riaology
Unit 2 · Lesson 1 of 11

Watch a secant become a tangent.

Two position observations give average velocity. Bringing their times closer reveals instantaneous velocity.

Change the input. Follow the mathematics.

What you'll be able to do
  • Set up the model and calculate its instantaneous velocity.
  • Connect the equation to the changing visual.
  • Explain the assumptions and identify a misleading interpretation.
Predict before exploring

What will change if you increase base time t? Predict the direction, then explain the response.

A model you can changeMove one control at a time.
Motion illustration: position 4. Velocity direction shown where applicable.Position 4 · arrow: velocity direction
Time (s) versus Position (m). Position, Secant, Tangent. Selected instantaneous velocity: 4 m/s. The dashed secant connects two observations. The tangent gives instantaneous velocity; shrink h to see the secant approach it.-10243.5497Position (m)Time (s)
PositionSecant · dashedTangent

instantaneous velocity(2+1)2−221=5→s′(2)=4 m/s\frac{(2+1)^2-2^2}{1}=5\to s'(2)=4\,\mathrm{m/s}

The dashed secant connects two observations. The tangent gives instantaneous velocity; shrink h to see the secant approach it.

Assumptions, units, and limits

A cart position s=t2\mathrm{s}=t^2 m. h is positive so the second secant point is later in time.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

01

From the situation to an equation

Subtract the two position values before dividing by elapsed time. Expand the squared expression to remove cancellation, then take the step-size limit. The derivative is a new function assigning slope to each base time.

s(t)=t2,s(t+h)−s(t)h=2t+h→s′(t)=2ts(t)=t^2,\quad\frac{s(t+h)-s(t)}h=2t+h\to s'(t)=2t
02

Keep the reasoning visible

s′(t)=lim⁡h→0s(t+h)−s(t)hs'(t)=\lim_{h\to0}\frac{s(t+h)-s(t)}h

The dashed secant connects two observations. The tangent gives instantaneous velocity; shrink h to see the secant approach it.

03

What makes this model useful?

A cart position s=t2\mathrm{s}=t^2 m. h is positive so the second secant point is later in time.

The secant slope at a finite step is an average rate, not automatically the instantaneous rate. Shrinking h changes the observation interval.

A worked example

Calculate the starting scenario

  1. Use base time t = 2 s; secant step h = 1 s.
  2. (2+1)2−221=5→s′(2)=4 m/s\frac{(2+1)^2-2^2}{1}=5\to s'(2)=4\,\mathrm{m/s}
  3. The dashed secant connects two observations. The tangent gives instantaneous velocity; shrink h to see the secant approach it. The checked quantity is instantaneous velocity.
Make the idea your own

Calculate. Compare. Explain.

Numeric checks allow 1% rounding error. Values near zero use a small absolute tolerance. Work through the setup before opening a solution.

Calculate and explain

Rebuild the starting calculation

Calculate the instantaneous velocity using base time t 2 s, secant step h 1 s. State the meaning of its sign and units.

Checked quantity: instantaneous velocity · m/s. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Subtract the two position values before dividing by elapsed time. Expand the squared expression to remove cancellation, then take the step-size limit. The derivative is a new function assigning slope to each base time.

Explore the visual solution
Motion illustration: position 4. Velocity direction shown where applicable.Position 4 · arrow: velocity direction
Time (s) versus Position (m). Position, Secant, Tangent. Selected instantaneous velocity: 4 m/s. The dashed secant connects two observations. The tangent gives instantaneous velocity; shrink h to see the secant approach it.-10243.5497Position (m)Time (s)
PositionSecant · dashedTangent

instantaneous velocity(2+1)2−221=5→s′(2)=4 m/s\frac{(2+1)^2-2^2}{1}=5\to s'(2)=4\,\mathrm{m/s}

The dashed secant connects two observations. The tangent gives instantaneous velocity; shrink h to see the secant approach it.

Assumptions, units, and limits

A cart position s=t2\mathrm{s}=t^2 m. h is positive so the second secant point is later in time.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Subtract the two position values before dividing by elapsed time. Expand the squared expression to remove cancellation, then take the step-size limit. The derivative is a new function assigning slope to each base time.

s(t)=t2,s(t+h)−s(t)h=2t+h→s′(t)=2ts(t)=t^2,\quad\frac{s(t+h)-s(t)}h=2t+h\to s'(t)=2t
Calculate and explain

Apply the idea to a new scenario

At base time t=3t=3 s and secant step h=0.5h=0.5 s, find instantaneous velocity and inspect the average velocity.

Checked quantity: instantaneous velocity · m/s. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Subtract the two position values before dividing by elapsed time. Expand the squared expression to remove cancellation, then take the step-size limit. The derivative is a new function assigning slope to each base time.

Explore the visual solution
Motion illustration: position 9. Velocity direction shown where applicable.Position 9 · arrow: velocity direction
Time (s) versus Position (m). Position, Secant, Tangent. Selected instantaneous velocity: 6 m/s. The dashed secant connects two observations. The tangent gives instantaneous velocity; shrink h to see the secant approach it.0024.53.5497Position (m)Time (s)
PositionSecant · dashedTangent

instantaneous velocity(3+0.5)2−320.5=6.5→s′(3)=6 m/s\frac{(3+0.5)^2-3^2}{0.5}=6.5\to s'(3)=6\,\mathrm{m/s}

The dashed secant connects two observations. The tangent gives instantaneous velocity; shrink h to see the secant approach it.

Assumptions, units, and limits

A cart position s=t2\mathrm{s}=t^2 m. h is positive so the second secant point is later in time.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Subtract the two position values before dividing by elapsed time. Expand the squared expression to remove cancellation, then take the step-size limit. The derivative is a new function assigning slope to each base time.

s(t)=t2,s(t+h)−s(t)h=2t+h→s′(t)=2ts(t)=t^2,\quad\frac{s(t+h)-s(t)}h=2t+h\to s'(t)=2t
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Motion illustration: position 4. Velocity direction shown where applicable.Position 4 · arrow: velocity direction
Time (s) versus Position (m). Position, Secant, Tangent. Selected instantaneous velocity: 4 m/s. The dashed secant connects two observations. The tangent gives instantaneous velocity; shrink h to see the secant approach it.-10243.5497Position (m)Time (s)
PositionSecant · dashedTangent

instantaneous velocity(2+1)2−221=5→s′(2)=4 m/s\frac{(2+1)^2-2^2}{1}=5\to s'(2)=4\,\mathrm{m/s}

The dashed secant connects two observations. The tangent gives instantaneous velocity; shrink h to see the secant approach it.

Assumptions, units, and limits

A cart position s=t2\mathrm{s}=t^2 m. h is positive so the second secant point is later in time.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. Subtract the two position values before dividing by elapsed time. Expand the squared expression to remove cancellation, then take the step-size limit. The derivative is a new function assigning slope to each base time.

Your learning path

Can you explain the connection?

Mark complete when you can set up the calculation and explain what the visual is showing. Opening a page never completes it automatically.

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