🎯⭐ INTERACTIVE LESSON

Position, Velocity, and Acceleration

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Position, Velocity, and Acceleration - Complete Interactive Lesson

Part 1: Position, Velocity & Acceleration

1D Kinematics — Position, Velocity & Acceleration

Part 1 of 7

In AP Physics C, kinematics is treated with the full power of calculus. We define the fundamental quantities as derivatives and integrals rather than simple ratios.

Definitions

Quantity	Symbol	Calculus Definition
Position	$x(t)$	Given or found by integration
Velocity	$v(t)$	$v = \frac{dx}{dt}$
Acceleration	$a(t)$	$a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$

Key Insight

Velocity is the rate of change of position, and acceleration is the rate of change of velocity:

$v(t) = \frac{dx}{dt}, \qquad a(t) = \frac{dv}{dt}$

These are instantaneous quantities — they describe motion at a single instant, not over an interval.

Average vs. Instantaneous Quantities

Average Velocity

$\bar{v} = \frac{\Delta x}{\Delta t} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}$

Acceleration as the Second Derivative

Since $a = \frac{dv}{dt}$ and $v = \frac{dx}{dt}$ , acceleration is the second derivative of position:

Graphical Connections

The derivative chain $x \to v \to a$ connects the three graphs:

From Graph	To Get	Operation
$x$ - $t$	$v(t)$

Part 2: Constant Acceleration Equations

1D Kinematics — Constant Acceleration Equations

Part 2 of 7

When acceleration is constant ( $a = \text{const}$ ), the calculus simplifies to a family of algebraic equations known as the kinematic equations.

Derivation from Calculus

Starting from $a(t) = a_0$ (constant):

Part 3: Free Fall

1D Kinematics — Free Fall

Part 3 of 7

Free fall is constant-acceleration motion with $a = -g$ (taking upward as positive), where $g \approx 9.8\text{ m/s}^2$ near Earth's surface.

Equations for Free Fall

General Form	Free-Fall Form

Part 4: Integration for Position

1D Kinematics — Integration for Position from Velocity

Part 4 of 7

One of the most important skills in AP Physics C is recovering position from a known velocity function using integration.

The Fundamental Relationship

$x(t) = x(t_0) + \int_{t_0}^{t} v(t')\,dt'$

Part 5: Differentiation for v & a

1D Kinematics — Differentiation for Velocity and Acceleration

Part 5 of 7

Given a position function $x(t)$ , we obtain velocity and acceleration through differentiation:

$v(t) = \frac{dx}{dt}, \qquad a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2}$

Part 6: Problem-Solving Workshop

1D Kinematics — Problem-Solving Workshop

Part 6 of 7

This workshop focuses on developing systematic problem-solving skills for AP Physics C kinematics problems. We'll work through progressively harder problems.

Problem-Solving Framework

Read carefully — identify what's given and what's asked.
Choose your approach — differentiation (given $x$ , find $v$ or $a$ ) or integration (given $v$ or $a$ , find ).

Part 7: Review & Applications

1D Kinematics — Review & Applications

Part 7 of 7 — Comprehensive Review

Formula Sheet

Relationship	Formula
Velocity from position	$v = \frac{dx}{dt}$
Acceleration from velocity

t2−t1x(t2)−x(t1)

This is the slope of the secant line on the $x$ - $t$ graph.

Instantaneous Velocity

$v(t) = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}$

This is the slope of the tangent line on the $x$ - $t$ graph.

