Two ships leave port at 4 p.m. One

is headed N38°E and is travelling at
11.5 km/h. The other is travelling at
13 km/h, with heading S47°E. How far
apart are the two ships at 6 p.m.?

i am in grade 11 :), studying ahead of the class :) its 8 pm here, this question is hard for me, i spent 10 minutes on it.

draw the diagram, figure each distance (velocity*time). time is 2 hours for each ship

Now you have a triangle, 2 known legs, and the included angle.

Find the opposite side: I recommend use the law of cosines.

The standard graph in polar coordinates has E at 0' ( o degree ) .

The makes N 38 E would be 90'- 38' = 52' and S 47 E at 2700'+ 47'= 314' , or -46 .
The difference in angle between the two is 53+46= 99'
At 6 PM , two hours have passed . This means the 1st ship went 2(11.5)=23 miles and the 2nd ship went 2(13)= 26 miles . Here , a= 23 , B=26 and A= 99'

This is the only way I can help , I am in 11th grade but I already did some excercise like this , not this one but some like it . I am a Haitian student , ( Haiti) so please maybe my English don't realy make you understand it because I do it in French at school but I translate it in English for you . This is the way they learn me to do it !!! Good luck Karen , be smart !!! Study hard !

Unless otherwise indicated, all angles

are measured CCW from +x-axis.

Ship #1:
d1 = 11.5km/h[52o] * 2h = 22km[52o].

Ship #2:
d2 = 13km/h[ 317o] * 2h = 26km[317o].

d2-d1 = 26[317o] - 22[52o].

X = 26*Cos317 - 22*Cos52) = 5.47 km.
Y = 26*sin317 - 22*sin52 = -35.1 km.

d2-d1 = Sqrt(X^2 + Y^2) = Distance apart.

To solve this problem, we can use the concept of vectors and displacement.

First, let's break down the information given:

Ship A:
- Heading: N38°E
- Speed: 11.5 km/h

Ship B:
- Heading: S47°E
- Speed: 13 km/h

We want to find the distance between the two ships at 6 p.m., which is 2 hours after they left port.

To get started, we need to calculate the displacements of both ships during this time.

The displacement of an object is a measure of its change in position. It is a vector quantity defined by magnitude and direction. In this case, we can think of the displacements of the ships as vectors that describe their movements.

To calculate the displacement of Ship A, we can use the following steps:

1. Determine the distance traveled by Ship A in 2 hours:
Distance = Speed x Time
Distance = 11.5 km/h x 2 h
Distance = 23 km

2. Determine the horizontal (east-west) and vertical (north-south) components of Ship A's displacement:
- The horizontal component is given by the equation: Distance x cos(angle)
Horizontal component = 23 km x cos(38°)
- The vertical component is given by the equation: Distance x sin(angle)
Vertical component = 23 km x sin(38°)

3. Repeat the same steps to calculate the displacement of Ship B:
- Distance traveled by Ship B = 13 km/h x 2 h
- Horizontal component of Ship B = Distance x cos(angle)
- Vertical component of Ship B = Distance x sin(angle)

Once we have the horizontal and vertical components of the displacements for both ships, we can calculate the distance between them using the Pythagorean theorem.

The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

So, the distance between the ships is given by the equation:

Distance = sqrt((Horizontal_A - Horizontal_B)^2 + (Vertical_A - Vertical_B)^2)

By substituting the calculated values, we can find the final answer for the distance between the two ships at 6 p.m.