Vector problem involving two ships

Question

A large commercial vessel, the Bigship, is sailing in perfectly calm seas at a speed (relative to the surrounding sea) of 25 km h−1 on a bearing of 40◦ . Unfortunately, it is low on fuel because Captain Bighead messed up his fuel calculations. A small fuel tanker, the Thinktank, is travelling in perfectly calm seas at a speed (relative to the surrounding sea) of 35 km h−1 on a bearing of 130◦ . Hence it should cross the path of the Bigship at right angles, allowing the shortest possible trip to carry out refueling at sea.

Take unit vectors i to point east and j to point north.

(a) Express in component form the velocity b of the Bigship and the velocity t of the Thinktank, both relative to the sea, giving the numerical values in km h−1 to one decimal place.

(b) A strong wind arises, and this has the effect of applying an additional velocity w to both vessels, of 10 km h−1 in the direction from west to east. Write the velocity w in component form, and hence express the resultant velocities of the Bigship and the Thinktank in component form, giving numerical values to one decimal place.

(c) Find the magnitude of the resultant velocity of each vessel. Give magnitudes in km h−1 to one decimal place.

(d) Hence find the actual angle between the courses of the two vessels, taking into account the effect of the wind. Give your answer to the nearest degree. (Hint: use the dot product of the resultant velocity vectors.)

(Sorry it's a long question. I'm so incredibly lost. I don't even know how to get started.)

Answer 1

This is a big question! Luckily, we can break it down into parts and work it through one piece at a time.

a) Each ship is moving through the sea on a diagonal. Assuming that 0° is due east (0° is almost always due east/to the right unless a problem specifies otherwise), that means that our Bigship is headed northeast, and the Thinktank is headed northwest. We have the speed of each ship, so we know that in one hour Bigship will go 25 km on its diagonal path, and Thinktank will go 35 km. What we want to do here is break down that diagonal path, that vector, into north/south and east/west components.

For instance, if Bigship goes on an angle of 40° at 25 km/hr, it will be going 19.2 km/hr to the east and 16.1 km/hr to the north. How did we figure that out?

Think about Bigship's path as the hypotenuse of a right triangle. The adjacent leg of the triangle is the total distance Bigship will travel east; the opposite leg of the triangle is the total distance Bigship will travel north. We know that H=25 km/hr, and θ=40°. We can use that information to calculate the length of both legs, and therefore the speed the ship is traveling north and east:

cos(40°) = adj/hyp

0.77 = adj/25

adj = 19.2 km/hr east

sin (40°) = opp/hyp

0.64 = opp/25

opp = 16.1 km/hr north

We can express this as a vector using the units given in the problem.

Bigship is on a vector of 19.2km/hr i + 16.1km/hr j.

Now that we've calculated the vector for Bigship, we can do the same thing for Thinktank:

cos(130°) = adj/hyp

-0.64 = adj/35

adj = -22.5 km/hr east

sin (130°) = opp/hyp

0.77 = opp/35

opp = 26.8 km/hr north

Thinktank is on an angle of 130°. Our value for sine is positive, and the ship heads north at 26.8 km/hr. On the other hand, our value for cosine is negative, so our ship heads east at -22.5 km/hr. In other words, it's heading west at 22.5 km/hr! Using the units given in the problem, we can write our second vector.

Thinktank is on a vector of -22.5 km/hr i + 26.8 km/hr j

.

b) Why did we go through all that work to turn these ship vectors into north and east components in the last problem? It's so we can do this next part!

Wind is blowing each ship at 10 km/hr to the west of their original vector. We can't do much with that information if all we know is the total speed of each ship and the angle they're traveling at, but now we know exactly how fast each ship is traveling east/west and north/south. We can add this new wind vector directly to the component-form vectors we just calculated.

Bigship:

(19.2km/hr i + 16.1km/hr j) -10.0km/hr i

9.2km/hr i + 16.1km/hr j

Thinktank:

(-22.5 km/hr i + 26.8 km/hr j) - 10.0km/hr i

-33.5km/hr i + 26.8km/hr j

.

c) Now we know how fast these ships are going north and east/west including that wind, but it's also useful to know how fast each ship is actually going along its diagonal path. We're going to turn these component vectors back into an overall speed. If you think back to the right triangles from part (a), we now have two legs of these new right triangles, and we need to find the hypotenuses.

We can do that easily using the Pythagorean theorem. For example, for Bigship:

A² = B² + C²

A² = (9.2 km/hr)² + (16.1 km/hr)²

A² = (84.64 + 259.21) km²/hr²

A² = 343.85 km²/hr²

A = 18.54 km/hr

Bigship is now moving along its diagonal path at about 18.5 km/hr

And for Thinktank:

A² = B² + C²

A² = (-33.5 km/hr)² + (26.8 km/hr)²

A² = (1122.25 + 718.24) km²/hr²

A² = 1840.49 km²/hr²

A = 42.91 km/hr

Thinktank is now moving along its diagonal path at about 42.9 km/hr

.

d) Lastly, we want to find out what angle these ships are moving compared to each other. We have plenty of information about these ships to answer that question; in fact, we can do it in a few different ways! Let's look at two.

METHOD 1: We can easily figure out each ship's angle compared to due east (which we've said is 0°) and then subtract those angles from each other.

We know that Bigship is moving east at 9.2 km/hr, north at 16.1 km/hr, and moving overall at 18.5 km/hr. Thinking about that right triangle again, it's easy to see that for Bigship:

sinθ = 16.1 / 18.5

cosθ = 9.2 / 18.5

tanθ = 16.1 / 9.2

We can use any one of these equations to calculate θ (the angle between Bigship's heading and due east).

θ = 60.5°

For Thinktank, we can use:

sinθ = 26.8 / 42.9

cosθ = -33.5 / 42.9

tanθ = 26.8 / -33.5

Using the first equation we get θ=38.7°, while the second and third equations give us θ=141.3°. That makes sense if you understand how trigonometry works on a coordinate plane, but if not, we can use common sense to figure out which of those numbers we want. Thinktank is traveling north and west, so it must be at more than a 90° angle away from due east. For Thinktank, θ=141.3°.

Therefore, we can find the angle between Bigship's line of travel and Thinktank's line of travel:

141.3°- 60.5° = 81° between the course of the two ships.

METHOD 2: We can use the dot product of Bigship and Thinktank's traveling vectors, like it suggests in the problem. The dot product is what happens when you multiply two vectors together, and there are two ways to calculate it:

a · b = |a| × |b| × cosθ

OR

a · b = (a_x × b_x) + (a_y × b_y)

We have almost all of these numbers ready at hand. |a| and |b| are the magnitudes of our two ships' vectors, or the speeds they're going along their diagonal paths. a_x and b_x are the east/west components of each vector, while a_y and b_y are the north/south components of each vector. Using this information, we know that:

|a| × |b| × cosθ = (a_x × b_x) + (a_y + b_y)

18.5 × 42.9 × cosθ = (9.2 × -33.5) + (16.1 × 26.8)

793.65cosθ = -308.2 + 431.48

cosθ = 0.155

θ = 81° between the courses of the two ships