One-Dimensional Kinematics: Key Concepts, Equations, and Example Problems

Overview

Focus on motion along a single axis (one-dimensional kinematics).
Quantities to track: position (distance along the axis), velocity (rate of change of position), and acceleration (rate of change of velocity).
Two primary constant-acceleration cases: zero acceleration (uniform motion) and constant nonzero acceleration.
In general, acceleration can vary with time, but for this course we restrict to the constant-acceleration or zero-acceleration cases.
Vertical free fall is a canonical constant-acceleration problem with acceleration g (neglecting air resistance): g ≈ 9.82 m/s extsuperscript{2}.
Air/fluid resistance can make acceleration non-constant; frictional forces also affect motion, but are not the focus here.
Real-world discussion: when a ball bounces, energy is dissipated (not perfectly elastic), which affects subsequent velocity after each bounce.

Time, Derivatives, and Functions of Time

Position as a function of time: $x(t)$ (often called the coordinate; can be a vector in higher dimensions, but here we discuss one axis).
Velocity is the first time derivative of position: $v(t) = \frac{dx}{dt}$ .
Acceleration is the first derivative of velocity (or the second derivative of position): $a(t) = \frac{dv}{dt} = \frac{d^2 x}{dt^2}$ .
Units (one-dimensional): position in meters (m); velocity in meters per second (m/s); acceleration in meters per second squared (m/s extsuperscript{2}).
Interpretation and visualization:
- Velocity is the slope (derivative) of the position vs time graph.
- Acceleration is the slope of the velocity vs time graph (and the curvature of the position vs time graph).
Conceptual note on increments:
- For a small time step (\Delta t), the change in position is (\Delta x) and the average velocity is approximately (\Delta x/\Delta t).
- In the limit (\Delta t \to 0), this becomes the derivative definitions above.
1D convention and sign:
- Choose a coordinate axis (positive direction). Distances traveled in the positive direction are positive; negative distances occur when moving opposite to the chosen positive direction.
- Vector magnitude can be taken as the positive quantity in 1D plots; sign encodes direction.

Constant Acceleration: Core Equations

When acceleration is constant, the motion is governed by a small set of universal equations:
- Position as a function of time:
 $x(t) = x0 + v0 t + \frac{1}{2} a t^2$
- Velocity as a function of time:
 $v(t) = v_0 + a t$
These formulas are the 1D kinematics equations for constant (a).
A time-free relation (eliminating time) often used:
- Starting from the two equations above, eliminate (t) to obtain
 $v^2 = v0^2 + 2 a (x - x0)$
 or, equivalently, solving for velocity directly:
 $v = \sqrt{v0^2 + 2 a (x - x0)}$
 (sign of (v) depends on the direction of motion relative to the sign convention for (v)).
Special case: if (a = 0), motion is uniform and
- $x(t) = x0 + v0 t$
- $v(t) = v_0$
These equations form the backbone of most 1D problems with constant acceleration (e.g., free fall, tossing motions with constant thrust, etc.).

Example: A Simple One-Dimensional Motion (Turtle)

Given: (x0 = 2\text{ m}), (v0 = 0.1\ \text{m/s}), (a = 0), (t = 10\text{ s}).
Find: position after 10 s, i.e., (x(10)).
Solution:
- Using the position equation with (a = 0):
 $x(10) = x0 + v0 \cdot t = 2 + 0.1 \times 10 = 3\text{ m}$
(Optional) Velocity after 10 s:
- $v(10) = v_0 + a t = 0.1 + 0 \times 10 = 0.1\text{ m/s}$
Key takeaway: with zero acceleration, position grows linearly with time.

Example: Free Fall from Height 10 m

Setup: An object starts from rest, falls a distance of 10 m with constant acceleration (a = g \approx 9.82\ \text{m/s}^2).
Given: initial height corresponds to (x0 = 0) at release, (v0 = 0).
Objective: final velocity after falling distance (x = 10) m (ignore air resistance).
Using the time-free relation:
- From (v^2 = v0^2 + 2 a (x - x0)):
 $v = \sqrt{v0^2 + 2 g (x - x0)} = \sqrt{0 + 2 \cdot 9.82 \cdot 10} \approx 14.0\ \text{m/s}$
Alternatively, time can be found from (x(t) = x0 + v0 t + \frac{1}{2} g t^2), but the velocity relation above is often quicker when time is not required.
Note on real-world caveats:
- In air, acceleration is not exactly constant; terminal velocity can be approached when drag balances gravity for long falls.
- Here we assumed vacuum-like conditions (no air resistance).

