Unit 4 · Topic 4.2

4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration

For a particle moving along a line, position, velocity and acceleration are linked by derivatives: v(t) = x′(t) and a(t) = v′(t). Signs tell the story. The sign of velocity gives the direction, and comparing the signs of velocity and acceleration tells you whether the particle is speeding up or slowing down.

Key terms

position
velocity
acceleration
speed
particle motion

Position, velocity, acceleration

Each one is the derivative of the one before it:

Position x(t) (or s(t)): where the particle is on the number line at time t.
Velocity v(t) = x′(t): how fast position changes, with a sign for direction. v > 0 means moving right (or up); v < 0 means moving left (or down); v = 0 means the particle is momentarily at rest.
Acceleration a(t) = v′(t) = x″(t): how fast velocity changes.
Speed = |v(t)|: how fast, with no direction. Speed is never negative.

Changing direction

A particle changes direction when its velocity changes sign, from positive to negative or the other way. v(t) = 0 alone isn't enough; the velocity might touch 0 and keep the same sign. To justify a direction change, show the sign of v on both sides of the time.

Speeding up or slowing down

Speed is increasing when velocity and acceleration have the same sign, and decreasing when they have opposite signs.

Why? If v > 0 and a > 0, velocity is positive and getting bigger, so speed grows. If v < 0 and a < 0, velocity is negative and getting more negative, so its size grows: the particle speeds up while moving left. If the signs differ, velocity is moving toward 0, so speed shrinks.

Velocity	Acceleration	The particle is
Positive	Positive	Moving right, speeding up
Positive	Negative	Moving right, slowing down
Negative	Negative	Moving left, speeding up
Negative	Positive	Moving left, slowing down

Reading motion from graphs

On a graph of v(t): the particle moves right when the graph is above the t-axis and left when it is below. Acceleration is the slope of the velocity graph. Speed is increasing when the graph is moving away from the t-axis, and decreasing when it is moving toward the axis.

On a graph of x(t): velocity is the slope. Where the position graph has a peak or valley, the particle turns around.

What you won't need here

Finding position from velocity, or total distance from a velocity formula, uses integrals and comes in Unit 8. In this topic, you start from position (or velocity) and differentiate.

Worked examples

Try each one yourself first, then open the solution.

Example 1
A full motion analysis
A particle moves along the x-axis with position x(t) = t³ − 6t² + 9t for t ≥ 0. (a) Find v(t) and a(t). (b) When is the particle moving left? (c) Is the particle speeding up or slowing down at t = 2.5? Justify.
Show the solutionHide the solution
1. Step 1: (a) v(t) = 3t² − 12t + 9 = 3(t − 1)(t − 3), and a(t) = 6t − 12.
2. Step 2: (b) v(t) = 0 at t = 1 and t = 3. Test the sign on each interval: v(0.5) = 3.75 > 0, v(2) = −3 < 0, v(4) = 9 > 0.
3. Step 3: So v(t) < 0 on 1 < t < 3. The particle moves left there.
4. Step 4: (c) v(2.5) = 18.75 − 30 + 9 = −2.25 < 0 and a(2.5) = 15 − 12 = 3 > 0.
5. Step 5: Velocity and acceleration have opposite signs.
Answer: (a) v(t) = 3t² − 12t + 9, a(t) = 6t − 12. (b) For 1 < t < 3. (c) Slowing down at t = 2.5, because v(2.5) < 0 and a(2.5) > 0 have opposite signs.
Example 2
Total distance from position values
For the same particle, x(t) = t³ − 6t² + 9t, find the total distance traveled from t = 0 to t = 4.
Show the solutionHide the solution
1. Step 1: The particle turns around at t = 1 and t = 3 (where v changes sign), so split the trip there.
2. Step 2: x(0) = 0, x(1) = 1 − 6 + 9 = 4, x(3) = 27 − 54 + 27 = 0, x(4) = 64 − 96 + 36 = 4.
3. Step 3: Leg distances: |4 − 0| = 4, then |0 − 4| = 4, then |4 − 0| = 4.
4. Step 4: Total = 4 + 4 + 4 = 12.
Answer: 12 units. (The displacement is x(4) − x(0) = 4, which is different from the distance.)
Example 3
Trap: negative acceleration doesn't always mean slowing down
At t = 5, a particle has v(5) = −4 m/s and a(5) = −2 m/s². A student says the particle is slowing down because the acceleration is negative. Is that right?
Show the solutionHide the solution
1. Step 1: Speeding up or slowing down depends on whether v and a have the same sign, not on the sign of a alone.
2. Step 2: Here v and a are both negative.
3. Step 3: The velocity is becoming more negative, from −4 toward −6 and beyond, so its size, the speed, is increasing.
Answer: No. The particle is speeding up at t = 5, because velocity and acceleration are both negative.

Common mistakes

Saying negative acceleration means slowing down. Compare the signs of v and a.
Claiming a direction change wherever v(t) = 0 without checking that v changes sign.
Mixing up displacement (final position minus starting position) with total distance (add the length of each leg).

On the exam

Particle motion is a classic free-response question. Typical parts: find when the particle changes direction, whether speed is increasing at a time, and the acceleration at a time. Justify with signs.
On calculator-active motion questions, you'll often be given v(t) and asked for a(t) at a point: use the calculator's numerical derivative and report three decimals.

Connected topics

Videos

Calculus AB/BC – 4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration
The AlgebrosWatch on YouTube (opens in a new tab)
Introduction to one-dimensional motion with calculus | AP Calculus AB | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)
AP Calculus AB 4.2 Rectilinear Motion: Left or Right, Velocity Increase or Decrease, Displacement
Math Teacher GOATWatch on YouTube (opens in a new tab)
❖ Position, Velocity, Acceleration using Derivatives ❖
Patrick JWatch on YouTube (opens in a new tab)
Calculus - Working with position, velocity, and acceleration
MySecretMathTutorWatch on YouTube (opens in a new tab)
Worked example: Motion problems with derivatives | AP Calculus AB | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)