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Engineering & Technology · Dynamics, Vibration & Mechanical Response

Position, velocity, acceleration and motion descriptions

A small guided carriage passes position marks faster and faster. The marks describe where it is, but time is needed to describe its motion. Before discussing the force that drives it, we can organise what position, velocity and acceleration each say about the same trajectory.

Use Forces, Energy & Simple Systems and Engineering Sketches, Models & Technical Communication. Be comfortable with signed quantities, ratios, squares and graph axes. Radians, frequency and each new symbol are introduced here. Sinusoidal response formulas are supplied and interpreted without requiring a differential-equation derivation. All numerical records and model parameters are original constructed cases, not equipment limits, operating instructions or dynamic qualification.

Back to the chapter

After reading, you can

Read position, velocity and acceleration in one reference frame and apply constant-acceleration relations only where justified.

01

Choose the frame and distinguish a path from a change

Let x be position along a straight guide, measured from an identified origin, with right positive. Displacement Δx is final minus initial position. Distance travelled is the accumulated path length. If the carriage reverses, its distance can exceed the magnitude of displacement. Neither quantity is a force or a measure of effort.

Average velocity over an interval is Δx/Δt. Its sign refers to the chosen axis. Speed is the magnitude of instantaneous velocity; average speed instead uses total path length divided by elapsed time. A negative position does not by itself mean leftward motion: a carriage can sit left of the origin while moving right.

02

Each graph answers a different rate question

Velocity describes the rate at which position changes. On a position-time plot, a steeper local slope means a larger velocity. Acceleration describes the rate of change of velocity, with units m/s². Positive acceleration can slow a left-moving object before making it reverse. “Accelerating” does not always mean increasing speed.

A horizontal velocity line indicates constant velocity, not rest unless the value is zero. A rising position curve can coexist with a constant acceleration line. Read the vertical quantity and units before interpreting a shape. Aligned time axes in the figure make the distinctions visible without pretending that metres, m/s and m/s² share one vertical scale.

03

Use a stated model rather than a formula by habit

For a constant acceleration a, starting at time zero with position x 0 and velocity v 0, the relations are v=v 0+at and x=x 0+v 0t+at²/2. The subscripts identify initial values. The squared-time term explains why equal time intervals can contain different displacements.

These expressions do not establish that an actual drive maintains constant acceleration. A change in load or control can require separate intervals or a varying-acceleration model. Position measurements at a few instants also leave motion between the samples unresolved. Begin with a model appropriate to the evidence, then ask separately what force would produce it.

Position,velocity and acceleration are different descriptionsThree aligned time plots show one constant-acceleration trajectory from one to sixteen metres over three seconds.One trajectory: x0 = 1 m, v0 = 2 m/s, a = 2 m/s²0123012301230816Position x / m048Velocity v / (m/s)012Acceleration a / (m/s²)Time / s; each plot has its own vertical units
For t 0, 1, 2, 3s, x=1, 4, 9, 16m; v=2, 4, 6, 8m/s; acceleration remains 2m/s². Position is quadratic, velocity linear and acceleration constant. Equal horizontal scales 150 drawing units/s align the three plots; vertical scales differ and are labelled.

Check the three linked descriptions

Take x 0=1 m, v 0=2 m/s and constant a=2 m/s². At t=0, 1, 2, 3 s, the positions are 1, 4, 9, 16 m. The velocities are 2, 4, 6, 8 m/s. Consecutive one-second displacements are 3, 5 and 7 m, so the corresponding interval-average velocities differ from the velocities at the interval endpoints.

Over the full 3 s, displacement is 16−1=15 m and average velocity 15/3=5 m/s. This also equals(2+8)/2 for this constant-acceleration case. Since velocity stays positive, the path length is 15 m too. That equality would need reconsideration if the trajectory reversed.

Try the reasoning

  1. Does x = −2 m mean the carriage moves left?

    Show the answer and reasoning

    No. It gives a location relative to the origin. Velocity supplies the direction of motion.

  2. What is the displacement from t = 1 to 2 s in this record?

    Show the answer and reasoning

    9−4=5 m, so the interval-average velocity is 5 m/s.

  3. Can the constant-a equations be assumed for any measured motion?

    Show the answer and reasoning

    No. Their acceleration assumption must match the motion or the stated approximation.

What to carry forward

Name the frame, quantity, units and time interval before interpreting a motion graph.

References

Teaching resources and research supporting this lesson’s explanations and examples.

  1. OpenStax · University Physics Volume 12 sources
    1. 3.1 Position, Displacement and Average VelocityOpenStax · University Physics Volume 1Teaching resourceopenstax.orgRead source
    2. 3.4 Motion with Constant AccelerationOpenStax · University Physics Volume 1Teaching resourceopenstax.orgRead source