Describing Motion: Kinematics in One Dimension – Page-by-Page Notes

Page 1

Key concepts

A car speedometer measures speed only; it does not provide information about direction, so it does not measure velocity.
If the velocity of an object is constant, both the speed and the direction of travel are constant. In that case, the average velocity equals the instantaneous velocity because the velocity vector is not changing.
There is no general relationship between the magnitude of speed and the magnitude of acceleration. An object can have a large constant speed but zero acceleration, or a small speed with positive acceleration (speed increasing). For freely falling bodies without air resistance, equal drops from rest at different times have the same acceleration (≈ $g \approx 9.80\ \mathrm{m/s^2}$) but different speeds.
If two objects travel in a straight line, they can have the same acceleration even if their speeds differ in magnitude, provided the change in velocity over the same time interval is the same.
A velocity can be positive while the acceleration is negative (e.g., moving north but slowing down, which corresponds to a negative acceleration in a north-positive coordinate system).
Velocity can be negative even when acceleration is positive, depending on the chosen positive direction. If the object travels left (negative velocity) but is slowing down (positive acceleration) in a rightward-positive frame, this situation can occur.

Basic definitions (notation)

Velocity: \mathbf{v} = \frac{\Delta \mathbf{x}}{\Delta t}
Acceleration: \mathbf{a} = \frac{\Delta \mathbf{v}}{\Delta t}

Note: In the same page, the text emphasizes that velocity (with direction) and speed (magnitude only) are different concepts, and that acceleration relates to changes in velocity, not just changes in speed.

Page 2

Positive direction conventions and sign examples

If north is defined as positive, an object traveling south and increasing speed has both negative velocity and negative acceleration.
If up is defined as positive, an object falling due to gravity has both negative velocity and negative acceleration.

Acceleration can decrease while velocity increases

Acceleration is the rate of change of velocity. It is possible for velocity to be increasing while its rate of change (acceleration) is decreasing.
Examples: a car speeds up from 40 to 50 in 1 s, then to 55 in the next second. The speed still increases, but the acceleration in the second second is smaller. With air resistance (downward positive), a falling object speeds up while its acceleration decreases as drag increases; terminal velocity occurs when acceleration approaches zero.

Relative motion and passing/catching up

If two cars start side by side with different speeds and accelerations, the faster one may pass, then the other may catch up later depending on the accelerations and speeds.

Special cases and averages

If air resistance is negligible, a freely falling object experiences a constant acceleration of $g \approx 9.80\ \mathrm{m/s^2}$, and its speed changes linearly with time. The time for speed to change from some positive value to another value is the same as the time for the reverse change, in the absence of drag.
When air resistance is present, acceleration decreases with speed, and the velocity approaches a terminal value where acceleration tends to zero.
Average speed notion (as opposed to average velocity) is the total distance traveled divided by total time. If a journey comprises segments at different speeds, the average speed is not simply the arithmetic mean of the speeds but depends on the times spent at each speed.

Examples and calculations

For a trip with segments at 70 km/h and 90 km/h over equal distances, the average speed evaluates to:
\bar{v} = \frac{2x}{\frac{x}{70}+\frac{x}{90}} = \frac{2}{\frac{1}{70}+\frac{1}{90}} \text{ km/h} = \approx 78.75 \text{ km/h} \approx 79 \text{ km/h}.
A rock thrown straight up has a constant downward acceleration (gravity) but its velocity passes through zero at the top.
A car traveling at constant velocity has zero acceleration.
A rock falling from a cliff (neglect air resistance) has constant acceleration; the distance fallen and time can be found from the kinematic equations.

Page 3

Describing Motion: Kinematics in One Dimension (conceptual)

The slope of a position–time graph gives velocity; a straight-line segment indicates constant velocity, while curvature indicates acceleration.
An object can have positive velocity with positive acceleration, velocity can be changing in direction, and the acceleration can be negative or positive depending on the frame of reference.

