(1 pt) At noon, ship A is 50 nautical miles due west of ship B. Ship A is sailing west at 15 knots and ship B is sailing north at 21 knots. How fast (in knots) is the distance between the ships changing at 4 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)

Answer 1

Relative to ship B's starting point, with t = 0 at noon, the vector position of ship A is
R1 = (-50 -15t) i
and the vector position of ship B is
R2 = 21 t j.
"i" is an east-pointing unit vector and "j" is a north-pointing unit vector.
At 4 PM, t = 4 hours.
The vector that separates R2 and R1 is
R2 - R1 = 21 t j + (50 + 15 t) i.
The rate of change of the separation is
d(R2-R1)/dt = 21 j + 15 i.
Note that is is independent of time.
The magnitude of that vector,
sqrt[(15^2 + 21^2],
is the separation speed that you want

Answer 2

let t hours be some time after noon
(so 4:00 pm is t=4)

so you have a right angled triangle with a vertical of 21t nautical miles and a horizontal of (15t + 50) nautical miles

let the distance between them, or the hypotenuse, be s nautical miles

s^2 = (21t)^2 + (15t+50)^2
2s(ds/dt) = 441t + 30(15t+50)

so when t=4
s^2 = 441(4^2) + 110^2
s = √19156

ds/dt = (441x4 + 30(110))/(2√16156)
= 18.29

so at 4:00 pm the distance between them is changing at 18.29 knots

check my arithmetic.

Answer 3

thanks for helping me but it keeps telling me the answer 18.29 knots is wrong

Answer 4

A tag boat travels at the rate of 15 knots how many hours will i take the boat to travel the distance of 42 nautical miles