Chapter 1: Physical Quantities and Vectors

What is Physics?

Physics is the study of the basic components of the universe and their interactions. Theories in physics must be verified by experimental measurements.

Physical Quantities

A physical quantity is something that can be measured and consists of a magnitude and a unit (e.g., 4.5 m, 70 km/h).

• Base Quantities: Basic building blocks (e.g., length, mass, time).
• Derived Quantities: Built from base quantities (e.g., area, volume, speed).

SI Units (International System of Units)

Dimension denotes the physical nature of a quantity (Length = L, Mass = M, Time = T). We use brackets \([ ]\) to denote the dimensions of a physical quantity.

Dimensional analysis can be used to check the homogeneity of an equation or to derive a relationship between physical quantities.

Example: Analysis of an Equation

Show that \(v = v_0 + at\) is dimensionally correct.

\[ [v] = [v_0] = \frac{L}{T} \]

\[ [at] = [a][t] = \frac{L}{T^2} \cdot T = \frac{L}{T} \]

Since all terms have the same dimension, the equation is dimensionally correct.

Vector Quantity: Specified by both a magnitude and a direction (e.g., velocity, force).

Scalar Quantity: Specified by a magnitude only (e.g., speed, mass, temperature).

Components of a Vector

A vector can be broken down into its x and y components using trigonometry.

\[ A_x = A \cos{\theta} \]

\[ A_y = A \sin{\theta} \]

The magnitude and direction of the resultant vector \(R\) from its components are:

\[ R = \sqrt{R_x^2 + R_y^2} \]

\[ \theta = \tan^{-1} \left( \frac{R_y}{R_x} \right) \]

Example: Take a Hike

A hiker walks 25.0 km at \(45.0^\circ\) south of east, then 40.0 km at \(60.0^\circ\) north of east. Find the total displacement.

Displacement A:
\( A_x = 25.0 \cos(-45.0^\circ) = 17.7 \) km
\( A_y = 25.0 \sin(-45.0^\circ) = -17.7 \) km

Displacement B:
\( B_x = 40.0 \cos(60.0^\circ) = 20.0 \) km
\( B_y = 40.0 \sin(60.0^\circ) = 34.6 \) km

Total Displacement R:
\( R_x = 17.7 + 20.0 = 37.7 \) km
\( R_y = -17.7 + 34.6 = 16.9 \) km

Magnitude and Direction:
\( R = \sqrt{37.7^2 + 16.9^2} = 41.3 \) km
\( \theta = \tan^{-1}(16.9 / 37.7) = 24.1^\circ \) north of east.

Unit conversions involve multiplying by a conversion factor, which is a fraction equal to 1.

Common Conversions:

• 1 mi = 1609 m
• 1 ft = 0.3048 m
• 1 in = 2.54 cm (exactly)

Example: Unit Conversion

Convert 15.0 inches to centimeters.

\[ 15.0 \text{ in} = (15.0 \text{ in}) \times \left( \frac{2.54 \text{ cm}}{1 \text{ in}} \right) = 38.1 \text{ cm} \]