AP Physics C: Mechanics — One-Dimensional Kinematics (Unit 1 Study Notes)

Position, Velocity, and Acceleration

Describing motion with a coordinate axis

In one-dimensional kinematics, you describe motion along a straight line (the $x$-axis). The key idea is that you can represent where an object is and how it moves using signed numbers. That sign matters: it tells you direction.

• Position is a location on the axis. You write it as position $x(t)$, emphasizing that position can change with time.
• A reference point (often $x = 0$) and a positive direction must be chosen before anything else makes sense. Once chosen, you keep them consistent.

A subtle but crucial point: kinematics cares about change in position, not absolute position. If you shift your origin, the physics of motion does not change—only the numerical values of $x$ do.

Displacement vs. distance

Displacement is the signed change in position:

$\Delta x = x_f - x_i$

Here, $x_i$ is initial position and $x_f$ is final position.

Displacement can be positive, negative, or zero. Distance is different: it is the total length traveled (always nonnegative). In 1D, students often confuse them because sometimes they match (motion in one direction only), and sometimes they do not (you go out and come back).

Why it matters: On AP Physics C, almost every kinematics equation uses displacement (signed), not distance. If you accidentally substitute distance into equations that expect displacement, your sign and direction will be wrong.

Average and instantaneous velocity

Velocity tells you how fast and in what direction position changes.

The average velocity over a time interval $\Delta t = t_f - t_i$ is:

$v_{avg} = \frac{\Delta x}{\Delta t}$

Average velocity depends only on the start and end positions (not on the path taken in between). This is why an object can travel a large distance but have $v_{avg} = 0$ if it ends where it started.

For AP Physics C, you also need the calculus definition: the instantaneous velocity is the derivative of position:

$v(t) = \frac{dx}{dt}$

How to interpret it:

• Velocity is the slope of the position vs. time graph at that instant.
• The sign of $v(t)$ tells direction: positive means moving in the positive $x$ direction; negative means moving in the negative direction.

A common conceptual trap is thinking “negative velocity” means “slowing down.” It doesn’t. Negative velocity only means the object is moving in the negative direction. Whether it is speeding up or slowing down requires comparing the signs of velocity and acceleration.

Average and instantaneous acceleration

Acceleration describes how velocity changes.

The average acceleration over an interval is:

$a_{avg} = \frac{\Delta v}{\Delta t}$

The instantaneous acceleration is:

$a(t) = \frac{dv}{dt}$

Combining with the velocity definition gives the second derivative relationship:

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

Why it matters: Acceleration is what connects forces to motion later in the course (Newton’s second law). Even in pure kinematics, acceleration tells you how motion evolves—whether velocity is constant, increasing, decreasing, or changing direction.

Reading motion from graphs (a core AP skill)

Graph interpretation is one of the highest-yield skills in 1D kinematics.

1. On an $x$ vs. $t$ graph:

   • Slope gives velocity: steeper slope means larger speed.
   • Curvature tells you about acceleration: if the slope is getting more positive, acceleration is positive.

2. On a $v$ vs. $t$ graph:

   • Slope gives acceleration.
   • The area under the curve (signed area) gives displacement:

$\Delta x = \int_{t_i}^{t_f} v(t)\,dt$

3. On an $a$ vs. $t$ graph:
   • The area under the curve gives change in velocity:

$\Delta v = \int_{t_i}^{t_f} a(t)\,dt$

Signed area matters. A region below the time axis contributes negative displacement or negative change in velocity.

Notation you may see (and how to translate it)

Different problems use different symbols for the same ideas. You should be bilingual in notation.

Worked example: displacement vs. distance

You start at $x = 2\,\text{m}$, walk to $x = 9\,\text{m}$, then return to $x = 5\,\text{m}$.

• Displacement:

$\Delta x = x_f - x_i = 5 - 2 = 3\,\text{m}$

• Distance traveled:
  • First leg: $9 - 2 = 7\,\text{m}$
  • Second leg: $9 - 5 = 4\,\text{m}$
  • Total distance: $7 + 4 = 11\,\text{m}$

You can see how displacement (net change) and distance (total travel) differ.