Misconceptions addressed

1) Physics problem-solving steps: including all appropriate methods and tools is necessary for solving problems (no single “right” method).
2) Positive acceleration does not always increase speed; if velocity and acceleration point in opposite directions, speed decreases.
3) If velocity and acceleration are opposite, the object slows to a stop; if acceleration remains constant, motion will reverse direction and continue.
4) In free fall, gravity provides a constant downward acceleration; velocity is changing through zero at the top of a trajectory.
5) Two rocks both accelerate under gravity, so their velocities increase at the same rate (neglecting air resistance).
6) When averaging speed for symmetric trips over equal distances but different speeds, more time is spent at the slower speed, so the average tends toward the slower speed.
7) In free-fall with no air resistance, objects dropped from rest have the same acceleration magnitude/direction regardless of their speeds.
8) When analyzing equations, signs must be carefully assigned; some problem statements show signs that need correcting (e.g., using –g for gravity, etc.).
9) Interpreting velocity–time and position–time graphs requires attention to slope and curvature; instantaneous velocity corresponds to slope of the x–t graph, while average velocity corresponds to a line's slope between two points.
10) The velocity–time graph slope corresponds to acceleration; the position–time graph slope corresponds to velocity.

Page 4

Problem-solving patterns (selected examples from the Solutions to Problems)

Problem 1: Distance traveled (displacement) can be found by rearranging the average-velocity relation. If given a time and velocity, displacement is x = v \Delta t. The example shows converting units carefully (e.g., seconds vs hours):
x = υ \cdot t.
Problem 2: The average speed equals total distance divided by total time; the setup shows the relation \bar{v} = \frac{d}{t}.
Problem 3: The average velocity is given by v_{\text{avg}} = \frac{\Delta x}{\Delta t}.
Problem 4: If a problem provides displacements and times, the average speed can be found from the total distance and total time, noting sign conventions.
Problem 6: The speed of sound example yields a numeric estimate of sound speed from a distance/time; the approximate rule derived is about 1 km every 3 seconds in that scenario.
Problem 7–12: Various applications of the same core equations (displacement from velocity-time data, average speed from total distance/time, etc.). The key method is to identify whether the given quantity is a velocity or a displacement, choose appropriate equation, and carry units consistently.

Common equations referenced across problems:
Average velocity: v_{\text{avg}} = \frac{\Delta x}{\Delta t}
Average speed: (as noted in the text) \bar{v}_{\text{speed}} = \frac{d}{t} (with caveats in the text discussion)
Displacement from distance and time: x = v t (for constant velocity) or more general x = x0 + v0 t + \tfrac{1}{2} a t^2.

Page 5

Typical equations used

For constant acceleration: x = x0 + v0 t + \tfrac{1}{2} a t^2,\quad v = v_0 + a t,\quad a = \text{const}.
Time of travel given average velocity: t = \frac{\Delta x}{v}.
If you know two legs: total distance = $d1 + d2$, total time = $t1 + t2$; average speed = total distance / total time.

Page 6

Additional problem solutions (highlights)

Relative motion problems (car vs. truck) show two equivalent ways to analyze: (i) in the ground frame, summing distances; (ii) in the moving frame of one object, considering relative velocity.
A problem about a listener using a sound-speed approach to determine times and distances shows applying: v = \frac{\Delta x}{\Delta t} for sound with no air resistance assumptions.
The examples emphasize careful unit conversions and significant figures.

Core takeaways

Relative motion can be solved by transforming to a reference frame where one object is at rest, then tracking the relative displacement and velocity.
When speeds are given in different units, convert to consistent units before applying kinematic formulas.

Page 7

Key ideas from the later problems

The car-train passing problem demonstrates the need to account for both the car’s travel and the train’s travel when determining pass times and distances.
The acceleration is obtained from velocity-time data using a = \dfrac{\Delta v}{\Delta t}. The displacement during a time interval [t1, t2] is found with the appropriate kinematic formula.
The concept of average acceleration over a time interval is illustrated in multiple problems, including sprinter-type problems where acceleration and time to reach a certain velocity are found.

Selected formulas seen

Average acceleration from velocity data: \bar{a} = \dfrac{\Delta v}{\Delta t} over a chosen interval.
Displacement from velocity-time data: \Delta x = \int v(t) dt; in constant-acceleration problems, the closed form is used: \Delta x = v_0 t + \tfrac{1}{2} a t^2.

Page 8

Notable examples

A sprinter with a known final velocity and displacement can have its average acceleration computed via the kinematic equations.
The problems often ask you to determine the time intervals at which velocity is increasing, decreasing, or constant by inspecting the slope of the graph.