Given $x(t) = t^3 - 6t$ :

Average velocity from $t = 1$ to $t = 3$ :

$\bar{v} = \frac{(27-18)-(1-6)}{3-1} = \frac{9+5}{2} = 7 \text{ m/s}$

Instantaneous velocity at $t = 2$ :

$v(t) = 3t^2 - 6 \implies v(2) = 12 - 6 = 6 \text{ m/s}$

$a(t) = \frac{d^2x}{dt^2}$

Condition	Motion Description
$v > 0$	Moving in the positive direction
$v < 0$	Moving in the negative direction
$a > 0, v > 0$	Speeding up (positive direction)
$a < 0, v > 0$	Slowing down (positive direction)
$a$ and $v$ same sign	Speeding up
$a$ and $v$ opposite sign	Slowing down

Given $x(t) = 2t^3 - 9t^2 + 12t$ :

$v(t) = 6t^2 - 18t + 12, \quad a(t) = 12t - 18$

At $t = 1$ : $v(1) = 0$ , $a(1) = -6$ . The particle is momentarily at rest and about to reverse direction.

The area under the $v$ - $t$ curve between $t_1$ and $t_2$ gives the displacement: $\Delta x = \int_{t_1}^{t_2} v(t)\,dt$

This is the fundamental theorem of calculus applied to motion!

$v(t) = \int a_0\,dt = a_0 t + C_1$

With initial condition $v(0) = v_0$ , we get $C_1 = v_0$ :

$\boxed{v(t) = v_0 + a_0 t}$

$x(t) = \int (v_0 + a_0 t)\,dt = v_0 t + \frac{1}{2}a_0 t^2 + C_2$

$\boxed{x(t) = x_0 + v_0 t + \frac{1}{2}a_0 t^2}$

The Complete Set of Kinematic Equations

For constant acceleration, five quantities are related: $x$ , $v$ , $v_0$ , $a$ , and $t$ .

Equation	Missing Variable
$v = v_0 + at$	$x$
$x = x_{0}$

The Time-Independent Equation

The third equation deserves special attention. It comes from eliminating $t$ :

From $v = v_0 + at$ , solve: $t = \frac{v - v_0}{a}$

Substitute into $x - x_0 = v_0 t + \frac{1}{2}at^2$ :

$v^2 = v_0^2 + 2a(x - x_0)$

This is equivalent to the work-energy theorem (as you'll see later in the course).

Problem-Solving Strategy

Step-by-Step Approach

Draw a diagram — label positive direction, origin, and key positions.
List knowns and unknowns — identify which of $x$ , $v_0$ , $v$ , $a$ , $t$ are given and which you need.
Choose the right equation — pick the one that contains your unknown and all your knowns.
Solve algebraically before substituting numbers.
Check units and reasonableness.

Worked Example

A train decelerates from $30$ m/s to $10$ m/s over $200$ m. Find the acceleration.

Known: $v_0 = 30$ m/s, $v = 10$ m/s, $\Delta x = 200$ m. .

Use $v^2 = v_0^2 + 2a\Delta x$ :

$100 = 900 + 2a(200)$ $-800 = 400a$ $a = -2 \text{ m/s}^2$

The negative sign confirms deceleration.

Multi-Step Problems

Many problems involve two phases of motion (e.g., acceleration then constant velocity, or two objects).

Example — Two-Phase Motion

A car accelerates from rest at $4\text{ m/s}^2$ for $5$ s, then travels at constant velocity. Find the total distance in $10$ s.

Phase 1 ( $0 \le t \le 5$ ): $x_1 = \frac{1}{2}(4)(25) = 50 \text{ m}$ $v_{\text{final}} = 0 + 4(5) = 20 \text{ m/s}$

Phase 2 ( $5 \le t \le 10$ ): $x_2 = 20 \times 5 = 100 \text{ m}$

Total: $x = 50 + 100 = 150$ m.

Air resistance is negligible.
$g$ is constant (valid near Earth's surface).
The only force is gravity.