Example: Train and Pole/Platform Problem (Eliminating Time)

Scenario 1 (pole): A train of length (l) passes a fixed pole. It takes 5 s for the entire train to pass the pole.
- Relationship: distance traveled from front of train reaching the pole to the rear passing is (l). With speed (v) (assumed constant),
  $l = v \cdot 5$ (Equation A)
Scenario 2 (platform): The same train passes a 100 m long platform. The time for the front to reach the start of the platform until the rear leaves the platform end is 30 s.
- Distance to cover for complete clearance: platform length plus train length, i.e., (100 + l).
- Therefore,
  $100 + l = v \cdot 30$ (Equation B)
Solve the two equations (A) and (B) for (v) and (l):
- From (A): (l = 5 v).
- Substitute into (B): (100 + 5 v = 30 v) → (100 = 25 v) → (v = 4\ \text{m/s}).
- Then (l = 5 v = 20\ \text{m}).
This problem illustrates how to combine two situations to eliminate time and solve for unknowns using the constant-velocity model.

Free-Fall Intuition: Acceleration, Velocity, and Energy Considerations

Between bounces (no contact with the ground), acceleration is constant and equal to (g) (positive in the chosen convention if you align the axis with the direction of gravity).
Velocity increases in the direction of gravity during free fall; after bounce, velocity reverses sign, with a new peak velocity that can be smaller due to energy loss.
The bounce is an impulse event, typically resulting in a reduction of kinetic energy; this is modeled by a coefficient of restitution (not explicitly used in the equations above but conceptually important).
Large accelerations can occur during bounce due to rapid reversal of velocity; the lecture mentions values like several g's (roller coasters show peaks of a few g; human tolerance is roughly 4–6 g for brief periods, with higher limits for trained individuals and shorter durations; estimates around 20 g or higher are extreme for humans).
Qualitative takeaway: the x-t graph shows a downward parabola during free-fall, the v-t graph shows a straight line with slope equal to (g); the bounce produces a sharp, nearly vertical change in velocity (a spike in the a-t graph).

Quick Reference: Fundamental Equations (Constant Acceleration)

Position:
$x(t) = x0 + v0 t + \frac{1}{2} a t^2$
Velocity:
$v(t) = v_0 + a t$
Velocity in terms of position (no time):
$v^2 = v0^2 + 2 a (x - x0)$
Solve for velocity when you know height (or distance) and acceleration:
$v = \sqrt{v0^2 + 2 a (x - x0)}$
Special case for vertical drop with initial velocity zero and starting at height (h):
$v = \sqrt{2 g h}$

Practice Problems (Additional Applications)

Practice 1: Turtle-like motion with acceleration 0
- Given: (x0 = 2\text{ m}), (v0 = 0.1\text{ m/s}), (a = 0), (t = 10\text{ s}).
- Compute: (x(10)) and optionally (v(10)).
- Answer: x(10) = 2 + 0.1 \times 10 = 3\text{ m};\quad v(10) = 0.1\text{ m/s}.$n- Practice 2: Free-fall from 10 m with initial rest
- Given: (x0 = 0), (v0 = 0), (x = 10) m, (g = 9.82\ \text{m/s}^2).
- Velocity on impact: v = \sqrt{2 g x} = \sqrt{2 \cdot 9.82 \cdot 10} \approx 14.0\ \text{m/s}.$$
Practice 3: Train problem (as above)
- Scenario 1: (t1 = 5\,\text{s}); length (l = v t1).
- Scenario 2: Platform length 100 m; (t2 = 30\,\text{s}); full clearance distance (100 + l) = (v t2).
- Solve: from Scenario 1, (l = 5 v). Substitute into Scenario 2: (100 + 5 v = 30 v) → (v = 4\ \text{m/s}); hence (l = 20\ \text{m}).

Practical Problem-Solving Workflow

Start with a clear choice of coordinate axis (define origin, positive direction).
Write down known quantities (initial position, initial velocity, acceleration, and time if given).
Choose the appropriate equation(s) based on whether you want time included or eliminated (use the three core equations above).
If time is not given, use the time-eliminated relation (e.g., (v^2 = v0^2 + 2 a (x - x0))).
If time is needed, use x(t) or v(t) equations to compute the desired quantity.
For problems with multiple stages or different contexts (e.g., bounces, platforms, poles), set up separate segments and use the appropriate boundary conditions at transitions (e.g., velocity sign change on bounce, energy loss).
Always check units and signs; interpret results with respect to your chosen coordinate system.

Conceptual Takeaways

The trio (x, v, a) are interrelated through calculus and constants of motion; constant acceleration provides a compact, predictive framework for motion along one axis.
Derivatives are powerful: velocity is the slope of x vs t; acceleration is the slope of v vs t.
When analyzing real-world motion, recognize the limits of the models (air resistance, bounce losses, and other non-idealities).
Simple experiments and thought experiments (e.g., a dropping ball or a moving turtle) help cement the link between the mathematics and the physical intuition.