Question

Q7). 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, ​

Answer 1

Let the time after which the distance between the two ships becomes shortest be 't' hours.

Then, for the time 't' hours, the ship A moves the distance '10t' km westwards, and since the ship B has the same velocity as that of A, the ship B also moves the same distance '10t' km but northwards.

So, since both the ships are initially 100 km away from each other, the ship A is '10t' km horizontally away from B and the ship B is '100 - 10t' km vertically away from A, after 't' hours.

Then the distance between the ships A and B, at the time 't', is,

$d=\sqrt{(10t)^2+(100-10t)^2}\\\\d=10\sqrt2\cdot\sqrt {t^2-10t+50}$

From this, we get that,

$t^2-10t+50\ \textgreater\ 0\\\\t^2-10t+25\ \textgreater\ -25\\\\(t-5)^2\textgreater\ -25$

Since the LHS is a square of something,

$(t-5)^2\geq0$

For the distance being the shortest,

$(t-5)^2=0\\\\\therefore\ \large \text{\underline {\underline { t=5 \ hours}}}$