Unit 1: Linear Motion – Study Notes for Physics with Calculus

Vectors & Motion Basics

Scalar vs. Vector Quantities

In physics, quantities are classified as either scalars or vectors based on whether they possess direction in addition to magnitude.

• Scalar: A quantity described only by magnitude (size or amount). Examples: mass, temperature, distance, speed.

• Vector: A quantity described by both magnitude and direction. Examples: displacement, velocity, acceleration, force.

• Notation: Vectors are often denoted with an arrow above the symbol (e.g., $\vec{v}$) or in boldface.

Distance vs. Displacement

Distance and displacement both describe how far an object moves, but with key differences:

• Distance: The total length of the path traveled, regardless of direction (scalar).

• Displacement: The straight-line change in position from initial to final point, including direction (vector).

• Example: Walking 3 m east, then 4 m west: distance = 7 m, displacement = 1 m west.

Speed vs. Velocity

Both speed and velocity measure how fast an object moves, but velocity includes direction.

• Speed: The rate at which distance is covered (scalar). $\text{Speed} = \frac{\text{distance}}{\text{time}}$

• Velocity: The rate at which displacement changes (vector). $\vec{v} = \frac{\Delta \vec{x}}{\Delta t}$

• Example: If you run around a track and end where you started, your average speed is nonzero, but your average velocity is zero.

Instantaneous vs. Average Quantities

Quantities like velocity and acceleration can be measured at a specific instant or averaged over a time interval.

• Average velocity: $\vec{v}_{\text{avg}} = \frac{\Delta \vec{x}}{\Delta t}$

• Instantaneous velocity: $\vec{v} = \lim_{\Delta t \to 0} \frac{\Delta \vec{x}}{\Delta t} = \frac{d\vec{x}}{dt}$

• Average acceleration: $\vec{a}_{\text{avg}} = \frac{\Delta \vec{v}}{\Delta t}$

• Instantaneous acceleration: $\vec{a} = \frac{d\vec{v}}{dt}$

Relative Motion

Relative motion describes how the velocity of an object appears different to observers in different reference frames.

• Relative velocity equation: $\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B$

• Example: If a train moves east at 20 m/s and a passenger walks west at 2 m/s relative to the train, their velocity relative to the ground is 18 m/s east.

Graphing Motion

Position vs. Time Graphs

These graphs show how an object's position changes over time.

• Slope: The slope at any point gives the object's velocity at that instant.

• Straight line: Constant velocity.

• Curved line: Changing velocity (acceleration).

Velocity vs. Time Graphs

These graphs display how velocity changes with time.

• Slope: The slope gives the object's acceleration.

• Area under curve: Represents displacement.

• Horizontal line: Constant velocity.

Acceleration vs. Time Graphs

These graphs show how acceleration varies with time.

• Area under curve: Represents change in velocity.

• Horizontal line: Constant acceleration.

Interpreting Slopes, Intercepts, and Areas

• Slope of position-time: Velocity

• Slope of velocity-time: Acceleration

• Area under velocity-time: Displacement

• Area under acceleration-time: Change in velocity

Calculations & Kinematics

Uniform (Constant) Acceleration

When acceleration is constant, motion can be analyzed using the kinematic equations.

• Acceleration: $a = \frac{\Delta v}{\Delta t}$

• Example: A car accelerates from rest at 2 m/s2 for 5 s. Final velocity: $v = at = (2)(5) = 10$ m/s.

Applying the Kinematic Equations

The kinematic equations relate displacement, velocity, acceleration, and time for constant acceleration:

• $v = v_0 + at$

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

• $v^2 = v_0^2 + 2a(x - x_0)$

• $x = x_0 + \frac{1}{2}(v_0 + v)t$

Where: $x$ = position, $v$ = velocity, $a$ = acceleration, $t$ = time, subscript 0 = initial value.

The G.U.E.S.S. Method for Problem Solving

To solve kinematics problems, use the following structured approach:

• G: Given – List all known values.

• U: Unknown – Identify what you are solving for.

• E: Equation – Write the relevant equation(s).

• S: Substitute – Plug in the known values.

• S: Solve – Complete the calculation and box your answer.

Example: A ball is dropped from rest from a height of 20 m. How long does it take to hit the ground? (Assume $g = 9.8$ m/s2)

• G: $x_0 = 0$, $x = 20$ m, $v_0 = 0$, $a = 9.8$ m/s2

• U: $t$

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

• S: $20 = 0 + 0 + \frac{1}{2}(9.8)t^2$

• S: $t^2 = \frac{40}{9.8} \Rightarrow t \approx 2.02$ s

Additional info: Mastery of these concepts is foundational for all subsequent topics in mechanics, including dynamics and energy.