Hello, Nikhiln25!

Did you make a sketch?

At noon, ship $\displaystyle A$ is 30 nautical miles due west of ship $\displaystyle B.$

Ship $\displaystyle A$ is sailing west at 16 knots and ship $\displaystyle B$ is sailing north at 24 knots.

How fast (in knots) is the distance between the ships changing at 3 PM? Code:

* B -
* | :
* | 24t :
* | :
D * * S :
* | 24t+72
* | :
* | 72 :
* | :
* | :
* - - - - * - - - * - - - - - * -
A 16t R 48 P 30 Q
: - - - - - - 16t+78 - - - - :

At noon, ship $\displaystyle A$ is at $\displaystyle P$ and ship $\displaystyle B$ is at $\displaystyle Q\!:\;PQ\,=\,30$

By 3 PM, ship $\displaystyle A$ has sailed 48 miles west to $\displaystyle R$,

. . and ship $\displaystyle B$ has sailed 72 miles north to $\displaystyle S.$

In the next t hours, ship $\displaystyle A$ sails $\displaystyle 16t$ miles west to $\displaystyle A$

. . and ship $\displaystyle B$ sails $\displaystyle 24t$ miles north to $\displaystyle B.$

Their distance is: $\displaystyle D = AB$, the hyptenuse of right triangle $\displaystyle ABQ$

. . with sides: .$\displaystyle 16t + 78$ and $\displaystyle 24t + 72$.

Hence: .$\displaystyle D \;=\;\sqrt{(16t+78)^2 + (24t + 72)^2} \;=\;\sqrt{832t^2 + 5952t + 11,\!268} $

Then: .$\displaystyle \dfrac{dD}{dt} \;=\;\dfrac{1}{2}\,\dfrac{1664t + 5952}{\sqrt{832t^2 + 5952t + 11,\!268}} \;=\;\dfrac{416t + 1488}{\sqrt{208t^2 + 1488t + 2817}}$

When $\displaystyle t=0\!:\;\;\dfrac{dD}{dt} \;=\;\dfrac{1488}{\sqrt{2817}} \;=\;\dfrac{496}{\sqrt{313}} \;=\;28.03557758$

Their distance is increasing at about $\displaystyle 28\text{ knots.}$