Kinematics in One Dimension

AP Physics C: Mechanics· AP Physics C: Mechanics CED — Kinematics· 14 min read

1. Core Definitions of 1D Kinematics★☆☆☆☆⏱ 3 min

Kinematics is the branch of mechanics that describes the motion of objects without reference to the forces that cause motion. One-dimensional kinematics restricts all motion to a single straight line, so all motion quantities have only a magnitude and a sign (positive/negative, indicating direction along the line).

📘 Definition

Displacement vs Distance Traveled

Displacement ( $Δ x$ ) is the net change in position: $Δ x = x_{f} - x_{i}$ , a signed quantity. Distance traveled is the total length of the path taken, always non-negative.

Example:

A ball thrown up and caught at launch has 0 displacement but non-zero distance traveled

📘 Definition

Average Velocity vs Average Speed

Average velocity is displacement divided by time interval, a signed quantity. Average speed is total distance traveled divided by time, always non-negative.

Example:

The throw-and-catch ball has 0 average velocity but non-zero average speed

2. Calculus Relationships for Motion Quantities★★☆☆☆⏱ 4 min

All kinematic quantities are defined by relationships between position, time, velocity, and acceleration. Average quantities are defined over a finite time interval $Δ t = t_{f} - t_{i}$ :

ar{v} = rac{ riangle x}{ riangle t} = rac{x_f - x_i}{t_f - t_i}, \quad ar{a} = rac{ riangle v}{ riangle t} = rac{v_f - v_i}{t_f - t_i}

Instantaneous quantities are the limit of average quantities as $Δ t$ approaches 0, resulting in the derivative definitions core to AP Physics C:

v(t) = rac{dx}{dt}, \quad a(t) = rac{dv}{dt} = rac{d^2x}{dt^2}

To reverse this relationship (get velocity from acceleration, position from velocity), use integration with the constant of integration fixed by initial conditions:

v(t) = v_0 + \\int_0^t a(\tau) d\tau, \quad x(t) = x_0 + \\int_0^t v(\tau) d\tau

📐 Worked Example

The position of a toy car moving along a straight track is given by $x (t) = 2 t^{3} - 12 t^{2} + 10 t + 4$ , where $x$ is in meters and $t$ in seconds for $t \geq 0$ . Find (a) the instantaneous velocity at $t = 2$ s, and (b) the average velocity between $t = 1$ s and $t = 3$ s.

1
For instantaneous velocity, use the derivative definition:
2
$latex: v(t) = \frac{dx}{dt} = 6t^2 - 24t + 10$
3
Substitute $t = 2$ s:
4
$latex: v(2) = 6(2^2) - 24(2) + 10 = -14 \, \text{m/s}$
5
The negative sign indicates velocity points in the negative x-direction. For average velocity, calculate position at the interval endpoints:
6
$latex: x(1) = 4 \, \text{m}, \quad x(3) = -20 \, \text{m}$
7
Calculate average velocity:
8
$latex: \bar{v} = \frac{x(3) - x(1)}{3 - 1} = \frac{-20 - 4}{2} = -12 \, \text{m/s}$

3. Constant Acceleration Kinematics★★☆☆☆⏱ 4 min

When acceleration is constant ( $a (t) = a$ for all $t$ ), the general integration rules simplify to a set of three widely used kinematic equations. These are extremely useful for problems like free fall, constant braking, etc., but are only valid when acceleration is constant.

\begin{aligned} v(t) &= v_0 + at \\ x(t) &= x_0 + v_0 t + \frac{1}{2} a t^2 \\ v^2 &= v_0^2 + 2a(x - x_0) \end{aligned}

📐 Worked Example

A ball is thrown straight upward from ground level with an initial speed of 24 m/s. Acceleration due to gravity is 9.8 m/s² downward. What is the maximum height the ball reaches before it begins falling back down?

1
Define the coordinate system: origin at ground level ( $x_{0} = 0$ ), positive x upward. Acceleration points downward, so:
2
$latex: a = -9.8 \, \text{m/s}^2$
3
At maximum height, the ball stops moving upward before reversing direction, so its instantaneous velocity is $v = 0$ at this point.
4
We know initial velocity, final velocity, and acceleration, so we use the time-independent kinematic equation to avoid solving for time first:
5
$latex: v^2 = v_0^2 + 2a(x - x_0)$
6
Substitute values and solve for maximum height $h$ :
7
$latex: 0^2 = (24)^2 + 2(-9.8)(h - 0) \implies h = \frac{576}{19.6} \approx 29.4 \, \text{m}$

4. Graphical Analysis of One-Dimensional Motion★★★☆☆⏱ 3 min

AP Physics C heavily tests the ability to interpret the three common motion graphs, with relationships that follow directly from calculus definitions:

On an $x$ vs $t$ graph: Slope at any point = instantaneous velocity; average slope between two points = average velocity.
On a $v$ vs $t$ graph: Slope at any point = instantaneous acceleration; net area under the graph between two times = displacement $Δ x$ .
On an $a$ vs $t$ graph: Area under the graph between two times = change in velocity $Δ v$ .

