need to know the knots in this question

Question

At noon, ship A is 20 nautical miles due west of ship B. Ship A is sailing west at 19 knots and ship B is sailing north at 16 knots. How fast (in knots) is the distance between the ships changing at 5 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)

Answer 1

In an xy cartesian system place ship B at the origin and ship A at the point on the x-axis (-20 ,0).

This is the position of the ships at noon.

Five hours later ship B is at ( 0, 80 ) and ship A at ( - 115,0 ).

Using the Pythagorean Theorem you can find that the distance of the two ships at 5 PM is

r = √( 115² + 80²)=140.089

At this very moment x² + y² = r²

Taking the derivative of the above equation with respect to time and of course applying the

chain Rule we get

d/dt ( x² + y² ) = d/dt ( r² )

2x ( dx/dt ) +2y ( dy/ dt) = 2r (dr/dt)

x ( dx/dt ) +y ( dy/ dt) = r (dr/dt), where dx/dt and dy/dt are the speeds of the two ships and dr/dt the rate of how fast the distance of the ships changing.

Then 115⋅19 + 80⋅16 = ( 140.089)⋅ (dr/dt)

hence dr/dt = 3465/ 140.089

dr/dt= 24.74 knots

Answer 2

Let x be the distance between the position of ship A and the original position of ship B at 12:00.

Let y be the distance between the position of ship B and it's original position at 12:00.

Given:

dx/dt = 19 knots

dy/dt = 16 knots

Let s = the distance between the two ships which is the hypotenuse of the triangle form by ship A, B and B's original position 12 12:00.

s²= x²+y²

Need to find ds/dt at 5:00 PM.

2sds/dt = 2xdx/dt+2ydy/dt

ds/dt = (xdx/dt +ydy/dt)/s

At 5:00 PM, x = 20 + 5•19 = 115 nm, y = 5•16 = 80 nm and s = √(115²+80²) = 161.32 nm

ds/dt = (115(19)+80(16))/(161.32) = (2185+1280)/161.32 = 21.48 knots