Question

g At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 17 knots and ship B is sailing north at 25 knots. How fast (in knots) is the distance between the ships changing at 4 PM

Answer 1

Answer:

$\triangle d=47.69 Knots$

Step-by-step explanation:

Distance of Ship A from B $d_1=10 West$

Speed of Ship A $V_a=17 knots West$

Speed of Ship A $V_b=25 knots North$

Generally the equation for Rate of distance change is mathematically given by

$\triangle d=\frac{1}{2\sqrt{(17t+10)^2+(25t)^2}}*\triangle t[17t+10^2+25t^2]$

$\triangle d=\frac{578+340+1250}{2\sqrt{(17t+10)^2+(25t)^2}}$

Therefore with

t=>4PM

We substitute

$\triangle d=\frac{882(4)+420+648(4)}{2\sqrt(21(4)+10^2)+(18(4))^2}$

$\triangle d=47.69 Knots$