Position, Velocity, and 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

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

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.

Example

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 $t = 2$:

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

Acceleration as the Second Derivative

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

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

Sign Conventions

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

Example

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.

Graphical Connections

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

From Graph	To Get	Operation
$x$-$t$	$v(t)$	Slope (derivative)
$v$-$t$	$a(t)$	Slope (derivative)
$v$-$t$	$\Delta x$	Area under curve (integral)
$a$-$t$	$\Delta v$	Area under curve (integral)

Key Principle

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!

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):

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

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

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

Integrating again:

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

With $x(0) = x_0$:

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

The Complete Set of Kinematic Equations

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

Equation	Missing Variable
$v = v_0 + at$	$x$
$x = x_0 + v_0 t + \frac{1}{2}at^2$	$v$ (final)
$v^2 = v_0^2 + 2a(x - x_0)$	$t$
$x = x_0 + \frac{1}{2}(v_0 + v)t$	$a$
$x = x_0 + vt - \frac{1}{2}at^2$	$v_0$

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

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

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.

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
$v = v_0 + at$	$v = v_0 - gt$
$y = y_0 + v_0 t + \frac{1}{2}at^2$	$y = y_0 + v_0 t - \frac{1}{2}gt^2$
$v^2 = v_0^2 + 2a\Delta y$	$v^2 = v_0^2 - 2g\Delta y$

Key Assumptions

• 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
• Total flight time: $6$ s
• Speed at landing: $30$ m/s (same as launch speed)

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.

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

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

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)
• $2 < t \le 3$: $v > 0$ (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.

1D Kinematics — Differentiation for Velocity and Acceleration

Part 5 of 7

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

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

Common Derivative Rules in Kinematics

$x(t)$	$v(t) = dx/dt$	$a(t) = dv/dt$
$At^n$	$nAt^{n-1}$	$n(n-1)At^{n-2}$
$A\sin(\omega t)$	$A\omega\cos(\omega t)$	$-A\omega^2\sin(\omega t)$
$Ae^{kt}$	$Ake^{kt}$	$Ak^2e^{kt}$
$A\ln(t)$	$A/t$	$-A/t^2$

Finding Turning Points and Direction Changes

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

Method

1. Find $v(t) = dx/dt$.
2. Solve $v(t) = 0$ for critical times.
3. 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
$0 < t < 2$	$-$	$+$	$-$	Slowing down
$t = 2$	$0$	$+$	$0$	Turning point
$t > 2$	$+$	$+$	$+$	Speeding up

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
$v(t)$	$-A\omega\sin(\omega t)$
$a(t)$	$-A\omega^2\cos(\omega t) = -\omega^2 x(t)$
$j(t)$	$A\omega^3\sin(\omega t)$

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.

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

1. Read carefully — identify what's given and what's asked.
2. Choose your approach — differentiation (given $x$, find $v$ or $a$) or integration (given $v$ or $a$, find $x$).
3. Apply initial conditions — use given values to determine constants of integration.
4. 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 $a(t) = 6t$ $m/s^{2}$. At $t = 10$ s, the engine cuts off and gravity takes over ($a = -10$ $m/s^{2}$). 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 = $\frac{4-(-4)}{4-0} = 2$)

$v = 0$ 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: $\int_0^2 |{-4+2t}|\,dt + \int_2^4 (2t-4)\,dt = \int_0^2(4-2t)\,dt + \int_2^4(2t-4)\,dt$ $= [4t-t^2]_0^2 + [t^2-4t]_2^4 = 4 + 4 = 8$ 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.

1D Kinematics — Review & Applications

Part 7 of 7 — Comprehensive Review

Formula Sheet

Relationship	Formula
Velocity from position	$v = \frac{dx}{dt}$
Acceleration from velocity	$a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$
Position from velocity	$x(t) = x_0 + \int_0^t v\,dt'$
Velocity from acceleration	$v(t) = v_0 + \int_0^t a\,dt'$
Constant accel: velocity	$v = v_0 + at$
Constant accel: position	$x = x_0 + v_0t + \frac{1}{2}at^2$
Constant accel: time-free	$v^2 = v_0^2 + 2a\Delta x$

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 $|a| = 8$ $m/s^{2}$) | |:---:|:---:| | $20$ m/s | $25$ m | | $30$ m/s | $56.25$ m | | $40$ m/s | $100$ 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 $t = 1$ and $t = 3$.

(b) Find the acceleration at each rest time.

$a(t) = 6t - 12$. $a(1) = -6$ $m/s^{2}$, $a(3) = 6$ $m/s^{2}$.

(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.