Objects Thrown Upward

When a ball is thrown upward with speed $v_0$ :

Maximum Height

At the peak, $v = 0$ :

$0 = v_0^2 - 2g\Delta y \implies \Delta y_{\max} = \frac{v_0^2}{2g}$

Time to Peak

$0 = v_0 - gt_{\text{peak}} \implies t_{\text{peak}} = \frac{v_0}{g}$

Total Flight Time (returning to launch height)

By symmetry: $t_{\text{total}} = 2t_{\text{peak}} = \frac{2v_0}{g}$

Worked Example

A ball is thrown upward at $30$ m/s from the ground ( $g = 10$ m/s $^2$ ):

Max height: $\frac{30^2}{2(10)} = 45$ m
Time to top: $\frac{30}{10} = 3$ s

Symmetry of Free Fall

Free-fall trajectories exhibit beautiful symmetry:

Property	Going Up	Coming Down
Speed at height $h$	$\sqrt{v_0^2 - 2gh}$	$\sqrt{v_0^2 - 2gh}$ (same!)
Time to reach height $h$	$t_1$	$t_{\text{total}} - t_1$
Acceleration	$-g$ (always)	$-g$ (always)

Calculus Proof of Symmetry

Position: $y(t) = v_0 t - \frac{1}{2}gt^2$

Setting $y(t) = y(t')$ for $t \neq t'$ :

$v_0 t - \frac{1}{2}gt^2 = v_0 t' - \frac{1}{2}gt'^2$

$v_0(t - t') = \frac{1}{2}g(t^2 - t'^2) = \frac{1}{2}g(t-t')(t+t')$

Dividing by $(t - t')$ : $t + t' = \frac{2v_0}{g}$

So the two times when the object is at the same height are symmetric about the midpoint $t = v_0/g$ .

Calculus Approach to Free Fall

Starting from Newton's second law for free fall:

$\frac{d^2y}{dt^2} = -g$

Integrate once (with initial condition $v(0) = v_0$ ):

$\frac{dy}{dt} = -gt + v_0$

Integrate again (with $y(0) = y_0$ ):

$y(t) = -\frac{1}{2}gt^2 + v_0 t + y_0$

When does $v = 0$ ?

$\frac{dy}{dt} = 0 \implies -gt + v_0 = 0 \implies t = \frac{v_0}{g}$

Checking with the second derivative test:

$\frac{d^2y}{dt^2} = -g < 0$

Since the second derivative is negative, $y$ has a maximum at $t = v_0/g$ . This confirms it's the peak height, not a minimum.

This is the antiderivative approach: position is the integral of velocity.

Displacement vs. Distance

Quantity	Formula	Meaning
Displacement	$\int_{t_1}^{t_2} v\,dt$	Net change in position (can be negative)
Distance	$\int_{t_1}^{t_2}	v

When velocity changes sign, displacement and distance differ!

Integration Techniques in Kinematics

Polynomial Velocity

If $v(t) = at^n$ , then:

$x(t) = \frac{a}{n+1}t^{n+1} + C$

Trigonometric Velocity

If $v(t) = A\cos(\omega t)$ :

$x(t) = \frac{A}{\omega}\sin(\omega t) + C$

Exponential Velocity

If $v(t) = v_0 e^{-kt}$ :

$x(t) = -\frac{v_0}{k}e^{-kt} + C$

Worked Example

A particle has $v(t) = 12\sin(3t)$ m/s and $x(0) = 0$ .

$x(t) = \int 12\sin(3t)\,dt = -4\cos(3t) + C$

From $x(0) = 0$ : $0 = -4\cos(0) + C = -4 + C \implies C = 4$ .

$x(t) = 4 - 4\cos(3t) = 4[1 - \cos(3t)]$

Computing Total Distance Traveled

When $v(t)$ changes sign, you must split the integral at the zeros.

Example

$v(t) = t^2 - 4$ for $0 \le t \le 3$ .

Step 1: Find where $v = 0$ : $t^2 = 4 \implies t = 2$ .

