Question

A ship leaves port at 1:00 P.M. and sails in the direction S35°E at a rate of 24 mi/hr. Another ship leaves port at 1:30 P.M. and sails in the direction S20°W at a rate of 18 mi/hr.
Approx. how far apart are the ships at 3:00pm

I made a picture and I dont know if I need to add some to the side to correspond with the changing times

Answer 1

What is the question asking , i think you left that out in the directions.

Answer 2

Oh sorry. Approx. how far apart are the ships at 3:00pm

Answer 3

I drew the south line just to aid us.
So we want to know how far apart are the two vectors

Answer 4

first lets find the magnitude of each vector.
Distance = rate * time

Answer 5

A ship leaves port at 1:00 P.M. and sails in the direction S35°E at a rate of 24 mi/hr. Another ship leaves port at 1:30 P.M. and sails in the direction S20°W at a rate of 18 mi/hr. Approx. how far apart are the ships at 3:00pm

the first vector
distance = 24 mi/hr * 2 hr = 48 mi

second vector
distance = 18 mi/hr * 1.5 hr = 27 mi

Answer 6

now we can use law of cosines for that triangle

Answer 7

Law of cosines : a^2 + b^2 - 2*a*b*cos(C) = c^2

27^2 + 48^2 - 2*27*48*cos(20 + 35 degrees) = x^2

Answer 8

then take the square root of the both sides
x = √ [27^2 + 48^2 - 2*27*48*cos(55 degrees)]

Answer 9

(24)(2) For the first vector, It wouldn't be something different because we started at 1:30 and we're going to 3?

Answer 10

That would be why perl only put "1.5 hr"

Answer 11

I get that for the second vector yes but the first one is at a different time

Answer 12

the first vector (boat) travels from 1 pm to 3 pm, so 2 hours.
the second vector (boat) travels from 1:30 pm to 3 pm, so 1.5 hours