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Chapter 3: Problem 35

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

Expert verified
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

01

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. \(\vec{A} = (2.00, 0)\) \(\vec{B} = (4.00\cos{(45°)}, -4.00\sin{(45°)})\)
02

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. \(\vec{F} = (6.00, 0)\)
03

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: \(\vec{C} = (0.717, 2.83)\)
04

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.

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