Step 2: Check signs:

$0 \le t < 2$ : $v < 0$ (moving left)
$2 < t \le 3$ : (moving right)

Step 3: Compute:

$\text{Displacement} = \int_0^3 (t^2-4)\,dt = \left[\frac{t^3}{3} - 4t\right]_0^3 = (9-12) = -3$

$\text{Distance} = \int_0^2 |t^2-4|\,dt + \int_2^3 (t^2-4)\,dt$

$= \int_0^2 (4-t^2)\,dt + \int_2^3 (t^2-4)\,dt$

$= \left[4t - \frac{t^3}{3}\right]_0^2 + \left[\frac{t^3}{3} - 4t\right]_2^3$

$= \left(8 - \frac{8}{3}\right) + \left(-3 - (-\frac{16}{3})\right) = \frac{16}{3} + \frac{7}{3} = \frac{23}{3}$

Using Definite Integrals from Data

On the AP exam, you may need to compute $\int v\,dt$ from a table or graph.

From a Table

$t$ (s)	$0$	$1$	$2$	$3$	$4$
$v$ (m/s)	$2$	$5$	$6$	$4$	$1$

Using the trapezoidal rule:

$\int_0^4 v\,dt \approx \frac{1}{2}\sum_{i} (v_i + v_{i+1})\Delta t$

$= \frac{1}{2}[(2+5) + (5+6) + (6+4) + (4+1)](1) = \frac{1}{2}(33) = 16.5 \text{ m}$

This gives an approximation of the displacement over the interval.

Common Derivative Rules in Kinematics

$x(t)$	$v(t) = dx/dt$	$a(t) = dv/dt$
$At^n$	$nAt^{n-1}$	$n(n-1)At^{n-2}$

Finding Turning Points and Direction Changes

A particle changes direction when $v(t) = 0$ and $v$ changes sign.

Method

Find $v(t) = dx/dt$ .
Solve $v(t) = 0$ for critical times.
Check sign of $v$ on either side (or use $a(t)$ at that point).

Worked Example

$x(t) = t^3 - 12t + 5$

$v(t) = 3t^2 - 12 = 3(t^2 - 4)$

$v = 0$ when $t = \pm 2$ . For $t > 0$ , the turning point is at $t = 2$ .

Check: $v(1) = 3 - 12 = -9 < 0$ , $v(3) = 27 - 12 = 15 > 0$ .

So the particle moves left for $0 < t < 2$ and right for $t > 2$ .

Position at turning point: $x(2) = 8 - 24 + 5 = -11$ .

Using the Second Derivative

At $t = 2$ : $a(2) = 6(2) = 12 > 0$ .

Since $v = 0$ and $a > 0$ , velocity is changing from negative to positive — confirming a direction reversal.

Speeding Up vs. Slowing Down

A particle is:

Speeding up when $v$ and $a$ have the same sign ( $v \cdot a > 0$ )
Slowing down when $v$ and $a$ have opposite signs ( $v \cdot a < 0$ )

Analysis Method

For $x(t) = t^3 - 12t$ :

$v(t) = 3t^2 - 12$ , $\quad a(t) = 6t$

Interval	$v$	$a$	$v \cdot a$	Motion
$0 < t < 2$

Key Insight

Speed is $|v(t)|$ . You can also check by computing $\frac{d}{dt}|v(t)|$ .

When $|v|$ is increasing, the particle speeds up. When $|v|$ is decreasing, it slows down.

Higher-Order Analysis: Jerk

The rate of change of acceleration is called jerk:

$j(t) = \frac{da}{dt} = \frac{d^3x}{dt^3}$

While rarely asked on the AP exam, understanding the derivative chain is important:

$x(t) \xrightarrow{d/dt} v(t) \xrightarrow{d/dt} a(t) \xrightarrow{d/dt} j(t)$

Example: Simple Harmonic Motion Preview

If $x(t) = A\cos(\omega t)$ :

Quantity	Expression
$v(t)$	$-A\omega\sin(\omega t)$
$a(t)$

Notice: $a(t) = -\omega^2 x(t)$ . This means acceleration is proportional to (and opposite in sign to) position — the hallmark of simple harmonic motion.

Apply initial conditions — use given values to determine constants of integration.

Verify — check units, signs, and limiting cases.

