Straight-Line Motion: Connecting Position, Velocity, and Acceleration
Later connection: apply the rate interpretation to calculated motion, direction, and changing speed.
Before this connection: Differentiation tools; First and second derivatives.
College Board Topic 4.2 · CED p. 83 (PDF p. 88) · CED effective Fall 2020; project course outline 2026–27.
Position s(t) tells you where a particle is on a line. Velocity v(t) = s′(t) tells you which direction it is moving and how fast its position changes. Acceleration a(t) = v′(t) = s″(t) tells you how the velocity changes. If position is in meters and time in seconds, velocity is in m/s and acceleration in m/s².
Speed is |v(t)|, so it never carries a direction. Where v ≠ 0, speed increases when v and a have the same sign, and decreases when they have opposite signs. This follows from differentiating |v| separately on v > 0 and v < 0: its derivative is a in the first case and −a in the second. At v = 0, inspect the behavior on both sides instead of blindly using this test.
A particle changes direction when velocity changes sign. A zero of v is only a candidate: v(t) = (t − 2)² touches zero but stays positive. On a position graph, v is the slope; on a velocity graph, a is the slope. Keep those two graph-reading jobs separate.
Let's work through it
Let s(t) = t³ − 6t² + 9t, with meters and seconds. Then v(t) = 3(t − 1)(t − 3) and a(t) = 6t − 12.
Why: Differentiate position once for velocity and again for acceleration.
At t = 2, v = −3 m/s and a = 0 m/s²; the particle moves left at speed 3 m/s.
Why: The sign of velocity supplies direction, while its absolute value supplies speed.
On (1,2), both v and a are negative, so speed increases. On (2,3), v is negative and a positive, so speed decreases.
Why: Same signs increase |v|; opposite signs decrease |v|.
The particle changes direction at t = 1 and t = 3.
Why: The factored velocity changes sign at both roots.
Try it from memory
Pause & retrieve
Try it before you peek.
Write your answer and the reason from memory. Then compare with the worked answer. Your self-assessment is a study decision, not an automatically graded score.
Question 1
If v = −4 and a = −2, is the particle speeding up?
See every recall question with its answer
These references preserve each prompt and its answer for direct evidence links. Try the interactive questions above before opening the worked reasoning.
Retrieval 1
If v = −4 and a = −2, is the particle speeding up?
Worked answer and reason
Yes: velocity and acceleration have the same sign.
The negative velocity is becoming more negative, so |v| grows.
Retrieval 2
Must a particle reverse direction whenever v = 0?
Worked answer and reason
No; velocity must change sign.
A zero can be a touch rather than a crossing.
Retrieval 3
A velocity graph is above the axis and slopes downward. What happens?
Worked answer and reason
The particle moves in the positive direction and slows down.
Height gives v > 0; slope gives a < 0.
Catch the tempting mistake
- Positive acceleration always means speeding up.
- Positive acceleration slows a particle moving in the negative direction until it reaches zero velocity.
- Position zero means the particle is stopped.
- Position zero means it passes the chosen origin; stopped means velocity zero.
Calculator workspace and limits
For calculus, set the angle selector to Radians. This scientific calculator evaluates arithmetic and functions. It does not graph arbitrary functions, find numerical roots, evaluate numerical derivatives, or compute definite integrals. Those tasks need an approved graphing tool; the May 2027 exam permits approved handheld graphing calculators and Bluebook’s built-in Desmos on calculator-required parts. A graph or decimal estimate helps you check a conclusion; your setup, hypotheses, units and justification still do the explaining.
Scientific calculator
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Put it to work
- Unit 4 problem 5 · No calculator · analytic
- Unit 4 problem 6 · No calculator · analytic
- Unit 4 problem 7 · No calculator · numerical
- Unit 4 problem 8 · No calculator · graphical
- Unit 4 problem 9 · No calculator · analytic
For the teacher
- Have learners point to a graph’s height and slope separately before assigning signs.
- At a rest instant, require a left/right sign chart rather than a claim based only on v = 0.
Glossary and deeper reading · 3 refreshers
These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.
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Velocity in calculus
In one-dimensional motion, velocity is the derivative of position with respect to time. Its sign records direction relative to the chosen positive axis. For example, s(t)=t²−4t meters gives velocity 2t−4 meters per second. Speed is the magnitude of velocity and therefore cannot be negative.
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Acceleration in calculus
Acceleration is the derivative of velocity with respect to time. In one-dimensional motion, it is also the second derivative of position. For example, s(t)=t² meters gives constant acceleration 2 meters per second squared. A negative acceleration can increase speed when the velocity is also negative.
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Derivative
A derivative gives the instantaneous rate of change of a function with respect to its input. At a point, it is the finite limit of the difference quotient when that limit exists. For example, the derivative of x² is 2x, so the tangent slope at x=3 is 6. A derivative function collects those rates across the inputs where the function is differentiable.