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In the Name of God بسم الله

A Physics Problem That Needs To Be Solved Asap

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Hasnain

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A Coast Guard cutter detects an unidentified ship at a distance of 19.4 km in the direction 15.6° east of north. The ship is traveling at 30 km/h on a course at 42.3° east of north. The Coast Guard wishes to send a speedboat to intercept the vessel and investigate it. If the speedboat travels 52.2 km/h, in what direction should it head? Express the direction as a compass bearing with respect to due north.

Answer:_____° east of north.

The correct answer is 30.6° east of north, if you don't get that please dont post anything.

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if the speed boat intercepts the other boat, then they are at the same location at the same time t. The vector representations of the intercept location for each boat is:

P = 19.4sin15.6i + 19.4cos15.6j + 30tsin42.3i + 30tcos42.3j

and for the speed boat is:

Q = 52.2tsinxi + 52.2tcosxj

Set the i and j vector coefficients to get two equations with two unknowns x and t.

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(bismillah)

(salam)

I might have solved your problem. The answer I get is slightly different to yours (33.43 degrees) but this may be due to incertitude.

I always like to begin classical mechanics problems with a sketch:

intro.JPG

The guard is at the point (0;0) and our problem begins when he sees a ship at a distance of 19.4 km (15.6 degrees to the east). This distance is shown by the red line. Since our problem begins here, we can say that t=0 (the time starts being counted at this point).

From this moment onwards we have a ship and a speedboat moving at the same time (the ship in blue, the speedboat in green, with respective speeds of 30 kmh and 52.2 kmh and respective angles of 42.3 degrees and x degrees). Our aim is to find x.

Let's first find out when the two objects meet.

Movement of ship: r0 + v0t = 19.4 + 30t

Movement of boat = 52.2t

When the two meet: 19.4 + 30t = 52.2t and so 19.4 = 22.2t and t = (19.4/22.2)

t (time when they meet) is approximately 0.874 hours.

With t we can find the distances covered by the ship and the speedboat:

Speedboat: 52.2t = (approximately) 45.62 km

Ship: 30t = (approx) 26.22 km

The rest is mainly trigonometry. We have the angle between the ship and the guard (initial point) and also the angle in which the ship travels, and thanks to these angles we calculate the vertical distance between the guard and the ship (final point), which should be the same as the vertical distance between the guard and the speedboat (final point). This allows us to calculate the angle between the guard (0;0) and the speedboat, which is what we are trying to find.

next.jpg

As you can see in the picture, we can identify three angles

Red: angle between North and initial position of the ship (known)

Blue: angle between North and direction (and final position) of the ship (known)

Green: angle between North and direction of the speedboat (unknown)

You can also see that the vertical distances are all the same.

This is where the trigonometry comes in.

The North axis is treated as the adjacent (because it touches the angle) and the lines showing the distance covered as the hypotenuse. This is based on the right-angled triangle (known to everyone who has covered trigonometry).

Cos = adj/hyp therefore adj = Cos * hyp

Total vertical distance (between guard and final position of ship/speedboat):

( Cos (15.6) * 19.4 ) + ( Cos (42.3) * 26.22 ) = (approx) 38.08 km

We have found the total vertical distance by adding the vertical distance between the guard and the initial position of the ship, and the verticle distance between this initial position and the final position of the ship. We could do this because we knew the angles in both cases.

We also know that this is the same as the verticle distance between the guard and the final position of the speedboat. Therefore:

Cos (x) * 45.62 = 38.08

Cos (x) = 38.08 / 45.62

x = arcos (38.08 / 45.62) = 33.41 degrees

I think this is the correct way to solve the problem. My final answer is a few degrees off (about 3 degrees off) this is probably due to approximations and rounding off).

Edited by Asadollah
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LAW OF SINES:

The angle your looking for, lets call it theta, is equal to phi + 15.6° where:

sin (phi) / 30 km/h = sin (180° - 42.3° +15.6°) / 52.2 km/h ------> or phi = 14.96°

Theta = phi + 15.6° = 14.96° + 15.6° = 30.56°

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(bismillah)

(salam)

I might have solved your problem. The answer I get is slightly different to yours (33.43 degrees) but this may be due to incertitude.

I always like to begin classical mechanics problems with a sketch:

intro.JPG

The guard is at the point (0;0) and our problem begins when he sees a ship at a distance of 19.4 km (15.6 degrees to the east). This distance is shown by the red line. Since our problem begins here, we can say that t=0 (the time starts being counted at this point).

From this moment onwards we have a ship and a speedboat moving at the same time (the ship in blue, the speedboat in green, with respective speeds of 30 kmh and 52.2 kmh and respective angles of 42.3 degrees and x degrees). Our aim is to find x.

Let's first find out when the two objects meet.

Movement of ship: r0 + v0t = 19.4 + 30t

Movement of boat = 52.2t

When the two meet: 19.4 + 30t = 52.2t and so 19.4 = 22.2t and t = (19.4/22.2)

t (time when they meet) is approximately 0.874 hours.

With t we can find the distances covered by the ship and the speedboat:

Speedboat: 52.2t = (approximately) 45.62 km

Ship: 30t = (approx) 26.22 km

The rest is mainly trigonometry. We have the angle between the ship and the guard (initial point) and also the angle in which the ship travels, and thanks to these angles we calculate the vertical distance between the guard and the ship (final point), which should be the same as the vertical distance between the guard and the speedboat (final point). This allows us to calculate the angle between the guard (0;0) and the speedboat, which is what we are trying to find.

next.jpg

As you can see in the picture, we can identify three angles

Red: angle between North and initial position of the ship (known)

Blue: angle between North and direction (and final position) of the ship (known)

Green: angle between North and direction of the speedboat (unknown)

You can also see that the vertical distances are all the same.

This is where the trigonometry comes in.

You were doing well with your attempted solution until this point. You assumed that the resulting triangle is a right triangle. Its not.

What you end up with is a triangle whose sides are 19.4 km, 26.22km and 45.62 km. Obviously, this is not a right triangle.

You do know one angle of the triangle, which is 180° - 42.3° + 15.6° = 137.7°.

Using the law of Sines:

45.62 km / sin 137.7° = 26.22 km / sin phi ------> or phi = 14.96°

This gives you the angle at which the speedboat takes off east of the 19.4 km line that you drew in your diagram, so you have to add 14.96° and 15.6° to get 30.56°, which is the angle East of North (where due North is the y-axis).

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(bismillah)

(salam)

Thanks for your reply brother Learned.

Actually, I didn't consider the resulting triangle as a right triangle. What I did was to consider three different right triangles:

-The initial distance and the North axis

-The distance travelled by the ship and the North axis

-The distance travelled by the speedboat and the North axis

Each of the above were the hypotenuse, and the North axis was the adjacent angle, and therefore the vertical distance between the guard and the meeting point of the two boats.

Perhaps the following diagram might explain it better:

plus.jpg

In any case, your method works very well. I hadn't though of the Law of Sines, but it's obviously the best approach in such a problem.

What's confusing me now is why my method didn't work out! Is it due to incertitude or is there a mistake somewhere?

Anyhow, at least the brother has his solution now!

PS:

The Law of Sines

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