Problem 34 - University Physics with Modern Physics

Question

Find the magnitude and direction of the vector represented by the following pairs of components: (a) A\(x\) = \(-\)8.60 cm, A\(y\) = 5.20 cm; (b) A\(x\) = \(-\)9.70 m, A\(y\) = -2.45 m; (c) A\(x\) = 7.75 km, A\(y\) = \(-\)2.70 km.

Step-by-Step Solution

Answer

(a) Magnitude: 10.05 cm, Direction: 148.8°; (b) Magnitude: 10.00 m, Direction: 194.1°; (c) Magnitude: 8.21 km, Direction: 340.8°.

Step 1: Calculate Magnitude of Vector (a)

The magnitude of a vector with components \(A_x\) and \(A_y\) is given by the formula: \(|\vec{A}| = \sqrt{A_x^2 + A_y^2}\). Substituting the given values for part (a), \(A_x = -8.60\) cm and \(A_y = 5.20\) cm, we compute:\[|\vec{A}| = \sqrt{(-8.60)^2 + (5.20)^2} = \sqrt{73.96 + 27.04} = \sqrt{101.00} = 10.05\text{ cm}\].

Step 2: Calculate Direction of Vector (a)

The direction (angle \(\theta\)) can be found using the tangent formula: \( \theta = \tan^{-1}\left(\frac{A_y}{A_x}\right) \). For part (a), \(A_x = -8.60\) cm and \(A_y = 5.20\) cm:\[\theta = \tan^{-1}\left(\frac{5.20}{-8.60}\right) \approx \tan^{-1}(-0.605) \approx -31.2^\circ\].Since the vector is in the second quadrant, \(\theta = 180^\circ - 31.2^\circ = 148.8^\circ\).

Step 3: Calculate Magnitude of Vector (b)

Similarly, for vector components given as \(A_x=-9.70\) m and \(A_y=-2.45\) m, the magnitude is:\[|\vec{A}| = \sqrt{(-9.70)^2 + (-2.45)^2} = \sqrt{94.09 + 6.0025} = \sqrt{100.0925} \approx 10.00\text{ m}\].

Step 4: Calculate Direction of Vector (b)

Using \(\theta = \tan^{-1}\left(\frac{A_y}{A_x}\right)\), substitute the given values for part (b):\[\theta = \tan^{-1}\left(\frac{-2.45}{-9.70}\right) \approx \tan^{-1}(0.252) \approx 14.1^\circ\].Since the vector is in the third quadrant, \(\theta = 180^\circ + 14.1^\circ = 194.1^\circ\).

Step 5: Calculate Magnitude of Vector (c)

For part (c), the vector components are \(A_x = 7.75\) km and \(A_y = -2.70\) km, so the magnitude is:\[|\vec{A}| = \sqrt{(7.75)^2 + (-2.70)^2} = \sqrt{60.0625 + 7.29} = \sqrt{67.3525} \approx 8.21\text{ km}\].

Step 6: Calculate Direction of Vector (c)

To find the direction using the tangent formula for part (c):\[\theta = \tan^{-1}\left(\frac{-2.70}{7.75}\right) \approx \tan^{-1}(-0.348) \approx -19.2^\circ\]. Since the vector lies in the fourth quadrant, the equivalent positive angle is \(360^\circ - 19.2^\circ = 340.8^\circ\).

Key Concepts

Magnitude of Vectors

The magnitude of a vector is a measure of its length, or how far it stretches from its starting point to its endpoint in space. This is often simply called the "size" of the vector and is denoted by \(|\vec{A}|\), where \(|\vec{A}|\) is the length of the vector \(\vec{A}\). To compute the magnitude, we use the Pythagorean theorem, which involves squaring the horizontal component \(A_x\), squaring the vertical component \(A_y\), adding those values, and taking the square root. For example, if \(A_x = -8.60\) cm and \(A_y = 5.20\) cm, the magnitude is calculated as: \[|\vec{A}| = \sqrt{(-8.60)^2 + (5.20)^2} = 10.05 \text{ cm}\].
The formula \[|\vec{A}| = \sqrt{A_x^2 + A_y^2}\] is universally applicable regardless of the physical dimensions of the problem, whether in centimeters, meters, or kilometers. It gives a single value that represents the vector's strength or speed, which is crucial in many physics problems.

Direction of Vectors

The direction of a vector tells you where it's pointing in space and can significantly affect how it's interpreted in physics or engineering problems. It's commonly expressed as an angle from a reference direction, often the positive x-axis. To determine the angle, we use the tangent function: \[ \theta = \tan^{-1}\left(\frac{A_y}{A_x}\right)\].
One needs to be careful with angles because vectors can lie in different quadrants of a coordinate system. For a vector with components \(A_x = -8.60\) cm and \(A_y = 5.20\) cm, the initially calculated angle is \(-31.2^\circ\). However, because the vector is actually in the second quadrant, the correct direction is \(148.8^\circ\), accounting for full 180 degrees rotation. Knowledge of the vector's direction is crucial. It allows you to fully understand the vector's orientation, which is just as important as understanding its magnitude.

Components of Vectors

Vectors are not just mere lines; they consist of parts or components. These components, often called \(A_x\) and \(A_y\), are the projections of the vector along the horizontal and vertical axes. Think of the components as shadows of the original vector cast along these axes.
The component along the x-axis, \(A_x\), tells us how much of the vector points in the horizontal direction, and the component along the y-axis, \(A_y\), tells us how much it points vertically. For instance, if a vector has components \(A_x = -9.70\) meters and \(A_y = -2.45\) meters, these literally mean the vector stretches leftward (since \(A_x\) is negative) and downward (since \(A_y\) is negative). Understanding the components is crucial because they allow for the analysis of the vector in terms of familiar and manageable x and y axes, making problems easier to tackle especially in physics, where forces and motions are often resolved along these directions.