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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/s² to m/s²: \(D_y = 6.5\, \text{cm}/\text{s}^2 * \frac{1 \text{m}}{100 \text{cm}} = 0.065 \text{m} / \text{s}^2\).
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

An African swallow carrying a very small coconut is flying horizontally with a speed of \(18 \mathrm{m} / \mathrm{s}\). (a) If it drops the coconut from a height of \(100 \mathrm{m}\) above the Earth, how long will it take before the coconut strikes the ground? (b) At what horizontal distance from the release point will the coconut strike the ground?
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Vector \(\overrightarrow{\mathbf{A}}\) is directed along the positive \(y\) -axis and has magnitude \(\sqrt{3.0}\) units. Vector \(\overrightarrow{\mathbf{B}}\) is directed along the negative \(x\) -axis and has magnitude 1.0 unit. (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 \(x\) - and \(y\) -components of \(\overrightarrow{\mathbf{B}}-\overrightarrow{\mathbf{A}} ?\)
A boat that can travel at \(4.0 \mathrm{km} / \mathrm{h}\) in still water crosses a river with a current of \(1.8 \mathrm{km} / \mathrm{h}\). At what angle must the boat be pointed upstream to travel straight across the river? In other words, in what direction is the velocity of the boat relative to the water?
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