Worked example: interpreting sign of acceleration

At an instant, an object has $v = -4\,\text{m/s}$ and $a = +2\,\text{m/s}^2$.

• The negative velocity means it’s moving in the negative $x$ direction.
• The positive acceleration means the velocity is becoming more positive over time.

Because acceleration points opposite the velocity, the object is slowing down at that instant. This is the clean rule:

• If $v$ and $a$ have the same sign, speed increases.
• If $v$ and $a$ have opposite signs, speed decreases.

Exam Focus

• Typical question patterns
  • Interpret slopes and areas on $x$-$t$, $v$-$t$, and $a$-$t$ graphs to find $v$, $a$, $\Delta x$, or $\Delta v$.
  • Distinguish displacement vs. distance and average vs. instantaneous quantities.
  • Determine whether an object is speeding up or slowing down based on signs of $v$ and $a$.
• Common mistakes
  • Treating negative velocity as “deceleration” automatically (direction and speeding up are separate ideas).
  • Forgetting that areas under graphs are signed (below the axis counts negative).
  • Mixing up average velocity $\Delta x/\Delta t$ with average speed (distance divided by time).

Motion with Constant Acceleration

What “constant acceleration” really means

Constant acceleration means $a(t)$ is the same value at all times in the interval you are analyzing. The direction can be positive or negative, but it does not change with time.

This model is powerful because it generates a small set of equations that connect position, velocity, acceleration, and time. Many real motions are approximately constant-acceleration over short intervals—most famously free fall near Earth’s surface (ignoring air resistance), where acceleration is roughly constant and downward.

Why it matters: These constant-acceleration relationships are some of the most frequently used tools on AP Physics C Mechanics, and they often serve as “building blocks” for more complicated, piecewise motions.

Building the kinematic equations from calculus (concept first)

If acceleration is constant, then by definition:

$a = \frac{dv}{dt}$

Integrating with respect to time gives velocity as a linear function of time:

$v(t) = v_0 + at$

Here, $v_0$ is the velocity at $t = 0$ (or whatever you choose as your start time).

Next, velocity is the derivative of position:

$v(t) = \frac{dx}{dt}$

Substitute $v(t) = v_0 + at$ and integrate again:

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

These aren’t just formulas to memorize—they express a story:

• Velocity changes steadily (linearly) because acceleration is constant.
• Position changes with a quadratic term because it accumulates the changing velocity.

The full set of constant-acceleration relations (and when to use them)

You’ll commonly use these four relationships in 1D constant-acceleration problems:

$v = v_0 + at$

$x = x_0 + v_0 t + \frac{1}{2}at^2$

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

$x - x_0 = \frac{1}{2}(v + v_0)t$

Variable meanings:

• $x$: position at time $t$
• $x_0$: initial position
• $v$: velocity at time $t$
• $v_0$: initial velocity
• $a$: constant acceleration
• $t$: elapsed time

How to choose: Pick the equation that contains the quantities you know and the quantity you need, while avoiding an extra unknown. For example, if time isn’t given and isn’t asked for, $v^2 = v_0^2 + 2a(x - x_0)$ often avoids needing $t$.

Physical meaning through graphs

Under constant acceleration:

• The $a$ vs. $t$ graph is a horizontal line.
• The $v$ vs. $t$ graph is a straight line with slope $a$. The area under it is displacement.
• The $x$ vs. $t$ graph is a parabola (concave up if $a>0$, concave down if $a<0$).

Seeing these shapes helps you catch errors. For example, if acceleration is constant and you draw a curved $v$ vs. $t$ graph, something is inconsistent.

Handling free fall carefully (sign conventions)

Near Earth, the gravitational acceleration magnitude is about $9.8\,\text{m/s}^2$, but on AP exams the sign depends on your coordinate choice:

• If you choose “up” as positive, then free-fall acceleration is $a = -g$.
• If you choose “down” as positive, then $a = +g$.

Neither is “more correct”—but mixing them mid-problem is a classic way to lose points.

