## anonymous 5 years ago Two ocean liners leave from the same port in Puerto Rico at 10:00 a.m. One travels at a bearing of N 51o W at 14 miles per hour, and the other travels at a bearing of S 56o W at 17 miles per hour. Approximate the distance between them at noon the same day. Round answer to two decimal places.

1. anonymous

is it 51 degree north of west

2. anonymous

?

3. anonymous

A. 20.33 miles B. 43.25 miles C. 28.20 miles D. 22.72 miles E. 37.19 miles

4. anonymous

5. anonymous

Find x and Y component of speed of each ship

6. anonymous

ok so ship a would be 28 miles and shib b would be 34 miles i think

7. anonymous

Here How I would solve it .. Ship 1 travel at 51 degree North of West Using standard coordinate system X component of speed is -cos(51)*14 Y component of speed is sin(51)*14

8. anonymous

This is just ship 1

9. anonymous

okay

10. anonymous

try this with ship 2

11. anonymous

so you will get speed of two ship

12. anonymous

Multiply x and Y component of each ship seperately to find position then just use distance formula

13. anonymous

ok so ship a is -97.49 ship b is 128.6 ?

14. anonymous

X position of Ship 1 at noon(12-10am) -cos(51)*14 mile per hours * 2hour -17.62 Y position of Ship 2 at noon sin(51)*14*2=21.76

15. anonymous

X position of Ship 2 -cos(56)*17*2=something Y position of Ship 2 -sin(56)*17*2=something Distance Formula $\sqrt(ship 1 (x) position - ship 2( x) position)^2+ship 1 (y) position - ship 2( y) position)^2)$

16. anonymous

X position of Ship 2 -cos(56)*17*2=-23.89 Y position of Ship 2 -sin(56)*17*2=17.73

17. anonymous

so then I just plug those into the formula?

18. anonymous

I got 19.01 for x position of ship 2

19. anonymous

hmmm ok you're probably right

20. anonymous

Yep

21. anonymous

can you see the distance formula in full

22. anonymous

no

23. anonymous

it is basically square root of (xposition of ship 1- xpos ship 2)^2 + (ypos ship 1-y pos ship 2)^2)

24. anonymous

Is it physics or math?

25. anonymous

math trig/pre calc

26. anonymous

Were you able to get answer

27. anonymous

?

28. anonymous

ok I have to go back to the begining and replug in stuff cause i messed something up hold on

29. anonymous

I will do the same thing

30. anonymous

ok thanks

31. anonymous

I got 45.73.....?

32. anonymous

33. anonymous

I read that problem wrong it is 51 degree west of north ---- not north of west so I have to switch sin to cos vice versa

34. anonymous

okay I have no clue what i did then

35. anonymous

Ship 1 position X: - sin(51)*14*2 -- A Y: cos(51)*14*2 --B Ship 2 position X: -sin(56)*17*2 -- C Y: -cos(56)*17*2 --D

36. anonymous

okay

37. anonymous

sqrt((A-C)^2+(B-D)^2)

38. anonymous

=37.19

39. anonymous

ok thank you!