Worked Problem 1: Multi-Phase Motion

Problem: A rocket launches vertically from rest. For $0 \le t \le 10$ s, its acceleration is $a(t) = 6t$ m/s². At $t = 10$ s, the engine cuts off and gravity takes over ( $a = -10$ m/s²). Find the maximum height.

Solution

Phase 1 ( $0 \le t \le 10$ ):

$v(t) = \int 6t\,dt = 3t^2 + C$

$v(0) = 0 \implies C = 0$ , so $v(t) = 3t^2$ .

At $t = 10$ : $v(10) = 300$ m/s.

$y(t) = \int 3t^2\,dt = t^3 + C'$

$y(0) = 0 \implies C' = 0$ , so $y(t) = t^3$ .

At $t = 10$ : $y(10) = 1000$ m.

Phase 2 ( $t > 10$ , let $\tau = t - 10$ ):

$v = 300 - 10\tau = 0 \implies \tau = 30 \text{ s}$

$\Delta y = 300(30) - \frac{1}{2}(10)(900) = 9000 - 4500 = 4500 \text{ m}$

Maximum height: $1000 + 4500 = 5500$ m.

Worked Problem 2: Graphs to Equations

Problem: The $v$ - $t$ graph shows a particle with:

$v(0) = -4$ m/s
$v$ increases linearly to $v(4) = 4$ m/s
$v$ remains constant at $4$ m/s for $4 \le t \le 6$

Find the total distance traveled and displacement from $t = 0$ to $t = 6$ .

Solution

Phase 1 ( $0 \le t \le 4$ ): $v(t) = -4 + 2t$ (slope = $\frac{4 - (- 4)}{4 - 0}$ )

$v = 0$ at $t = 2$ s (direction change).

Displacement Phase 1: $\int_0^4 (-4+2t)\,dt = [-4t + t^2]_0^4 = -16 + 16 = 0$

Distance Phase 1: $\int_0^2 |{-4+2t}|\,dt + \int_2^4 (2t-4)\,dt = \int_0^2(4-2t)\,dt + \int_2^4(2t-4)\,dt$ m

Phase 2 ( $4 \le t \le 6$ ): $v = 4$ m/s (constant)

Displacement = Distance = $4 \times 2 = 8$ m

Totals: Displacement $= 0 + 8 = 8$ m, Distance $= 8 + 8 = 16$ m.

a = \frac{dv}{dt} = \frac{d^2x}{dt^2}

Key Concepts Summary

Differentiation: $x \to v \to a$
Integration: $a \to v \to x$ (with initial conditions!)
Displacement $\neq$ distance when velocity changes sign
Speeding up: $v$ and $a$ same sign
Turning points: $v = 0$ and $v$ changes sign

Application: Braking Distance

A car traveling at speed $v_0$ brakes with constant deceleration $a$ (where $a < 0$ ). The stopping distance is:

$v^2 = v_0^2 + 2a \cdot d \implies 0 = v_0^2 + 2ad$

$d = \frac{v_0^2}{2|a|}$

Key Insight

Stopping distance is proportional to $v_0^2$ . Doubling the speed quadruples the stopping distance.

| Speed | Stopping Distance (if $|a| = 8$ m/s²) | |:---:|:---:| | $20$ m/s | $25$ m | | $30$ m/s | $56.25$ m | | $40$ m/s | m |

With Reaction Time

If the driver takes time $t_r$ to react:

$d_{\text{total}} = v_0 t_r + \frac{v_0^2}{2|a|}$

This combines linear and quadratic dependence on $v_0$ .

AP-Style Problem

Problem: A particle moves along the $x$ -axis with velocity $v(t) = 3t^2 - 12t + 9$ for $t \ge 0$ . At $t = 0$ , the particle is at position $x = 2$ .

(a) Find all times when the particle is at rest.

$v(t) = 3(t^2 - 4t + 3) = 3(t-1)(t-3) = 0$ at and .

(b) Find the acceleration at each rest time.