Page 9

Interpreting velocity-time and position-time graphs

The instantaneous velocity at a given time is the slope of the position–time graph at that time.
The instantaneous speed is the magnitude of velocity; negative velocity indicates direction opposite to the defined positive direction.
The slope of a velocity–time graph gives the acceleration; constant slope implies constant acceleration; flat slope implies zero acceleration.

Examples and applications

Given a velocity–time graph, determine the time intervals where the velocity is increasing or decreasing.
For a position–time graph, identify the times when the velocity is maximum, zero, or negative, and relate these to the corresponding slopes.

Page 10

Common approach

Identify the interval of constant acceleration or constant velocity.
Use the appropriate kinematic eqns to compute any unknowns (time, displacement, velocity) for each interval.
Sum displacements and times across intervals to obtain total results.

Page 11

Advanced multi-interval motion problems

Exercises involve pieces of motion with different accelerations, stopping distances, and times, followed by solving for unknowns by equating total displacements or total times across segments.
Some problems consider passenger or pedestrian motion, finishing with the requirement that two times or distances coincide to achieve a simultaneous finish.

Core equations used

For a segment with constant acceleration: x = x0 + v0 t + \tfrac{1}{2} a t^2,\quad v = v_0 + a t.
The total motion may require solving a quadratic in time for a given final displacement.

Page 12

Additional problem-solving examples and sign-checks

The two-vehicle pursuit examples illustrate setting displacements equal to solve for the pursuit time; one can also analyze in the reference frame of one vehicle to simplify.
A typical approach is to set the car’s displacement equal to the sum of the train’s displacement and any fixed offsets, then solve for time.
There are numerous worked-out steps showing how to compute the acceleration from velocity-time data, and how to use it to obtain displacement at specific times.

Key formula references

Distance traveled under constant acceleration: \Delta x = v_0 t + \tfrac{1}{2} a t^2.
Final velocity under constant acceleration: v = v_0 + a t.

Page 13

Projectile and vertical motion (gravity) focus

For vertical motion with upward positive, gravity is $a = -g$ ($g \approx 9.80\ \mathrm{m/s^2}$).
The displacement at any time can be found with the standard equations; at the peak of a trajectory, velocity is zero even though acceleration remains nonzero.
Height reached and time of flight can be found by solving the quadratic equations for vertical motion.

Examples discussed

Ball thrown upward: initial velocity, final height, and time to reach the top are computed using the vertical motion equations.
Additional problems illustrate calculating time to return to the original height and the height reached on the way up or down.

Page 14

Page 15

Rocket problem and multiple stages

A rocket problem considers multiple stages: burn to maximum height, coast, and fall back.
Calculations involve determining: (a) the velocity at fuel exhaustion, (b) the time to reach the maximum height, (c) coast time, (d) total time to reach max altitude, (e) velocity on return to Earth, and (f) total time of flight.
The core equations remain the one-dimensional kinematic equations with constant or piecewise-constant acceleration for each stage.

Key results (illustrative)

If you know the initial speed, acceleration, and distance, you can compute the velocity at a given distance using: v^2 = v0^2 + 2 a (x - x0).
Total flight time is the sum of stage times.

Page 16

Balloon and pelican problems; multi-part analysis

Several subparts cover a parachute/jump scenario, including height attained, time in air, and the acceleration during different phases.
A pelican dive problem analyzes three phases: acceleration to a maximum speed, coasting, and diving/impact with water, each with its own acceleration and time relations.

Equations emphasized

Vertical motion with given displacements uses: y = y0 + v0 t + \tfrac{1}{2} a t^2.
Times for up and down phases can be found by solving the above equations for the appropriate y positions.

Page 17

Ballistics and vertical motion (window/height problems)

Several problems involve computing the velocity at a given height, time to reach a given height, and the time to fall to, and then past, a window.
Sign conventions can change depending on whether upward or downward is chosen as positive; in each problem, it is important to maintain consistency.
The top of the motion has velocity zero; the time to go from throw to top and from top to landing can be computed separately and added.

Example equations used

For upward motion with initial velocity $v0$ and final height $y$: v^2 = v0^2 - 2 g y.
Time to reach a height y: solve y = v_0 t - \tfrac{1}{2} g t^2. for t.

