Monday

September 1, 2014

September 1, 2014

Posted by **Reanna** on Tuesday, March 27, 2012 at 7:41pm.

- Calculus -
**Reiny**, Tuesday, March 27, 2012 at 9:21pmMake a sketch to get a right angled triangle

let the time passes since noon be t hrs

let the noon position of ship B be the origin O

AO = 10 + 16t miles

OB = 24t miles

AB^2 = AO^2 + OB^2

= (10+16t)^2 + (24t)^2

2 AB d(AB)/dt = 2(10+16t)(16) + 2(24t)((24)

when t = 3 ( at 3:00 pm)

AB^2 = (58)^2 + 72^2 = 8548

AB = √8548

d(AB)/dt = (32(58)+ 48(72) )/(2√8548)

= 28.72

at that moment they are separating at 28.7 knots

**Related Questions**

calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...

calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...

calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...

calculus - At noon, ship A is 30 nautical miles due west of ship B. Ship A is ...

CALCULUS - At noon, ship A is 40 nautical miles due west of ship B. Ship A is ...

Calculus - At noon, ship A is 20 nautical miles due west of ship B. Ship A is ...

Calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...

Calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...

calculus 1 - At noon, ship A is 20 nautical miles due west of ship B. Ship A is ...

calculus 1 - At noon, ship A is 30 nautical miles due west of ship B. Ship A is ...