Worked problem: braking to a stop

A car moving at $v_0 = 20\,\text{m/s}$ brakes with constant acceleration $a = -5\,\text{m/s}^2$. How long until it stops, and how far does it travel during braking?

Step 1: Time to stop
Stopping means final velocity $v = 0$. Use

$v = v_0 + at$

Substitute:

$0 = 20 + (-5)t$

Solve:

$t = 4\,\text{s}$

Step 2: Braking distance
Use the position equation (taking $x_0 = 0$ for convenience):

$x = x_0 + v_0 t + \frac{1}{2}at^2$

$x = 0 + 20(4) + \frac{1}{2}(-5)(4^2)$

$x = 80 - 40 = 40\,\text{m}$

The displacement is positive because the car continues moving in the positive direction while slowing.

Worked problem: vertical throw and “turning point”

You throw a ball upward with $v_0 = 15\,\text{m/s}$ from height $x_0 = 1.5\,\text{m}$. Take upward as positive and use $a = -9.8\,\text{m/s}^2$. Find the maximum height.

At maximum height, the velocity is momentarily zero: $v = 0$. Use the time-free equation:

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

Substitute:

$0 = 15^2 + 2(-9.8)(x - 1.5)$

Compute:

$0 = 225 - 19.6(x - 1.5)$

$19.6(x - 1.5) = 225$

$x - 1.5 = \frac{225}{19.6}$

$x \approx 1.5 + 11.48 = 12.98\,\text{m}$

Common misconception: Students sometimes think acceleration is zero at the top because velocity is zero. But gravity is still acting, so acceleration remains $-9.8\,\text{m/s}^2$ throughout the flight (ignoring air resistance).

Exam Focus

• Typical question patterns
  • Use kinematic equations to relate $x$, $v$, $a$, and $t$ for a single constant-acceleration interval (free fall, braking, launch straight up).
  • Multi-part questions: find time to reach a condition (stop, max height), then use that to find displacement or speed.
  • Connect graphs to equations: identify $a$ from slope of $v$-$t$ or displacement from area.
• Common mistakes
  • Plugging in $g = 9.8$ without a sign, then “fixing” signs later (choose a coordinate direction first).
  • Assuming acceleration becomes zero when velocity is zero at a turning point.
  • Using constant-acceleration equations in situations where acceleration is not constant (often hidden in word problems involving drag, springs, or changing forces).

Motion with Non-Constant Acceleration (Calculus)

When constant-acceleration formulas stop working

Many real situations have acceleration that changes with time, position, or velocity—for example:

• A car whose acceleration varies as the driver shifts gears.
• Motion with air resistance (acceleration depends on velocity).
• Any situation where the net force changes with position (you’ll see this later with springs and energy).

In those cases, the constant-acceleration equations are not valid. Instead, you return to the definitions of velocity and acceleration as derivatives and use calculus to connect them.

The core calculus relationships

The definitions always hold:

$v(t) = \frac{dx}{dt}$

$a(t) = \frac{dv}{dt}$

If you know acceleration as a function of time, you can integrate to get velocity:

$v(t) = v(t_0) + \int_{t_0}^{t} a(\tau)\,d\tau$

Then integrate velocity to get position:

$x(t) = x(t_0) + \int_{t_0}^{t} v(\tau)\,d\tau$

This “acceleration integrates to velocity integrates to position” chain is the backbone of non-constant kinematics.

Why it matters: AP Physics C expects you to be comfortable moving between derivatives, integrals, and graphs. Even when a problem can be done algebraically, calculus often provides the cleanest reasoning and reduces memorization.

Graph-based integration: area under curves

Even if you don’t have an explicit formula, you can use areas:

• From an $a$ vs. $t$ graph, the signed area gives $\Delta v$.
• From a $v$ vs. $t$ graph, the signed area gives $\Delta x$.

This comes directly from the integral meaning: integrals accumulate rate of change over time.