Page 18

More data‑rich problem work and unit considerations

The page emphasizes sign conventions, unit consistency, and interpreting graphs for velocity and acceleration.
Many problems compare different gear configurations (in a hypothetical vehicle or machine) using the same core equations.
A common theme across pages is translating a real-world scenario into a sequence of one-dimensional motions (with constant acceleration in each segment) and applying kinematic equations to each segment.

Recap of core equations on this page

One-dimensional motion with constant acceleration: x = x0 + v0 t + \tfrac{1}{2} a t^2,\quad v = v_0 + a t,\quad a = \text{constant}.

Page 19

Data interpretation of graphs

Several items revolve around reading velocity vs. time and position vs. time graphs to extract instantaneous velocity, constant velocity intervals, and constant accelerations.
Instantaneous velocity at $t$ is the slope of the tangent to the position–time graph at $t$; instantaneous velocity at a time on a velocity–time graph is the value of $v$ there.
The instantaneous acceleration is the slope of the velocity–time graph.

Practice insights

When the velocity–time graph shows a flat region, acceleration is zero; when the velocity–time graph has a constant slope, the acceleration is constant.
The average velocity over a period is the total displacement divided by total time, i.e., \bar{v} = \frac{\Delta x}{\Delta t}.

Page 20

Conceptual and applied problems (summary)

The text emphasizes the inverse relationship between acceleration and displacement when considering limits on acceleration (e.g., Moon vs Earth) and discusses how displacement scales with acceleration for a fixed initial velocity.
A note about g-forces: the displacement is proportional to the inverse of acceleration when initial velocity is fixed.
A calculation example uses a scenario with balloon, rocket, or other projectiles to illustrate how height and velocity change with time.

Takeaways

When acceleration is reduced, displacement for a given time can increase if initial velocity remains the same.
In many problems, you may compare two measurements to obtain a height or a time difference by using the kinematic equations.

Page 21

Notable equations and ideas

If you need to know the time to overtake, you can set the car’s displacement equal to the train’s displacement plus any fixed offsets, then solve for time.
The total displacement in a multi-segment problem is the sum of segment displacements.

Page 22

Further problems involving height, energy-like considerations, and sound travel

Problems include combining vertical motion with sound travel to compute the height of a tower, the time for the sound to travel back, and the resulting distance fallen.
A typical method is to first compute the height from stone/ballistics equations, then use sound-time to close the problem with a time constraint.

Core formulae to remember

Vertical motion with gravity: y = y0 + v0 t - \tfrac{1}{2} g t^2. (signs depend on chosen positive direction)
The speed of sound problem uses: v{sound} = \frac{\Delta x}{\Delta t}, and the total time includes the travel time of the stone and the sound, i.e., T = t{stone} + t_{sound}.

Page 23

Diver/ballistic trajectories and heights

Various items cover calculating velocity at impact, time of flight to a given height, and total distance traveled for vertical throws and drops.
Some subtasks explore the effect of choosing different sign conventions on the solutions and interpret the meaning of negative velocities.

Example results and methods

Using the velocity–time relation for vertical motion: v^2 = v0^2 - 2 g (y - y0) in appropriate form depending on orientation.
Time to reach a given height can be obtained by solving a quadratic in $t$ from the position equation.

Page 24

Train/stop-light style problems and more

This page includes problems that combine acceleration, speed, and distance with practical constraints like stopping distance and reaction times at stoplights.
The solutions emphasize: (i) segmentation of motion into phases, (ii) using the standard kinematic equations in each phase, and (iii) summing across phases to check constraints such as whether a car can clear an intersection in time.

Key relationships exercised

Acceleration- and distance-based constraints in braking scenarios: x = v_0 t + \tfrac{1}{2} a t^2.
The need to ensure final speeds are physically reasonable (nonnegative speeds, depending on the scenario) and to verify the time intervals against given deadlines or light timings.

Page 25

Continuous scanning of the problem set

The problems cover a wide range of one-dimensional motion scenarios: braking, accelerating, passing, inter-station travel, and other kinematic challenges.
The solutions consistently use the standard equations of motion with careful sign conventions and unit consistency.

Highlights

The same core toolkit is used across diverse problems: (i) displacement from velocity, (ii) velocity from acceleration, (iii) time from distance, and (iv) total time as the sum of interval times.

Page 26

More multi-interval motion and passing scenarios

A set of problems considers passing maneuvers, where both vehicles have initial velocities and accelerations, and the pass must avoid collision by ensuring total displacements stay within a limit.
Time to pass is found by equating the displacements of the passing car and the approaching car and solving for time.

