Trig Word Problem?

Question

A ship is moving due west at 8 MPH. You are in a speed boat at a distance of 10 miles from the ship. Your bearing as seen from the ship is 130 degrees. You need to catch up with the ship. So you take off at a speed of 16 MPH and a bearing of _____? degrees, and you reach the ship in _____? minutes. FYI Bearing Starts at North = 0 degrees, and east =90 degrees and move clockwise

Answer 1

OK, this is a bear of a problem. Let's say the ship is, at time = 0, at the origin, moving west, that is, in the negative x direction. Your bearing, in speed boat, is 130 degrees --- that's in the fourth quadrant, 40 degrees below the positive x-axis. You are 10 miles away in that direction, so your coordinates at t = 0 are (10 cos(40), -10sin(40)). At time t = 0, you are going to take off in your speed boat, aiming up and to the left, toward some point on the negative x-axis where you expect to rendezvous the boat. Let's say that the meeting point is a distance D to the left of the origin, and T is the time from start until meeting. The equation of motion for the ship simply says D = 8T, that's it. For the speed boat, distance = 16T, but we have to figure out that distance. From starting point to ending point, the vertical distance is still 10sin(40), but the horizontal distance is now [10 cos(40) + D]. Use the Pythagorean theorem to combine these to get a total distance for the speed boat ---- distance^2 = 100[sin(40)]^2 + [10 cos(40) + D]^2 = 100[sin(40)]^2 + 100[cos(40)]^2 + 20Dcos(40) + D^2 = 100 + 20Dcos(40) + D^2. (Notice, we got a nice simplification there from the Pythagorean identity [sin(A)]^2 + [cos(A)]^2 = 1.) This equation is fine as is --- now, we can substitute to eliminate the variable T. We know that D = 8T, and (speed boat distance) = 16T, so (speed boat distance) = 2D. This means (speed boat distance)^2 = 4D^2. Set that equal to the equation above. 4D^2 = 100 + 20Dcos(40) + D^2 ---- This is a quadratic in the variable D, so move everything to one side of the equation, equal to zero. 3D^2 - 20Dcos(40) - 100 = 0. There's no factoring this bad boy: we need the quadratic formula. The positive root is D = 8.86645233 miles. Divide that by 8 to get T, the time until the intersection: T = 1.108306541 hours. Now, to find the bearing of the speed boat's course --- it's 10sin(40) in the y-direction, and [D + 10cos(40)] in the x-direction. The reference angle will be arctan(y/x) = 21.25276275 = 21 deg, 15 min, 9.946 sec. That's an angle above the x-axis in the second quadrant, so for a bearing, we'd have to go around 270 degrees to get to where it starts. Bearing equals that number plus 270, which is 291.25276275 degrees = 291 deg, 15 min, 9.946 sec. Actually, I don't think steering in a speed boat is accurate to an arcsecond. :-)

Answer 2

Not much of a speed boat if that's all the faster it goes.

The ship is moving due west (bearing 270°) at 8 mph.
The speed boat is at bearing 130° from the initial position of the ship.
The speed boat is initially 10 miles from the ship.

t = time required to meet the ship.

We have a triangle.
Side a = 10 = distance between the initial positions of the
ship and speed boat.
Side b = 8t = side sailed by the boat
Side c = 16t = side covered by the speed boat

The angle between sides "a" and "b" = 270 - 130 = 140°

Use the Law of Cosines to find side c.

c² = a² + b² = 2ab cosC

(16t)² = 10² + (8t)² - 2(10)(8t) cos140°

256t² = 100 + 64t² - 160t*cos140°

192t² + 160t*cos140° - 100 = 0

48t² + 40t*cos140° - 25 = 0

t = {-40*cos140° ± √[(40*cos140°)² - 4*48*(-25)]} / (2*48)

t = {-40*cos140° ± √(1600*cos²140° + 4800)} / 96

t = {-40*cos140° ± 40√(cos²140° + 3)} / 96

t = {-5*cos140° ± 5√(cos²140° + 3)} / 12

t ≈ -0.4699361, 1.1083065
The negative solution is rejected because distance must be positive.

t ≈ 1.1083065

16t = 17.732905 miles

Time to meet ship = t hours = 1.1083065 hours

Time to meet ship = 1 hour, 6 min, 29.903548 sec
_____________________

To find the bearing the speed boat took to meet the ship use the Law of Sines.

c = 16t = 17.732905 miles
b = 8t = 8.8664523 miles
a = 10 miles


c / sinC = b / sinB

sinB = (b/c) sinC = (16t/(8t)) sinC = (1/2)sin140° ≈ 0.3213938

B = arcsin[(1/2)sin140°] ≈ 18.747237°

So the bearing of the speed boat is

130° + 180° - 18.747237° = 291.25276°

Answer 3

Speed boat catches ship in t*60 minutes, where

(16t)^2 = (8t)^2 + 10^2 - 2*8t*10*cos(140 deg)