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Chapter 2: Problem 4

(a) What are the components of a vector \(\overrightarrow{\mathbf{a}}\) in the \(x y\) plane if its direction is \(252^{\circ}\) counterclockwise from the positive \(x\) axis and its magnitude is \(7.34\) units? \((b)\) The \(x\) component of a certain vector is \(-25\) units and the \(y\) component is \(+43\) units. What are the magnitude of the vector and the angle between its direction and the positive \(x\) axis?

Short Answer

Expert verified
The components of the first vector are approximately \(a_x = -3.67\) and \(a_y = -6.67\). The magnitude of the second vector is approximately 49.39 units and its direction/angle from the positive x-axis is approximately 211 degrees (or 3.68 radians) in the counterclockwise direction.

Step by step solution

01

Find vector components

The components of a vector can be calculated using the formula \(a_x = r\cos(\theta)\) for the x-component and \(a_y = r\sin(\theta)\) for the y-component. Here, \(r\) is the magnitude (7.34 units) and \(\Theta\) is the angle from the positive x-axis (252 degrees). However, the angle should be converted to radians first: \(252^{\circ} = 252\times\frac{\pi}{180} = 4.4\) rad. Hence, the components are \[a_x = 7.34\cos(4.4)\] \[a_y = 7.34\sin(4.4)\]
02

Calculate the magnitude and direction of vector

The magnitude of a vector is calculated as \(r = \sqrt{x^2 + y^2}\), where \(x\) and \(y\) are the x, y components of the vector respectively. Substituting the given values: \[ r = \sqrt{(-25)^2 + (+43)^2}\] Now, the direction or angle \(\Theta\) from positive x-axis can be calculated using the formula \(\Theta = \tan^{-1}(\frac{y}{x})\). Hence substituting the given values, the direction is \(\Theta = \tan^{-1}(\frac{43}{-25})\). Please note because x is negative we need to add \(\pi\) (or 180 degrees) to the arctan result. This is necessary due to the properties of arctan function.

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Most popular questions from this chapter

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