Graphical Analysis of Motion - Page-by-Page Notes

Topic: Graphical Analysis of Motion.
Key idea: Use graphs to study motion, typically with perpendicular axes (horizontal x-axis and vertical y-axis) to represent relationships between quantities like displacement, velocity, and time.

Un-scrambled key terms and their definitions:
- 1. Velocity: tells how fast or slow a moving object is.
- 2. Displacement: the shortest distance between two points.
- 3. Acceleration: the rate of change of an object's velocity.
- 4. Graph: a pictorial representation of numbers.
- 5. Cartesian coordinate system: also known as the x–y plane and is used for plotting.
Additional related ideas (based on the scrambled words visible):
- Points: locations in the coordinate plane used to specify positions.
- Coordinate system/Cartesian coordinates system: framework for plotting points with pairs (x, y).

General concept: A graph, like a picture, is worth a thousand words.
Slopes and general relationships:
- Graphs in this text have perpendicular axes (one horizontal, one vertical).
- Slopes encode rates of change (e.g., velocity, acceleration) depending on the plotted variables (x vs t, v vs t, etc.).

Slope-intercept form of a line:
- The equation of a straight line is:
  $y = mx + b$
- Here, $m$ is the slope (rise over run) and $b$ is the y-intercept (the value of $y$ when $x=0$).
Key terms:
- Intercept: the point where the graph crosses an axis; for $y = mx + b$, the y-intercept is $b$.
- Rise: change in $y$ (Δy).
- Run: change in $x$ (Δx).
Relationship:
- Slope $m$ can be computed as
 $m = \frac{Δy}{Δx} = \frac{y2 - y1}{x2 - x1}.$
- The line’s equation can also be written as $y = m x + b$ with $b = y - m x$ for any point $(x,y)$ on the line.

Examine the displacement graph for Jet Car (Displacement, $x$, in meters) versus time ($t$ in seconds).
Observations:
- The displacement increases from $x=0$ to a higher value (the graph shows up to about 3500 m) over a time interval up to about 40 s.
- The line’s slope represents velocity: $v = \frac{Δx}{Δt}$.
- The graph is used to extract velocities at two points (denoted by points on the line, e.g., P and Q) which are then used to obtain instantaneous velocities via tangents in the next graph.
Conceptual takeaway: The steeper the line (greater slope), the greater the velocity; a straight-line x–t graph indicates constant velocity over that interval.

Statement (a): The slope of an $x$ vs. $t$ graph is velocity.
How this is used:
- This is shown at two points on the graph; the instantaneous velocities at those points are plotted in the next graph.
- Fact: Instantaneous velocity at any point equals the slope of the tangent to the $x$ vs. $t$ curve at that point.
Mathematical reminder:
- If the $x$–$t$ relation is differentiable, instantaneous velocity is $v = \frac{dx}{dt}$, which for a straight-line segment reduces to the constant slope value.

Examine the velocity graph (Velocity, $v$ in m/s vs Time, $t$ in s).
Key observation:
- The slope of the $v$ vs. $t$ graph is constant over this portion of motion, indicating constant acceleration.
What the constant slope means:
- If $v(t)$ is linear, then $a = \frac{dv}{dt}$ is constant.
- The graph shows velocity changing at a steady rate with time, i.e., uniform acceleration.

Reiteration (b): The slope of the $v$ vs. $t$ graph is constant for this part of the motion, indicating constant acceleration.
Implication:
- A straight line in a velocity-time graph corresponds to a constant acceleration value $a$ over the plotted interval.

Examine the acceleration graph for the Jet Car: Acceleration, $a$ (in m/s²) vs Time, $t$.
Key point (c): The acceleration has the constant value of $5.0\ \mathrm{m/s^2}$ over the time interval plotted.
Practical takeaway:
- For the duration shown, the jet car experiences a steady push that increases its velocity at a constant rate.

Explicit statement: The acceleration is constant with value $a = 5.0\ \mathrm{m/s^2}$ over the plotted time interval.
Consequence:
- With constant acceleration, the kinematic equations can be applied directly to relate $x$, $v$, $a$, and $t$ over that interval.

The four kinematic equations that describe an object's motion:
The Kinematic Equations (overview):
- These equations relate displacement, velocity, acceleration, and time for motion with constant acceleration.

The four key equations (with definitions):
- Constants and variables:
- $d$ = distance (displacement)
- $t$ = time
- $a$ = acceleration
- $v_i$ = initial velocity
- $v_f$ = final velocity
- Equations (one form each):
 $vf = vi + a t$
 $d = vi t + frac{1}{2} a t^2$ $vf^2 = vi^2 + 2 a d$ $d = frac{(vi + v_f)}{2} \, t$
Notes:
- Any one of these can be used depending on which quantities are known and which are unknown.
- All four assume constant acceleration over the interval considered.

(No new content is provided in the transcript for this page.)
Summary reminder: The pages collectively introduce graph-based analysis of motion, connect slopes to physical quantities (velocity and acceleration), and present the fundamental kinematic framework for uniformly accelerated motion.

Key formulas to remember (LaTeX):

Slope of a line: $m = \frac{Δy}{Δx} = \frac{y2 - y1}{x2 - x1}$
Line equation: $y = m x + b$
Slope-intercept components:
- Intercept: $b = y - m x$
- Slope: $m = \frac{Δy}{Δx}$
Displacement as a function of time with constant acceleration: $d = v_i t + \frac{1}{2} a t^2$
Final velocity with constant acceleration: $vf = vi + a t$
Final velocity from displacement: $vf^2 = vi^2 + 2 a d$
Average velocity over a time interval with constant acceleration (distance-time form): $d = \frac{vi + vf}{2} \, t$
Velocity as a function of time (derivative perspective): $v = \frac{dx}{dt}$
Acceleration as a function of time (derivative perspective): $a = \frac{dv}{dt}$
Notes on graphs:
- The slope of an $x$ vs. $t$ graph equals velocity: $v = \frac{dx}{dt}$
- The slope of a $v$ vs. $t$ graph equals acceleration: $a = \frac{dv}{dt}$