Question

Two ships leave a harbor at the same time. One ship travels on a bearing S11 degrees W at 12 mph. The second ship travels on a bearing N75 degrees E at 9 mph. How far apart will the ships be after 3 hours?

Answer 1

Ship #1: 12mi/h[11o] W. of S. = 12mi/h[191o] CW,
Ship #2: 9mi/h[75o] E. of N. = 9mi/h[75o] CW.

d = 36mi[191o] - 27mi[75o],
X = 36*sin191 - 27*sin75 = -32.9 mi,
Y = 36*Cos191 - 27*Cos75 = -42.33 mi,
d^2 =(-39.9)^2 + ( - 42.33)^2 = 3383.84,
d = 58.2 miles apart.

Answer 2

My answer is this :
The V1 represent 12mph[S11W], and V2 represent 9mph [N75E]
Then we can write A and B as follows:

From the definition of velocity we have,

V1 = d1/t
V2 = d2/t
d1 = V1 * t
d2 = V2 * t
After 3hours, the ship 1 and ship 2 will travel,

d1 = 12mph[S11W] * 3h = 36m [S11W]
d2 = 9mph[N75E] *3h= 27m [N75E]

Now the two displacements, which are vectors, can be rewrite in the following x and y coordinate forms:

d1 = (36m * cos259, 36sin259)
d2 = (27m * cos15, 27 * sin15)

Now, the distance between them now can be easily calculated using a distance formula,

d = sqrt((x2-x1)^2 + (y2 - y1)^2))
=sqrt((36cos259 -27cos15)^2 + (36sin259 -27sin15)^2)
=53. 63947715 m

Thus, the two ship will be apart 54m after three hours.


V1 = (12mph* cos 15, 12mph * sin15)
V2 = (9mph * cos259, 9mph * sin 259)

Answer 3

I made a sketch saw that I had a triangle with sides 36 and 27 and a contained angle
of 116°
So, using the cosine law: let the distance between them be x
x^2 = 36^2 + 27^2 - 2(36)(27)cos116
= 2877.19...
x = appr 53.64 miles, which agrees with Daniel's answer.

Answer 4

Correction: Replace 39.9 with 32.9:
d^2 = (-32.9)^2 + (-42.33)^2 = 2874.24,
d = 53.61 miles which agrees with Daniel and Reiny's answer.