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

`\triangle d=47.69 Knots`

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=(1)/(2√((17t+10)^2+(25t)^2))*\triangle t[17t+10^2+25t^2]`

`\triangle d=(578+340+1250)/(2√((17t+10)^2+(25t)^2))`

Therefore with

t=>4PM

We substitute

`\triangle d=(882(4)+420+648(4))/(2\sqrt(21(4)+10^2)+(18(4))^2)`

`\triangle d=47.69 Knots`