Unit 2 · Lesson 10 of 11

Position, velocity, and acceleration tell different stories.

A cart can move backward while accelerating forward. The signs of its velocity and acceleration determine whether its speed rises or falls.

Change the input. Follow the mathematics.

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

What will change if you increase time? 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, Velocity (numeric overlay). Selected signed acceleration: 6 m/s². A negative velocity can coexist with positive acceleration. Speed grows when velocity and acceleration have the same sign.-40102244Position (m)Time (s)
PositionVelocity (numeric overlay) · dashed

signed accelerationv(2)=0 m/s,a(2)=6 m/s2v(2)=0\,\mathrm{m/s},\quad a(2)=6\,\mathrm{m/s^2}

A negative velocity can coexist with positive acceleration. Speed grows when velocity and acceleration have the same sign.

Assumptions, units, and limits

A prescribed one-dimensional trajectory; signed velocity differs from speed.

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

01

From the situation to an equation

Differentiate position once for velocity and again for acceleration. Speed is the magnitude of velocity. When velocity crosses zero the cart can reverse; acceleration need not vanish at that instant.

s=t3−3t2,v=3t2−6t,a=6t−6s=t^3-3t^2,\quad v=3t^2-6t,\quad a=6t-6
02

Keep the reasoning visible

v=s′,a=v′=s′′,speed=∣v∣v=s',\quad a=v'=s'',\quad\text{speed}=|v|

A negative velocity can coexist with positive acceleration. Speed grows when velocity and acceleration have the same sign.

03

What makes this model useful?

A prescribed one-dimensional trajectory; signed velocity differs from speed.

Positive acceleration does not always mean increasing speed. Compare the signs of velocity and acceleration.

A worked example

Calculate the starting scenario

  1. Use time = 2 s.
  2. v(2)=0 m/s,a(2)=6 m/s2v(2)=0\,\mathrm{m/s},\quad a(2)=6\,\mathrm{m/s^2}
  3. A negative velocity can coexist with positive acceleration. Speed grows when velocity and acceleration have the same sign. The checked quantity is signed acceleration.
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 signed acceleration using time 2 s. State the meaning of its sign and units.

Checked quantity: signed acceleration · m/s². Follow the definition for finite-cutoff and numerical-model questions.

Hint

Differentiate position once for velocity and again for acceleration. Speed is the magnitude of velocity. When velocity crosses zero the cart can reverse; acceleration need not vanish at that instant.

Explore the visual solution
Motion illustration: position -4. Velocity direction shown where applicable.Position -4 · arrow: velocity direction
Time (s) versus Position (m). Position, Velocity (numeric overlay). Selected signed acceleration: 6 m/s². A negative velocity can coexist with positive acceleration. Speed grows when velocity and acceleration have the same sign.-40102244Position (m)Time (s)
PositionVelocity (numeric overlay) · dashed

signed accelerationv(2)=0 m/s,a(2)=6 m/s2v(2)=0\,\mathrm{m/s},\quad a(2)=6\,\mathrm{m/s^2}

A negative velocity can coexist with positive acceleration. Speed grows when velocity and acceleration have the same sign.

Assumptions, units, and limits

A prescribed one-dimensional trajectory; signed velocity differs from speed.

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.

Differentiate position once for velocity and again for acceleration. Speed is the magnitude of velocity. When velocity crosses zero the cart can reverse; acceleration need not vanish at that instant.

s=t3−3t2,v=3t2−6t,a=6t−6s=t^3-3t^2,\quad v=3t^2-6t,\quad a=6t-6
Calculate and explain

Apply the idea to a new scenario

At t=1t=1 s, find acceleration and inspect the signed velocity for s=t3−3\mathrm{s}=t^3-3t².

Checked quantity: signed acceleration · m/s². Follow the definition for finite-cutoff and numerical-model questions.

Hint

Differentiate position once for velocity and again for acceleration. Speed is the magnitude of velocity. When velocity crosses zero the cart can reverse; acceleration need not vanish at that instant.

Explore the visual solution
Motion illustration: position -2. Velocity direction shown where applicable.Position -2 · arrow: velocity direction
Time (s) versus Position (m). Position, Velocity (numeric overlay). Selected signed acceleration: 0 m/s². A negative velocity can coexist with positive acceleration. Speed grows when velocity and acceleration have the same sign.-40102244Position (m)Time (s)
PositionVelocity (numeric overlay) · dashed

signed accelerationv(1)=−3 m/s,a(1)=0 m/s2v(1)=-3\,\mathrm{m/s},\quad a(1)=0\,\mathrm{m/s^2}

A negative velocity can coexist with positive acceleration. Speed grows when velocity and acceleration have the same sign.

Assumptions, units, and limits

A prescribed one-dimensional trajectory; signed velocity differs from speed.

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.

Differentiate position once for velocity and again for acceleration. Speed is the magnitude of velocity. When velocity crosses zero the cart can reverse; acceleration need not vanish at that instant.

s=t3−3t2,v=3t2−6t,a=6t−6s=t^3-3t^2,\quad v=3t^2-6t,\quad a=6t-6
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, Velocity (numeric overlay). Selected signed acceleration: 6 m/s². A negative velocity can coexist with positive acceleration. Speed grows when velocity and acceleration have the same sign.-40102244Position (m)Time (s)
PositionVelocity (numeric overlay) · dashed

signed accelerationv(2)=0 m/s,a(2)=6 m/s2v(2)=0\,\mathrm{m/s},\quad a(2)=6\,\mathrm{m/s^2}

A negative velocity can coexist with positive acceleration. Speed grows when velocity and acceleration have the same sign.

Assumptions, units, and limits

A prescribed one-dimensional trajectory; signed velocity differs from speed.

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

The reasoning. Differentiate position once for velocity and again for acceleration. Speed is the magnitude of velocity. When velocity crosses zero the cart can reverse; acceleration need not vanish at that instant.

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.

Completion is saved on this browser. This is separate from account practice progress.