Question

During a jaunt on your sailboat, you sail \(2.00 \mathrm{~km}\) east, \(4.00 \mathrm{~km}\) southeast, and an additional distance in an unknown direction. Your final position is \(6.00 \mathrm{~km}\) directly east of the starting point. Find the magnitude and direction of the third leg of your journey.

Short answer

Answer: The magnitude of the third leg of the journey is 2.95 km, and its direction is 75.9° counterclockwise from east.

Step-by-step solution

Define the known vectors in terms of components

Displacement vector 1 (\(\vec{A}\)) is \(2.00\) km east, which translates to \(x\)-component of \(2.00\) km and \(y\)-component of \(0\) km. Displacement vector 2 (\(\vec{B}\)) is \(4.00\) km southeast, which translates to \(x\)-component of \(4.00\cos{(45°)}\) km and \(y\)-component of \((-1)4.00\sin{(45°)}\) km. \(\vec{A} = (2.00, 0)\) \(\vec{B} = (4.00\cos{(45°)}, -4.00\sin{(45°)})\)

Define the final position vector

The final position vector (\(\vec{F}\)) is \(6.00\) km directly east of the starting point, which translates to an \(x\)-component of \(6.00\) km and a \(y\)-component of \(0\) km. \(\vec{F} = (6.00, 0)\)

Use vector addition to find the third displacement vector

We can use the vector addition equation \(\vec{F} = \vec{A} + \vec{B} + \vec{C}\), where \(\vec{C}\) is the third leg of the journey. Rearranging for \(\vec{C}\), we get: \(\vec{C} = \vec{F} - \vec{A} - \vec{B}\) Calculating the \(x\) and \(y\) components of \(\vec{C}\): \(x_C = x_F - x_A - x_B = 6.00 - 2.00 - 4.00\cos{(45°)} = 6.00 - 2.00 - 4.00(0.707) = 0.717\,\text{km}\) \(y_C = y_F - y_A - y_B = 0 - 0 - (-4.00\sin{(45°)}) = 4.00(0.707) = 2.83\,\text{km}\) So the third displacement vector \(\vec{C}\) is: \(\vec{C} = (0.717, 2.83)\)

Calculate the magnitude and direction of the third displacement vector

We can find the magnitude of \(\vec{C}\) using the Pythagorean theorem: \(|\vec{C}| = \sqrt{x_C^2 + y_C^2} = \sqrt{(0.717)^2 + (2.83)^2} = 2.95 \,\text{km}\) Now, we find the angle between \(\vec{C}\) and the positive \(x\)-axis: \(\theta = \tan^{-1}{\left(\dfrac{y_C}{x_C}\right)} = \tan^{-1}{\left(\dfrac{2.83}{0.717}\right)} = 75.9°\) The magnitude of the third leg of the journey is \(2.95\) km and its direction is \(75.9°\) counterclockwise from east.

Key concepts

Displacement Vectors

Displacement vectors are a way to describe movement in both direction and magnitude. When you're navigating, like our sailboat example, each direction you travel can be thought of as a vector. These vectors point in the direction of movement and the length of the arrow represents the distance traveled. In the boat exercise, the journey had three legs, or vectors: one going east, the second southeast, and a mysterious third leg. By understanding each displacement vector's x and y components, we can figure out the resultant journey. It's like connecting the dots of your route on a map, visualizing a combination of movements in different directions that eventually take you to your final position.

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in geometry, expressing the relationship between the sides of a right-angled triangle. For any triangle with a right angle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides: \(c^2 = a^2 + b^2\). In the context of our sailboat problem, once the components of the third vector are known, we need to find its magnitude, which is the length of \(\vec{C}\). Here, this is done by treating the vector components like the sides of a right triangle, thus applying: \(\sqrt{x^2 + y^2}\). This technique allows us to compute the total distance traveled during the third leg, piecing the journey together into one complete picture.

Trigonometry

Trigonometry helps us work with angles and sides in triangles, especially when vectors are involved. Using angles, we can break down vectors into their x and y components, like the sailboat's southeast vector in our example. By knowing the angle of travel (45° for southeast), we use cosine and sine functions to find the exact horizontal (x) and vertical (y) distances. Cosine helps us with the x-component \(x = r \cdot \cos{\theta}\), and sine to find the y-component \(y = r \cdot \sin{\theta}\). Inversely, we use the arc tangent function \(\tan^{-1}\) to find the direction of the third vector from its components. This gives us a precise angle, providing a complete directional understanding of the boat's path.

Component Form

Any vector can be broken into components based on directions typically aligned with a coordinate axis. Component form simplifies complex movement into manageable parts, using x (horizontal) and y (vertical) values. In our problem, the first vector is purely horizontal, identified as \((2, 0)\), while the second vector is dissected using trigonometric relationships given its southeast direction. Converting vectors like \(\vec{B} = (4\cos{45°}, -4\sin{45°})\) allows for easy manipulation when performing vector addition or subtraction. This way, analyzing multiple movements and adding them together becomes straightforward, ultimately revealing the unknown third vector \(\vec{C}\). By summing these components, we piece together how various movements result in a new position.