Vectors and Kinematics: Study Notes for College Physics

Math Review and Unit Conversions

Unit Conversions

Unit conversions are essential in physics to ensure all quantities are in compatible units before performing calculations.

• Speed Conversion: To convert from km/min to m/s, use the relationships: 1 km = 1000 m, 1 min = 60 s.

• Time Conversion: To convert days to seconds, use: 1 day = 24 hours, 1 hour = 3600 seconds.

Example:

• Convert 94 km/min to m/s: $94 \frac{\text{km}}{\text{min}} \times \frac{1000\ \text{m}}{1\ \text{km}} \times \frac{1\ \text{min}}{60\ \text{s}} = 1566.67\ \text{m/s}$

• Convert 5 days to seconds: $5\ \text{days} \times 24\ \text{hr/day} \times 3600\ \text{s/hr} = 432,000\ \text{s}$

Vectors

Vector Representation and Operations

Vectors are quantities that have both magnitude and direction. They are often represented in component form as \( \vec{A} = (A_x, A_y) \).

• Vector Addition: Add corresponding components: $\vec{C} = \vec{A} + \vec{B} = (A_x + B_x, A_y + B_y)$

• Magnitude of a Vector: $|\vec{C}| = \sqrt{C_x^2 + C_y^2}$

• Direction (Angle) of a Vector: $\theta = \arctan\left(\frac{C_y}{C_x}\right)$

Example: If \( \vec{A} = (6, 4) \) and \( \vec{B} = (6, -2) \), then:

• \( \vec{A} + \vec{B} = (12, 2) \)

• Magnitude: $\sqrt{12^2 + 2^2} = \sqrt{148} \approx 12.17$

• Direction: $\arctan\left(\frac{2}{12}\right) \approx 9.46^\circ$ above the x-axis

Vector Components

• Given a vector \( \vec{A} = (A_x, A_y) \), the x-component is \( A_x \), and the y-component is \( A_y \).

• For a vector sum or difference, apply the operation to each component.

Example: If \( \vec{A} = (5.6, 15.7) \), \( \vec{B} = (14.3, -18.4) \), and \( \vec{A} - \vec{B} + 5.5\vec{C} = 0 \), solve for components of \( \vec{C} \).

Displacement

• Displacement is a vector quantity representing the change in position.

• For multiple movements along a straight line, sum the signed distances (east = positive, west = negative, etc.).

Example: 40 m east, 25 m west, 10 m east: Net displacement = $40 - 25 + 10 = 25$ m east.

1D Motion / Kinematics

Uniform Acceleration

When an object moves with constant acceleration, the following kinematic equations apply:

• $v = v_0 + at$

• $x = x_0 + v_0 t + \frac{1}{2} a t^2$

• $v^2 = v_0^2 + 2a(x - x_0)$

Definitions:

• Acceleration (a): The rate of change of velocity, in m/s2.

• Average velocity: $\bar{v} = \frac{\Delta x}{\Delta t}$

Example: A motorcycle decelerates from 90 km/h to rest in 5 s. Find acceleration:

• Convert 90 km/h to m/s: $90 \times \frac{1000}{3600} = 25$ m/s

• $a = \frac{v - v_0}{t} = \frac{0 - 25}{5} = -5$ m/s2

2D Kinematics and Projectile Motion

Projectile Motion

Projectile motion involves two-dimensional motion under constant acceleration due to gravity (usually $g = 9.8$ m/s2 downward).

• Horizontal Component: $v_{0x} = v_0 \cos \theta$

• Vertical Component: $v_{0y} = v_0 \sin \theta$

• Time of Flight: $t = \frac{2 v_{0y}}{g}$ (if launched and lands at same height)

• Maximum Height: $h_{max} = \frac{v_{0y}^2}{2g}$

• Range: $R = v_{0x} \times t_{total}$

Example: A ball is kicked at 18 m/s at 42° above the horizontal.

• Horizontal: $v_{0x} = 18 \cos 42^\circ$

• Vertical: $v_{0y} = 18 \sin 42^\circ$

Solving for Maximum Height and Time

• At maximum height, vertical velocity is zero.

• Use $v_y = v_{0y} - gt$ to solve for time to max height.

• Total time in air is twice the time to max height (if landing at same vertical level).

Advanced Vector Operations

Linear Combinations of Vectors

Vectors can be scaled and combined:

• For $\vec{D} = 2\vec{A} - \vec{B}$, calculate each component: $D_x = 2A_x - B_x$, $D_y = 2A_y - B_y$.

• Magnitude: $|\vec{D}| = \sqrt{D_x^2 + D_y^2}$

• Angle: $\theta = \arctan\left(\frac{D_y}{D_x}\right)$

Example Table:

Operation	x-component	y-component
$2\vec{A} - \vec{B}$	$2A_x - B_x$	$2A_y - B_y$
$3\vec{A} - \vec{B}$	$3A_x - B_x$	$3A_y - B_y$

Summary Table: Key Kinematic Equations

Equation	Description
$v = v_0 + at$	Final velocity after time t
$x = x_0 + v_0 t + \frac{1}{2} a t^2$	Position after time t
$v^2 = v_0^2 + 2a(x - x_0)$	Relates velocity and displacement
$\bar{v} = \frac{\Delta x}{\Delta t}$	Average velocity

Applications and Problem-Solving Tips

• Always draw a diagram and label vectors and directions.

• Break vectors into components before adding or subtracting.

• Check units for consistency throughout calculations.

• For projectile problems, treat horizontal and vertical motions separately.

Additional info: These notes cover the main concepts and problem types found in introductory college physics on vectors, 1D and 2D kinematics, and projectile motion, as reflected in the provided questions.