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Chapter 3: Problem 14

Vector \(\overrightarrow{\mathbf{A}}\) has magnitude 4.0 units; vector \(\overrightarrow{\mathbf{B}}\) has magnitude 6.0 units. The angle between \(\overrightarrow{\mathbf{A}}\) and \(\overrightarrow{\mathbf{B}}\) is \(60.0^{\circ} .\) What is the magnitude of \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}} ?\)

Short Answer

Expert verified
Answer: The magnitude of the sum \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) is approximately \(8.72\) units.

Step by step solution

01

Understand the Law of Cosines for Vectors

The Law of Cosines for vectors states that the magnitude of the sum of two vectors can be calculated using the magnitudes of the vectors and the angle between them. The formula is: $$||\overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}}|| = \sqrt{||\overrightarrow{\mathbf{A}}||^{2} + ||\overrightarrow{\mathbf{B}}||^{2} + 2||\overrightarrow{\mathbf{A}}||||\overrightarrow{\mathbf{B}}||\cos{\theta}}$$ Where \(||\overrightarrow{\mathbf{A}}||\) is the magnitude of vector \(\overrightarrow{\mathbf{A}}\), \(||\overrightarrow{\mathbf{B}}||\) is the magnitude of vector \(\overrightarrow{\mathbf{B}}\), and \(\theta\) is the angle between \(\overrightarrow{\mathbf{A}}\) and \(\overrightarrow{\mathbf{B}}\).
02

Insert the given values into the formula

Now you can insert the given values into the Law of Cosines formula: $$||\overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}}|| = \sqrt{4^{2} + 6^{2} + 2(4)(6)\cos{60^{\circ}}}$$
03

Calculate the cosine of the angle

First, calculate the cosine of the given angle: $$\cos{60^{\circ}} = 0.5$$
04

Perform the calculation

Now that you have the cosine value, substitute it back into the formula and perform the calculation: $$||\overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}}|| = \sqrt{4^{2} + 6^{2} + 2(4)(6)(0.5)}$$ Simplify the expression: $$||\overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}}|| = \sqrt{16 + 36 + 24}$$ $$||\overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}}|| = \sqrt{76}$$
05

Final Answer

Therefore, the magnitude of the sum \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) is: $$||\overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}}|| = \sqrt{76} \approx 8.72$$ units.

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Most popular questions from this chapter

In each of these, the \(x\) - and \(y\) -components of a vector are given. Find the magnitude and direction of the vector. (a) $A_{x}=-5.0 \mathrm{m} / \mathrm{s}, A_{y}=+8.0 \mathrm{m} / \mathrm{s} .\( (b) \)B_{x}=+120 \mathrm{m}$ $B_{y}=-60.0 \mathrm{m} .(\mathrm{c}) C_{x}=-13.7 \mathrm{m} / \mathrm{s}, C_{y}=-8.8 \mathrm{m} / \mathrm{s} .(\mathrm{d}) D_{x}=$ \(2.3 \mathrm{m} / \mathrm{s}^{2}, D_{y}=6.5 \mathrm{cm} / \mathrm{s}^{2}\)
Displacement vector \(\overrightarrow{\mathbf{A}}\) is directed to the west and has magnitude \(2.56 \mathrm{km} .\) A second displacement vector is also directed to the west and has magnitude \(7.44 \mathrm{km}\) (a) What are the magnitude and direction of \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}} ?\) (b) What are the magnitude and direction of \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}} ?\) (c) What are the magnitude and direction of \(\overrightarrow{\mathbf{B}}-\overrightarrow{\mathbf{A}} ?\)
Two cars are driving toward each other on a straight and level road in Alaska. The BMW is traveling at \(100.0 \mathrm{km} / \mathrm{h}\) north and the VW is traveling at \(42 \mathrm{km} / \mathrm{h}\) south, both velocities measured relative to the road. At a certain instant, the distance between the cars is \(10.0 \mathrm{km} .\) Approximately how long will it take from that instant for the two cars to meet? [Hint: Consider a reference frame in which one of the cars is at rest. \(]\)
A baseball is thrown horizontally from a height of \(9.60 \mathrm{m}\) above the ground with a speed of \(30.0 \mathrm{m} / \mathrm{s}\) Where is the ball after 1.40 s has elapsed?
A helicopter is flying horizontally at \(8.0 \mathrm{m} / \mathrm{s}\) and an altitude of \(18 \mathrm{m}\) when a package of emergency medical supplies is ejected horizontally backward with a speed of \(12 \mathrm{m} / \mathrm{s}\) relative to the helicopter. Ignoring air resistance, what is the horizontal distance between the package and the helicopter when the package hits the ground?
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