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?
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.
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.
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)
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.
To find the distance between the two ships after 3 hours, we can use the concept of vectors and trigonometry.
First, let's label the two ships as Ship 1 (traveling on a bearing S11 degrees W) and Ship 2 (traveling on a bearing N75 degrees E).
We need to convert the bearings into compass headings:
- Ship 1: S11 degrees W = 270 degrees - 11 degrees = 259 degrees
- Ship 2: N75 degrees E = 360 degrees - 75 degrees = 285 degrees
Now, we can find the horizontal and vertical components of each ship's velocity using trigonometric functions.
For Ship 1:
- The horizontal component (Westward direction) = 12 mph * cos(259 degrees)
- The vertical component (Southward direction) = 12 mph * sin(259 degrees)
For Ship 2:
- The horizontal component (Eastward direction) = 9 mph * cos(285 degrees)
- The vertical component (Northward direction) = 9 mph * sin(285 degrees)
Next, we need to calculate the distance traveled by each ship after 3 hours, using the formula distance = speed * time.
For Ship 1:
- Distance traveled = 12 mph * 3 hours
For Ship 2:
- Distance traveled = 9 mph * 3 hours
Now, we can calculate the horizontal and vertical distances between the two ships.
The horizontal distance (Dh) = (Distance traveled by Ship 1 * horizontal component of Ship 1) - (Distance traveled by Ship 2 * horizontal component of Ship 2)
The vertical distance (Dv) = (Distance traveled by Ship 2 * vertical component of Ship 2) - (Distance traveled by Ship 1 * vertical component of Ship 1)
Finally, we can use the Pythagorean theorem to find the distance between the ships after 3 hours. The formula is: distance = sqrt(Dh^2 + Dv^2).
Let's plug in the values and calculate it.