If acceleration depends on position: using the chain rule

Sometimes you are told $a$ as a function of $x$, not time. Then you can’t directly integrate $a(t)$. A very useful identity comes from the chain rule:

Start with:

$a = \frac{dv}{dt}$

Write $dv/dt$ as $dv/dx$ times $dx/dt$:

$\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt}$

But $dx/dt = v$, so:

$a = v\frac{dv}{dx}$

This is powerful because it connects acceleration, velocity, and position without time.

Worked problem: time-dependent acceleration

A particle moves in 1D with acceleration

$a(t) = 6t$

Given $v(0) = 2\,\text{m/s}$ and $x(0) = -1\,\text{m}$, find $v(t)$ and $x(t)$.

Step 1: Integrate acceleration to get velocity
Use

$v(t) = v(0) + \int_{0}^{t} 6\tau\,d\tau$

Compute the integral:

$\int_{0}^{t} 6\tau\,d\tau = 6\cdot \frac{t^2}{2} = 3t^2$

So

$v(t) = 2 + 3t^2$

Step 2: Integrate velocity to get position

$x(t) = x(0) + \int_{0}^{t} (2 + 3\tau^2)\,d\tau$

Compute:

$\int_{0}^{t} 2\,d\tau = 2t$

$\int_{0}^{t} 3\tau^2\,d\tau = 3\cdot \frac{t^3}{3} = t^3$

Thus

$x(t) = -1 + 2t + t^3$

A good self-check is differentiating: $dx/dt = 2 + 3t^2$ matches $v(t)$, and differentiating again gives $6t$.

Worked problem: acceleration as a function of position

A particle has acceleration

$a(x) = 4x$

and at $x = 0$ its speed is $v_0 = 3\,\text{m/s}$ (take that as the instant where you define the initial condition). Find $v$ as a function of $x$.

Step 1: Use the chain-rule identity

$a = v\frac{dv}{dx}$

Substitute $a(x)=4x$:

$4x = v\frac{dv}{dx}$

Step 2: Separate variables and integrate
Rewrite as:

$v\,dv = 4x\,dx$

Integrate from the initial condition $x=0, v=3$ to a general position $x, v$:

$\int_{3}^{v} u\,du = \int_{0}^{x} 4s\,ds$

Compute both sides:

$\frac{1}{2}(v^2 - 3^2) = 2x^2$

So

$v^2 = 9 + 4x^2$

If you want speed, you take the positive root:

$v = \sqrt{9 + 4x^2}$

If you want velocity (including direction), the sign depends on the direction of motion; the equation for $v^2$ alone does not encode whether the particle is moving left or right.

Common pitfall: Forgetting that integrating can introduce a sign ambiguity when you take square roots. AP questions sometimes probe whether you understand that speed and velocity are different.

Piecewise motion: using calculus with intervals

Many AP problems quietly become “piecewise” even in 1D. For instance, acceleration might be:

• One function for $0 \le t \le 2$
• Another function after $t = 2$

In that case, you integrate on each interval and use the end conditions of one interval as the initial conditions of the next. Conceptually, you’re ensuring that position and velocity remain continuous at the transition time unless the problem explicitly introduces an impulse-like event (which is not typical in basic kinematics).

Connecting back to constant acceleration

Constant acceleration is just the special case where $a(t)=a$ is constant. Then

$v(t) = v_0 + \int_{0}^{t} a\,d\tau = v_0 + at$

and you recover the usual equations. This is a useful perspective: if you remember how to integrate, you can rebuild kinematics instead of relying purely on memorization.

Exam Focus

• Typical question patterns
  • Given $a(t)$ or a graph of $a$ vs. $t$, integrate to find $v(t)$, then integrate again (or use areas) to find displacement.
  • Given $v(t)$, differentiate to find $a(t)$, or integrate to find $x(t)$.
  • Use $a = v(dv/dx)$ when acceleration is given as a function of position to relate speed and position without solving explicitly for time.
• Common mistakes
  • Using constant-acceleration equations even though $a$ varies (a red flag is that the problem gives $a(t)$, $a(x)$, or a non-horizontal $a$ vs. $t$ graph).
  • Dropping initial conditions when integrating (you need constants of integration or definite integrals with bounds).
  • Confusing “area under curve” with “area magnitude”—remember areas are signed and can cancel.