A Coast Guard cutter detects an unidentified ship at a distance of 17.0 km in th

Question

A Coast Guard cutter detects an unidentified ship at a distance of 17.0 km in the direction 15.0° east of north. The ship is traveling at 23.0 km/h on a course at 40.0° east of north. The Coast Guard wishes to send a speedboat to intercept and investigate the vessel.

(a) If the speedboat travels at 54.0 km/h, in what direction should it head? Express the direction as a compass bearing with respect to due north.

I have gotten this far:

x: 17 cos 75o + 23 cos 50o t = 52 cos sb t

y: 17 sin 75o + 23 sin 50o t = 52 sin sb t

Explanation / Answer

The unidentified ship starts at a distance of 20 km, at 15º east of north,
with a velocity of 26 km/h, at 40º east of north.
These (r,) polar coordinates can be transformed (x,y) coordinates by using the following formulas.
x = r sin
y = r cos

So the ship unidentified starts at position ( 5.18x + 19.3y ) km
with a velocity of ( 16.7 x + 19.9y) km/h

The speedboat starts at corrdinates (0,0), and has an initial velocity of (50.0 , ) in polar coordinates. In cartesian, this becomes ( 50sin , 50cos )..

At a given time t from the start, the x-coordinate of the ship will be 5.18 + 16.7t, while that of the speedboat will be 50sin*t.
The y-coordinate of the ship will be 19.3 + 19.9t while that of the speedboat will be 50cos*t.

We want the solution for which the equations
5.18 + 16.7t = 50sin*t
19.3 + 19.9t = 50cos*t
are both satisfied. This is a system of two independdant equations with two unknowns, which can be solved.

Taking the first equation, we have
t = 5.18 / (50sin - 16.7)
Plugging it into the second one gets us
19.3 = (50cos - 19.9)t
19.3 = (50cos - 19.9) * 5.18 / (50sin - 16.7)
3.732 = (50cos - 19.9) / (50sin - 16.7)
3.732 = (cos - 0.398) / (sin - 0.334)
3.732 * (sin - 0.334) = (cos - 0.398)
3.732 sin - 1.247 = cos - 0.398
3.732 sin = cos + 0.849

Let's take the square of each side
13.93 sin^2 = cos^2 + 1.70 cos + 0.721
Knowing that [sin^2 + cos^2 = 1], we have
13.93 (1 - cos^2) = cos^2 + 1.70 cos + 0.721
13.93 - 13.93 cos^2 = cos^2 + 1.70 cos + 0.721
14.93 cos^2 + 1.70 cos - 13.21 = 0

This is a simple quadratic equation in cos.
The solutions are
cos = 0.885
cos = -0.999
acos (0.885) = 27.7º
acos (-0.999) = 177º (clearly not what we're looking for)

So it seems the direction in which to head is about 27.7º east of north

As in the start, this direction can be expresed in (x,y) coordinates by using
x = r sin ; y = r cos
so it becomes; v = (23.2x + 44.3y) km/h

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