Problem Statement: A ship A is moving westward with a speed of 10km/s. and ship B 100km south of A is moving northward with speed of 10km/hr. The time after which distance between them becomes shortest is. 1) 5hr 2) 5√2 hr 3) 10√2 hr 4) 0 hr. (aipmt 2015)

Solution:
Let's assume the position of ship A as (x,0) and the position of ship B as (0,y).
The velocity of ship A is 10 km/s westward, which can be written as (-10,0) km/s.
The velocity of ship B is 10 km/hr northward, which can be written as (0,10) km/hr.
Let's assume that after time 't', the distance between the two ships is minimum and their positions are (x',y') and (0,y'+100).
The relative position of ship A with respect to ship B after time 't' can be written as (x'-0, y'+100-0) = (x', y'+100).
The relative velocity of ship A with respect to ship B can be written as (-10, 10) km/hr.
The distance between the two ships after time 't' can be written as:
distance = ((x'-0)^2 + (y'+100-0)^2)^(1/2)
distance = (x'^2 + (y'+100)^2)^(1/2)

To find the minimum distance between the two ships, we need to differentiate the distance equation with respect to 't' and equate it to zero.
d(distance)/dt = 0
=> 2x'(dx'/dt) + 2(y'+100)(dy'/dt) = 0
=> x'(dx'/dt) + (y'+100)(dy'/dt) = 0

Substituting the values of x', y' and the relative velocity, we get:
x'*(-10) + (y'+100)*10 = 0
=> -10x' + 10y' + 1000 = 0
=> x' - y' = 100

We also know that the velocity of ship A is constant and is (-10,0) km/s.
So, we can write:
x' = -10t + x
y' = 10t + y + 100

Substituting the values of x' and y' in the equation x' - y' = 100, we get:
-10t + x - 10t - y - 100 = 100
=> -20t + x - y = 200

Solving these two equations simultaneously, we get:
x = 500 km
y = -400 km

Substituting the values of x and y in the equations x' = -10t + x and y' = 10t + y + 100, we get:
x' = -10t + 500
y' = 10t - 300

Substituting these values in the equation distance = (x'^2 + (y'+100)^2)^(1/2), we get:
distance = (10000t^2 - 20000