SCIENCE · INVESTIGATE AND EXPLAIN

Read a force-versus-speed curve without calling it a time history

Choose downward as positive for a falling point in still fluid. The stipulated drag magnitude is kv² upward for nonnegative downward speed v, and gravity is 10 m/s². The parameter k has units kg/m, so kv² has force units. This is one assigned resistance law, not a universal description of all fluids, shapes or speeds.

Learning outcome: Hold k fixed and compare masses 1,2,3 kg; then double k at one mass. Predict terminal-speed ratios before checking. Retain the force balance and a state above equilibrium, explicitly naming the graph’s horizontal variable.

Starting recap

Draw recipient-specific interactions, resolve signed components and identify the domain of each force/contact model before substituting.

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Try

At m=2 kg and k=0.5 kg/m, vt=√40≈6.32456 m/s. At v=vt/2 acceleration is 7.5 m/s² downward; at 2vt it is−30 m/s² in the downward convention.

Open the experiment

Predict

At twice the terminal downward speed in this model, the object:

Your prediction and reason

Experiment

Hold k fixed and compare masses 1,2,3 kg; then double k at one mass. Predict terminal-speed ratios before checking. Retain the force balance and a state above equilibrium, explicitly naming the graph’s horizontal variable.

Compare force balance above and below terminal speed

Worked teaching model · synthetic values · Downward is positive, gravity 10 m/s², still fluid, drag magnitude k v² upward for nonnegative downward speed. Only algebraic force-versus-speed states are shown, NOT a solved time history or real fall experiment.

Falling object mass: 2 kg Quadratic drag constant: 0.5 kg/m

Reset keeps your prediction, captures and practice history.

Algebraic acceleration at a given downward speed

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: net-force state (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Algebraic acceleration at a given downward speed

net-force state

downward speed (m/s)	downward acceleration (m/s²)
0	10
0.316228	9.975
0.632456	9.9
0.948683	9.775
1.26491	9.6
1.58114	9.375
1.89737	9.1
2.21359	8.775
2.52982	8.4
2.84605	7.975
3.16228	7.5
3.47851	6.975
3.79473	6.4
4.11096	5.775
4.42719	5.1
4.74342	4.375
5.05964	3.6
5.37587	2.775
5.6921	1.9
6.00833	0.975
6.32456	-1.77636e-15
6.64078	-1.025
6.95701	-2.1
7.27324	-3.225
7.58947	-4.4
7.90569	-5.625
8.22192	-6.9
8.53815	-8.225
8.85438	-9.6
9.17061	-11.025
9.48683	-12.5
9.80306	-14.025
10.1193	-15.6
10.4355	-17.225
10.7517	-18.9
11.068	-20.625
11.3842	-22.4
11.7004	-24.225
12.0167	-26.1
12.3329	-28.025
12.6491	-30

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
Quantity	Value	Unit
Terminal downward speed	6.32456	m/s
Downward acceleration at half terminal speed	7.5	m/s²
Downward acceleration at terminal speed	0	m/s²
Downward acceleration at twice terminal speed	-30	m/s²
Weight	20	N
Drag at terminal speed	20	N

Terminal speed is a force balance, not a universal maximum allowed initial speed.

For v≥0, downward acceleration is g−kv²/m and terminal speed sqrt(mg/k). At twice terminal speed the upward resistance exceeds weight, so downward motion slows; this does not instantly reverse velocity. The graph’s horizontal coordinate is speed, not time. No exact approach-time formula or universal drag coefficient is implied.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

Terminal speed is a force balance, not a universal maximum allowed initial speed.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Define the recipient and model

Build the algebraic relationship

Newton’s law gives downward acceleration g−kv²/m at a selected speed. Terminal speed occurs where that expression is zero: vt=√(mg/k). At half vt, drag is one quarter of weight and downward acceleration is 0.75g. At twice vt, drag is four times weight and downward acceleration is−3g, meaning the downward motion slows.

Check the changed state

Terminal speed is an equilibrium value, not a universal bound on initial conditions. An object introduced with downward speed greater than vt can decelerate toward it. Negative downward acceleration does not instantly make velocity upward. A constant-speed state can still have two nonzero forces whose resultant is zero.

Interpret evidence and limits

The graph plots acceleration against speed. It does not show speed against time and does not supply the elapsed time needed to reach any state. Solving that transient would require additional mathematics not assumed in this algebra category. Changes in fluid density, shape, buoyancy or the resistance law would require a new model rather than reuse of a fitted k without evidence.

Worked original packet

At m=2 kg and k=0.5 kg/m, vt=√40≈6.32456 m/s. At v=vt/2 acceleration is 7.5 m/s² downward; at 2vt it is−30 m/s² in the downward convention.

Bounded investigation

Changed-case numerical practice

Try the changed case before feedback

For m=1.8 kg, g=10 m/s² and k=0.2 kg/m, find terminal downward speed in m/s.

Numerical result in the requested units

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and boundaries

Original prose, data and diagrams checked 30 October 2026 against OpenStax College Physics 2e. This is an algebra-based virtual evidence exercise, not a physical experiment, equipment certification, transport recommendation or independent mastery result. Save assumptions and a competing explanation with the course evidence; supervised practical requirements remain external.

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Public practice checks a specific claim. It does not establish independent mastery.

Why does a force-balance calculation not determine a travel time?

Your explanation and limitations

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Previous: Separate apparent weight, mass and support contact Next: Reject a circular path that needs a pulling surface

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