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LESSON 05 / 16 · TOPIC 1.2

From a secant to a derivative

You will be able to: Interpret a derivative as a limiting slope and differentiate a simple position function.

Calculus-based kinematicsFree study resourceReview editionTeacher review pending

How do you find velocity at one instant?

A tracking camera records a cart whose position is x(t) = (1 m/s²)t². Between 2 s and 3 s it moves 5 m. But the velocity at exactly 2 s is 4 m/s, not 5 m/s.

A useful starting point: Average velocity, speed and acceleration →

Words and symbols before equations

Function x(t)
A rule that assigns a position to each time.
Secant
A line through two points on a curve; its slope gives an interval average.
Tangent
The limiting direction of a smooth curve at one point.
Derivative dx/dt
Instantaneous rate of change of position with time, measured in m/s.
Power rule
d(tⁿ)/dt = ntⁿ⁻¹; constant multipliers remain, constant terms differentiate to zero.
Secant approaches tangent on x(t) = t²position (m)time (s)0-411.527312.5418Orange: secant · purple: tangent
Read this model snapshot. At t = 2 s: tangent slope 4 m/s. With h = 1 s, secant slope 5 m/s. Difference 1 m/s.
What this picture assumes

Exact smooth function x = (1 m/s²)t². Orange secant uses t and t+h; purple tangent is at t. This is a mathematical model without measurement noise.

Read the picture in three steps

  1. Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
  2. At t = 2 s: tangent slope 4 m/s. With h = 1 s, secant slope 5 m/s. Difference 1 m/s.
  3. Check what the picture assumes below. Use the Explore task to predict one change before moving a control.

Connect the picture to the physics

Use a short positive interval h after time t. For x = kt², the slope is [k(t+h)² − kt²]/h = 2kt + kh. As h approaches zero, the extra kh approaches zero, leaving v(t) = 2kt. This is what the derivative means.

The notation dx/dt names a limiting rate, not an ordinary finite distance divided by a literal zero. You can use the power rule once you understand the slope interpretation. For x = x₀ + bt + ct², v = b + 2ct. The constant x₀ changes height on the graph, not its slope.

Acceleration is the derivative of velocity: a = dv/dt = d²x/dt². Here a = 2c, with units m/s². A derivative must use the complete expression before inserting the requested time; differentiating a single already-evaluated position number discards the time dependence.

Average and instantaneous velocity
PropertyAverageInstantaneous
TimeAn intervalOne instant
GraphSecant slopeTangent slope
RelationΔx/Δtdx/dt

A worked example, step by step

For x(t) = 2 m + (3 m/s)t + (1 m/s²)t², find v and a at t = 2 s.

  1. Differentiate each term: the constant 2 m gives zero.
  2. v(t) = 3 m/s + (2 m/s²)t.
  3. At 2 s, v = 3 + 4 = +7 m/s.
  4. Differentiate again: a = +2 m/s² at every time. Positive velocity and acceleration mean speed is increasing here.
Common mix-up

The value of x/t is not generally dx/dt. Average from the origin and instantaneous slope answer different questions.

CHECK THE IDEA

Does adding 20 m to every position change velocity?

Compare with an explanation

No. The graph shifts vertically, but all slopes are unchanged.

Now investigate one change Explore →

Predict. Change one thing. Explain.

Hold t = 2 s. Shrink h from 1 s toward 0.05 s and compare the orange secant slope with the purple tangent slope. Explain the remaining difference.

On narrow screens, swipe or scroll diagrams sideways to read all labels.

Secant approaches tangent on x(t) = t²position (m)time (s)0-411.527312.5418Orange: secant · purple: tangent

At t = 2 s: tangent slope 4 m/s. With h = 1 s, secant slope 5 m/s. Difference 1 m/s.

Compare slopesm/s · same scale for all bars0Secant Δx/h5Tangent dx/dt4

Exact smooth function x = (1 m/s²)t². Orange secant uses t and t+h; purple tangent is at t. This is a mathematical model without measurement noise.

Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant position, velocity, acceleration or integral relationship to justify your prediction.

Apply the idea to a fresh problem Practice →

Show what you understand.

Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.

1. If x = (2 m/s³)t³, v(t) is…

Show answer and reasoning

(6 m/s³)t². The power rule multiplies by three and lowers the exponent to two.

2. A horizontal tangent to x(t) means…

Show answer and reasoning

v = 0. The position graph’s slope is instantaneous velocity, regardless of its height.

Original written challenge

4 points · self-check · not an official AP question

For x(t) = 1 m − (4 m/s)t + (2 m/s²)t², derive v(t), derive a(t), find the time when v = 0, and describe the direction before and after that time.

This response is not submitted or saved. Copy it before leaving.

Compare with the answer and four-point rubric
  1. 1 point: v(t) = −4 m/s + (4 m/s²)t.
  2. 1 point: a(t) = +4 m/s².
  3. 1 point: v = 0 at t = 1 s.
  4. 1 point: For 0 ≤ t < 1 s velocity is negative; after 1 s it is positive, so this is a reversal.

Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.

Recall the ideas without notes Review →

Retrieve it before you reveal it.

RECALL 1What does a secant slope give?

Average velocity over its time interval.

RECALL 2What does a tangent slope give?

Instantaneous velocity.

RECALL 3Why differentiate before substituting time?

The time-dependent rule is needed to determine its rate of change.

Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.

From a secant to a derivative

  • v(t) = dx/dt; a(t) = dv/dt.
  • d(Ctⁿ)/dt = nCtⁿ⁻¹.
  • Derivative units divide the original quantity’s units by seconds.

Remember: The value of x/t is not generally dx/dt. Average from the origin and instantaneous slope answer different questions.

Conditions: Exact smooth function x = (1 m/s²)t². Orange secant uses t and t+h; purple tangent is at t. This is a mathematical model without measurement noise.

Refresh Kid · AP Physics C: Mechanics Unit 1 (official Unit 1) · Objectives 1.2.C · Review edition

Framework, scope and review status

Mapped to College Board CED, Topic 1.2, objectives 1.2.C. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. This is Mechanics Unit 1: Kinematics. Quantitative motion problems stay in one or two dimensions; the optional spatial view clarifies vector notation. Students should study calculus alongside this course. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.

Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.

Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.

Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.

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