`
Velocity of ship A=`-25hatikm*h^(-1)` and that of ship `B=25hatjkm*h^(-1)` .
`therefore` After a time of t hours, the positions of the ships are,
`vecr_(A)=0+(-25hati)t=-25thatikm`
`vecr_(B)=-70hatj+(25hatj)t=(25t70)hatjkm`
position of ship B relative to ship A,
`vecr=vecr_(B)-vecr_(A)=25thati+(25t-70)hatj`
The distance between them is r=`|vecr|`. So,
`r^2=(25t)^2+(25t-70)^2`
`=625t^2+625t^2-3500t+4900`
`50(25t^2-70t+98)`
`=50[(5t-7)^ 2-7^2+98]`
`=50[(5t-7)^2+49]`
When the distance r between the ships is minimum, `r^2` is also minimum .
This happens when `(5t-7)^2` , a squared quantity is minimum , i.e., zero.
`therefore 5t-7=0`
or, `t=5/7h=84 min =1h24 min`
At this instant of time,
`r^2=50[0+49]=50xx49=25xx49xx2`
`therefore` The distance of closest approach between the two ships,
`r_(min)=sqrt(25xx49xx2)=35sqrt(2) km`