/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 In each of these, the \(x\) - an... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 21

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}$ $B_{y}=-60.0 \mathrm{m} .(\mathrm{c}) C_{x}=-13.7 \mathrm{m} / \mathrm{s}, C_{y}=-8.8 \mathrm{m} / \mathrm{s} .(\mathrm{d}) D_{x}=$ \(2.3 \mathrm{m} / \mathrm{s}^{2}, D_{y}=6.5 \mathrm{cm} / \mathrm{s}^{2}\)

Short Answer

Expert verified
Short answer: For the given cases, we found the following magnitudes and directions for each vector: (a) Magnitude: \(9.43\, \text{m}/\text{s}\), Direction: \(122.0^\circ\) (b) Magnitude: \(134.16\, \text{m}\), Direction: \(333.4^\circ\) (c) Magnitude: \(16.28\, \text{m}/\text{s}\), Direction: \(212.6^\circ\) (d) Magnitude: \(2.30\, \text{m}/\text{s}^2\), Direction: \(1.6^\circ\)

Step by step solution

01

Magnitude calculation

Calculate the magnitude using the Pythagorean theorem: \(A=\sqrt{A_{x}^2 + A_{y}^2} = \sqrt{(-5)^2 + 8^2} = \sqrt{25 + 64} = \sqrt{89} \approx 9.43 \mathrm{m}/\mathrm{s}\)
02

Direction calculation

Calculate the angle using the arctangent function: \(\theta = \arctan{\frac{A_y}{A_x}} = \arctan{\frac{8}{-5}} \approx -58.0^\circ\). Since the angle is in the second quadrant, add \(180^\circ\): \(\theta = -58.0^\circ + 180^\circ = 122.0^\circ\). Case (b) \(B_{x}=+120 \mathrm{m}, B_{y}=-60.0 \mathrm{m} .\)
03

Magnitude calculation

Calculate the magnitude using the Pythagorean theorem: \(B=\sqrt{B_{x}^2 + B_{y}^2} = \sqrt{120^2 + (-60)^2} = \sqrt{14400 + 3600} = \sqrt{18000} = 134.16\mathrm{m}\)
04

Direction calculation

Calculate the angle using the arctangent function: \(\theta = \arctan{\frac{B_y}{B_x}} = \arctan{\frac{-60}{120}} \approx -26.6^\circ\). Since the angle is in the fourth quadrant, add \(360^\circ\): \(\theta = -26.6^\circ + 360^\circ = 333.4^\circ\) Case (c) \(C_{x}=-13.7 \mathrm{m} / \mathrm{s}, C_{y}=-8.8 \mathrm{m} / \mathrm{s}\)
05

Magnitude calculation

Calculate the magnitude using the Pythagorean theorem: \(C=\sqrt{C_{x}^2 + C_{y}^2} = \sqrt{(-13.7)^2 + (-8.8)^2} = \sqrt{187.69 + 77.44} = \sqrt{265.13} \approx 16.28\mathrm{m}/\mathrm{s}\).
06

Direction calculation

Calculate the angle using the arctangent function: \(\theta = \arctan{\frac{C_y}{C_x}} = \arctan{\frac{-8.8}{-13.7}} \approx 32.6^\circ\). Since the angle is in the third quadrant, add \(180^\circ\): \(\theta = 32.6^\circ + 180^\circ = 212.6^\circ\) Case (d) \(D_{x}=2.3 \mathrm{m} / \mathrm{s}^{2}, D_{y}=6.5 \mathrm{cm} / \mathrm{s}^{2}\) First, we need to have the same units. Convert \(D_y\) from cm/s虏 to m/s虏: \(D_y = 6.5\, \text{cm}/\text{s}^2 * \frac{1 \text{m}}{100 \text{cm}} = 0.065 \text{m} / \text{s}^2\).
07

Magnitude calculation

Calculate the magnitude using the Pythagorean theorem: \(D=\sqrt{D_{x}^2 + D_{y}^2} = \sqrt{(2.3)^2 + (0.065)^2} = \sqrt{5.29 + 0.004225} = \sqrt{5.294225} \approx 2.30 \mathrm{m}/\mathrm{s^2}\)
08

Direction calculation

Calculate the angle using the arctangent function: \(\theta = \arctan{\frac{D_y}{D_x}} = \arctan{\frac{0.065}{2.3}} \approx 1.6^\circ\). Since the angle is in the first quadrant, it remains the same.

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Most popular questions from this chapter

Geoffrey drives from his home town due east at \(90.0 \mathrm{km} / \mathrm{h}\) for 80.0 min. After visiting a friend for 15.0 min, he drives in a direction \(30.0^{\circ}\) south of west at \(76.0 \mathrm{km} / \mathrm{h}\) for 45.0 min to visit another friend. (a) How far is it to his home from the second town? (b) If it takes him 45.0 min to drive directly home, what is his average velocity on the third leg of the trip? (c) What is his average velocity during the first two legs of his trip? (d) What is his average velocity over the entire trip? (e) What is his average speed during the entire trip if he spent 55.0 min visiting the second friend?
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A particle has a constant acceleration of \(5.0 \mathrm{m} / \mathrm{s}^{2}\) to the east. At time \(t=0,\) it is \(2.0 \mathrm{m}\) east of the origin and its velocity is \(20 \mathrm{m} / \mathrm{s}\) north. What are the components of its position vector at \(t=2.0 \mathrm{s} ?\)
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