$a(t) = 6t - 12$ . $a(1) = -6$ m/s², $a(3) = 6$ m/s².

(c) Find the position at $t = 3$ .

$x(3) = 2 + \int_0^3(3t^2 - 12t + 9)\,dt = 2 + [t^3 - 6t^2 + 9t]_0^3 = 2 + (27-54+27) = 2$

(d) Find the total distance from $t = 0$ to $t = 3$ .

Split at $t = 1$ : distance $= |x(1) - x(0)| + |x(3) - x(1)|$

$x(1) = 2 + [t^3-6t^2+9t]_0^1 = 2 + 4 = 6$

Distance $= |6-2| + |2-6| = 4 + 4 = 8$ m.

Topic Complete!

You've mastered 1D Kinematics for AP Physics C:

Part	Topic	Status
1	Position, velocity, acceleration	✅
2	Constant acceleration equations	✅
3	Free fall	✅
4	Integration for position	✅
5	Differentiation for velocity/acceleration	✅
6	Problem-solving workshop	✅
7	Review & applications	✅

AP Exam Tip: On free-response problems, always show your calculus work explicitly. Write $v = dx/dt$ or $x = \int v\,dt$ before evaluating — the setup earns points even if arithmetic is wrong.

x = x_{0} + v_{0} t + \frac{1}{2} a t^{2}

Total flight time:

6

Speed at landing:

30

m/s (same as launch speed)

- A ω^{2} cos (ω t) = - ω^{2} x (t)

\frac{4 - ( - 4 )}{4 - 0} = 2

= [4t-t^2]_0^2 + [t^2-4t]_2^4 = 4 + 4 = 8

Position, Velocity, and Acceleration

Position, Velocity, and Acceleration - Complete Interactive Lesson

Part 1: Position, Velocity & Acceleration

1D Kinematics — Position, Velocity & Acceleration

Definitions

Key Insight

Average vs. Instantaneous Quantities

Average Velocity

Acceleration as the Second Derivative

Graphical Connections

Part 2: Constant Acceleration Equations

1D Kinematics — Constant Acceleration Equations

Derivation from Calculus

Part 3: Free Fall

1D Kinematics — Free Fall

Equations for Free Fall

Part 4: Integration for Position

1D Kinematics — Integration for Position from Velocity

The Fundamental Relationship

Part 5: Differentiation for v & a

1D Kinematics — Differentiation for Velocity and Acceleration

Part 6: Problem-Solving Workshop

1D Kinematics — Problem-Solving Workshop

Problem-Solving Framework

Part 7: Review & Applications

1D Kinematics — Review & Applications

Formula Sheet

Instantaneous Velocity

Example

Sign Conventions

Example

Key Principle

The Complete Set of Kinematic Equations

The Time-Independent Equation

Problem-Solving Strategy

Step-by-Step Approach

Worked Example

Multi-Step Problems

Example — Two-Phase Motion

Key Assumptions

Objects Thrown Upward

Maximum Height

Time to Peak

Total Flight Time (returning to launch height)

Worked Example

Symmetry of Free Fall

Calculus Proof of Symmetry

Calculus Approach to Free Fall

When does v=0v = 0v=0?

Checking with the second derivative test:

Displacement vs. Distance

Integration Techniques in Kinematics

Polynomial Velocity

Trigonometric Velocity

Exponential Velocity

Worked Example

Computing Total Distance Traveled

Example

Using Definite Integrals from Data

From a Table

Common Derivative Rules in Kinematics

Finding Turning Points and Direction Changes

Method

Worked Example

Using the Second Derivative

Speeding Up vs. Slowing Down

Analysis Method

Key Insight

Higher-Order Analysis: Jerk

Example: Simple Harmonic Motion Preview

Worked Problem 1: Multi-Phase Motion

Solution

Worked Problem 2: Graphs to Equations

Solution

Key Concepts Summary

Application: Braking Distance

Key Insight

With Reaction Time

AP-Style Problem

Topic Complete!

When does $v = 0$ ?