Practical approach

Compute the displacement of each car during the passing window, ensuring the sum does not exceed the allowed clearance.
If the computed total distance exceeds the allowed limit, the pass is not safe.

Page 27

Bicycles and velocity comparisons

The pages discuss when one bicycle’s velocity equals another, based on the instantaneous slopes of their position–time graphs.
The more concave graph indicates greater acceleration; a flat slope indicates zero acceleration.
The times of crossing correspond to the instants where the displacements are equal.

Additional insights

The instantaneous velocity is the slope of the x–t graph at a given time; the graph with the steeper slope at a crossing indicates the higher instantaneous velocity at that moment.

Page 28

Key takeaways

The same fundamental kinematic equations apply, but the sign conventions and the segmenting into distinct phases are crucial for correct solutions.

Page 29

Final practice problems and search/learn solutions

A final set of problems includes: (a) visual interpretation of velocity–time graphs, (b) the timing of overtaking or catching up, (c) identifying when velocity is zero, (d) interpreting the slopes and concavity of graphs to infer acceleration, (e) average velocity vs average speed concepts, and (f) dynamics of hitting a moving target.

Useful tips

To analyze a problem, split the motion into phases with constant acceleration, compute each phase, then combine the results to obtain total displacement or total time.
Use the slope method to read instantaneous velocity and acceleration from graphs when provided.

Page 30

Additional search-and-learn and solutions overview

The last pages reiterate how to set up the motion equations, compute time to overtake, and use relative motion concepts when one object is moving with respect to another (e.g., police vs. speeder).
The content emphasizes that solving such problems typically involves:
- Choosing a convenient origin and sign convention,
- Identifying segments of motion with constant acceleration,
- Applying the kinematic equations to each segment,
- Summing displacements and times to obtain the final answer.

Key equations repeatedly used

One-dimensional constant-acceleration equations:
x = x0 + v0 t + \tfrac{1}{2} a t^2,\quad v = v0 + a t,\quad v^2 = v0^2 + 2 a (x - x_0).$$
Relative motion and problem-solving in a moving frame can simplify calculations by transforming to a frame where one object is at rest.

If you’d like, I can convert these notes into a more compact study guide with a single cohesive outline (and keep page-by-page headings as requested), or I can expand any page with more detailed verbatim problem steps and exact numeric examples from the transcript.

Describing Motion: Kinematics in One Dimension – Page-by-Page Notes

Page 1

Key concepts

Basic definitions (notation)

Page 2

Positive direction conventions and sign examples

Acceleration can decrease while velocity increases

Relative motion and passing/catching up

Special cases and averages

Examples and calculations

Page 3

Describing Motion: Kinematics in One Dimension (conceptual)

Misconceptions addressed

Page 4

Problem-solving patterns (selected examples from the Solutions to Problems)

Page 5

More problem patterns and results

Typical equations used

Page 6

Additional problem solutions (highlights)

Core takeaways

Page 7

Key ideas from the later problems

Selected formulas seen

Page 8

More solutions and sign conventions

Notable examples

Page 9

Interpreting velocity-time and position-time graphs

Examples and applications

Page 10

More problem patterns and calculations

Common approach

Page 11

Advanced multi-interval motion problems

Core equations used

Page 12

Additional problem-solving examples and sign-checks

Key formula references

Page 13

Projectile and vertical motion (gravity) focus

Examples discussed

Page 14

More vertical motion problems

Page 15

Rocket problem and multiple stages

Key results (illustrative)

Page 16

Balloon and pelican problems; multi-part analysis

Equations emphasized

Page 17

Ballistics and vertical motion (window/height problems)

Example equations used

Page 18

More data‑rich problem work and unit considerations

Recap of core equations on this page

Page 19

Data interpretation of graphs

Practice insights

Page 20

Conceptual and applied problems (summary)

Takeaways

Page 21

More applied problems and strategies

Notable equations and ideas

Page 22

Further problems involving height, energy-like considerations, and sound travel

Core formulae to remember

Page 23

Diver/ballistic trajectories and heights

Example results and methods

Page 24

Train/stop-light style problems and more

Key relationships exercised

Page 25

Continuous scanning of the problem set

Highlights

Page 26

More multi-interval motion and passing scenarios

Practical approach