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Vector \(\vec{A}\) is \(3.00\) units in length and points along the positive x-axis. Vector \(\overrightarrow{\mathbf{B}}\) is \(4.00\) units in length and points along the negative y-axis. Use graphical methods to find the magnitude and direction of the vectors (a) \(\vec{A}+\vec{B}\) and (b) \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}\).

Short Answer

Expert verified
\(\vec{A} + \vec{B} = 5.00\) units, \(306.87^\circ\) and \(\vec{A} - \vec{B} = 5.00\) units, \(53.13^\circ\).

Step by step solution

01

Vector Addition

Vector \(\vec{A} + \vec{B}\) can be found by adding them head-to-tail. First place vector \(\vec{A}\) along the x-axis, then place vector \(\vec{B}\) such that its tail is at the head of \(\vec{A}\). The vector from the tail of \(\vec{A}\) to the head of \(\vec{B}\) is \(\vec{A} + \vec{B}\).
02

Magnitude and Direction for Vector Addition

The magnitude \(|\vec{A} + \vec{B}|\) can be found using the Pythagorean theorem, \(|\vec{A} + \vec{B}| = \sqrt{(3)^2 + (-4)^2} = 5\). The direction can be found using trigonometry, \(\tan^{-1}(\frac{|B|}{|A|}) = \tan^{-1}(\frac{4}{3}) = 53.13^\circ\) in the fourth quadrant, or \(360^\circ - 53.13^\circ = 306.87^\circ\).
03

Vector Subtraction

Vector \(\vec{A} - \vec{B}\) can be found by reversing the direction of \(\vec{B}\) and then adding it to \(\vec{A}\) as in vector addition.
04

Magnitude and Direction for Vector Subtraction

The magnitude \(|\vec{A} - \vec{B}|\) can be found using the Pythagorean theorem, \(|\vec{A} - \vec{B}| = \sqrt{(3)^2 + (4)^2} = 5\). The direction can be found using trigonometry, \(\tan^{-1}(\frac{|B|}{|A|}) = \tan^{-1}(\frac{4}{3}) = 53.13^\circ\) in the first quadrant.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Magnitude
Understanding the magnitude of a vector is essential for analyzing motion and forces in physics. The magnitude, often referred to as the length or size of a vector, measures how far the vector 'reaches' in a particular direction. In our example, vector \( \vec{A} \) is given as 3.00 units, and vector \( \vec{B} \) is 4.00 units. To visualize it, imagine you are walking 3 steps east and then 4 steps south; the total distance you covered in each direction corresponds to the magnitudes of \( \vec{A} \) and \( \vec{B} \) respectively.

Even though vectors can be in two or three dimensions, determining their magnitude in a two-dimensional plane involves a straightforward application of the Pythagorean theorem—a method you will encounter in the next section.
Vector Direction
The direction of a vector is equally as important as its magnitude. It's the 'where' that complements the 'how much' of the magnitude. Direction can be described using various units, such as degrees, radians, or even by reference to the cardinal points (north, east, south, and west). For instance, in our exercise, \( \vec{A} \) is pointing along the positive x-axis, which is considered to be 0 degrees or pointing due east.

On the other hand, \( \vec{B} \) points along the negative y-axis, which corresponds to 270 degrees—or due south. Knowing both the magnitude and the direction enables us to accurately describe the vector's position and to understand vector addition and subtraction, which lead to resultant vectors with their unique magnitudes and directions.
Pythagorean Theorem
The Pythagorean theorem is a cornerstone of geometry, critical to understanding vectors in physics. It relates the lengths of the sides of a right-angled triangle. Applied to vector analysis, we often use it to find the resultant magnitude when we add or subtract vectors, as done in our exercise.

The formula is: \[ a^2 + b^2 = c^2 \], where \(a\) and \(b\) are the lengths of the two shorter sides of the triangle, and \(c\) is the length of the longest side, or the hypotenuse. When \( \vec{A} \) and \( \vec{B} \) are added head-to-tail, they form a right-angled triangle with their resultant vector. Calculating this resultant magnitude involves squaring the individual vector lengths, adding them together, and then taking the square root of the sum—directly applying the Pythagorean theorem.
Trigonometry in Physics
Trigonometry, the branch of mathematics dealing with the properties of triangles, is indispensable in physics for solving problems involving vectors. It provides tools for calculating unknown angles or side lengths in right-angled triangles. This is particularly useful when we need to find the direction of a resultant vector after addition or subtraction.

For our vectors, we use the tangent function, which is the ratio of the opposite side to the adjacent side, to figure out the angle of the direction. For instance, the direction angle \( \theta \) of the resultant vector can be found using the equation \( \tan(\theta) = \frac{opposite}{adjacent} \), and solving for \( \theta \) with an inverse tangent function \( \tan^{-1} \). Trigonometry provides us a clear angle of direction from the reference axis, which, combined with the magnitude, gives a complete description of the vector's position and quantity.

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Most popular questions from this chapter

A chinook salmon has a maximum underwater speed of \(3.58 \mathrm{~m} / \mathrm{s}\), but it can jump out of water with a spced of \(6.26 \mathrm{~m} / \mathrm{s}\). To move upstream past a waterfall, the salmon does not need to jump to the top of the fall. but only to a point in the fall where the water speed is less than \(3.58 \mathrm{~m} / \mathrm{s}\); it can then swim up the fall for the remaining distance. Because the salmon must make forward progress in the water, let's assume it can swim to the top if the water speed is \(3.00 \mathrm{~m} / \mathrm{s}\). If water has a speed of \(1.50 \mathrm{~m} / \mathrm{s}\) as it passes over a ledge, how far below the ledge will the water be moving with a speed of \(9.00 \mathrm{~m} / \mathrm{s}^{2}\) (Note that water undergoes projectile motion once it leaves the ledge.) If the salmon is able to jump vertically upward from the base of the fall, what is the maximum height of waterfall that the salmon can clear?

A student decides to measure the muzzle velocity of a pellet shot from his gun. He points the gun horizontally. He places a target on a vertical wall a distance \(x\) away from the gun. The pellet hits the target a vertical distance \(y\) below the gun. (a) Show that the position of the pellet when traveling through the air is given by \(y=A x^{2}\), where \(A\) is a constant. (b) Express the constant \(A\) in terms of the initial (muzzle) velocity and the free-fall acceleration. (c) If \(x=3.00 \mathrm{~m}\) and \(y=0.210 \mathrm{~m}\), what is the initial speed of the pellet?

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Two canoeists in identical canoes exert the same effort paddling and hence maintain the same speed relaLive to the water. One paddles directly upstream (and moves upstream), whereas the other paddles directly downstream. With downstream as the positive direction, an observer on shore determines the velocities of the two canoes to be \(-1.2 \mathrm{~m} / \mathrm{s}\) and \(+2.9 \mathrm{~m} / \mathrm{s}\), respectively. (a) What is the speed of the water relative to the shore? (b) What is the speed of each canoe relative to the water?

A figure skater glides along a circular path of radius \(5.00 \mathrm{~m}\). If she coasts around one half of the circle, find (a) the magnitude of the displacement vector and (b) what distance she skated. (c) What is the magnitude of the displacement if she skates all the way around the circle?

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