📐 Worked Example

The velocity of a car accelerating from rest along a straight highway follows $v (t) = 2 t$ for $0 \leq t \leq 10$ s, and $v (t) = 40 - 2 t$ for $10 < t \leq 20$ s, where $v$ is in m/s. What is the total displacement of the car between $t = 0$ and $t = 20$ s?

1
By definition, displacement equals the net area under the $v (t)$ curve between the two times.
2
The $v (t)$ curve forms a triangle with base 20 s and maximum height 20 m/s. Calculate area directly:
3
$latex: \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \times 20 = 200 \, \text{m}$
4
Integration confirms the result:
5
$latex: \triangle x = \\int_0^{10} 2t dt + \\int_{10}^{20} (40-2t) dt = 100 + 100 = 200 \, \text{m}$

5. Additional Practice Worked Examples★★★☆☆⏱ 4 min

✓ Quick check

Test your understanding of average acceleration:

The position of an object moving along the x-axis is given by $x (t) = t^{3} - 3 t^{2}$ , where $x$ is in meters and $t$ is in seconds. What is the average acceleration between $t = 1$ s and $t = 3$ s?
- 0 m/s²
- 3 m/s²
- 6 m/s²
- 12 m/s²
Reveal answer
6 m/s² —
Average acceleration is $Δ v /Δ t$ . Differentiate to get $v (t) = 3 t^{2} - 6 t$ , so $v (1) = - 3$ m/s, $v (3) = 9$ m/s, $Δ v = 12$ m/s, $Δ t = 2$ s, giving 6 m/s².

📐 Worked Example

A high-speed passenger train is traveling at 80 m/s along a straight track when the driver spots a broken rail 1500 m ahead and triggers emergency braking. Emergency braking gives the train a constant deceleration of 2.5 m/s². Determine if the train stops before reaching the broken rail, and if so, how far the train is from the broken rail when it comes to rest.

1
Define origin at the train's position when braking is triggered, positive x toward the broken rail. Initial conditions: $x_{0} = 0$ , $v_{0} = 80$ m/s, $a = - 2.5$ m/s². When stopped, $v = 0$ .
2
Use the time-independent constant acceleration equation:
3
$latex: v^2 = v_0^2 + 2a(x - x_0) \implies 0 = 80^2 + 2(-2.5)x \implies x = 1280 \, \text{m}$
4
Calculate distance from the broken rail:
5
$latex: 1500 - 1280 = 220 \, \text{m}$
6
The train stops 220 m before the broken rail, avoiding collision.

6. Common Pitfalls

Wrong move:

Using constant-acceleration kinematic equations when acceleration is non-constant.

Why:

Students default to memorized simple equations even when acceleration is given as a function of time or position.

Correct move:

Always check if acceleration is constant before using these equations; if not, use integration of $a (t)$ to find $v (t)$ and $x (t)$ .

Wrong move:

Using displacement instead of total distance traveled to calculate average speed.

Why:

Students confuse displacement (net change in position) with distance (total path length).

Correct move:

Always calculate average speed as total distance traveled divided by time, never displacement divided by time.

Wrong move:

Forgetting to add initial velocity/position when integrating acceleration/velocity.

Why:

Students forget the constant of integration is fixed by initial conditions, not zero by default.

Correct move:

Always add the initial condition when writing the result of an integral for kinematics.

Wrong move:

Mixing up slope and area for motion graphs, e.g., taking area under a position vs time graph as displacement.

Why:

Students confuse the order of differentiation between position, velocity, and acceleration.

Correct move:

Remember: slope of lower = higher (x→v→a, so slope of x is v, slope of v is a), area of higher = lower (area of a is Δv, area of v is Δx).

Wrong move:

Assuming gravity is always positive regardless of coordinate system direction.

Why:

Students memorize gravity as 9.8 m/s² and forget to adjust the sign to match the coordinate system.

Correct move:

Always assign the sign of acceleration based on direction relative to your defined positive axis before plugging into equations.

7. Quick Reference Cheatsheet

Category	Formula	Notes
Average Velocity	$\overset{v}{ˉ} = \frac{Δ x}{Δ t} = \frac{x _{f} - x _{i}}{t _{f} - t _{i}}$	Displacement over time, direction-sensitive
Average Acceleration	$\overset{a}{ˉ} = \frac{Δ v}{Δ t} = \frac{v _{f} - v _{i}}{t _{f} - t _{i}}$	Change in velocity over time
Instantaneous Velocity	$v (t) = \frac{d x}{d t}$	Derivative of position with respect to time
Instantaneous Acceleration	$a (t) = \frac{d v}{d t} = \frac{d ^{2} x}{d t ^{2}}$	Derivative of velocity, second derivative of position
General Velocity from Acceleration	$v (t) = v_{0} + \int_{0}^{t} a (τ) d τ$	Works for any acceleration, constant or non-constant
General Position from Velocity	$x (t) = x_{0} + \int_{0}^{t} v (τ) d τ$	Works for any velocity, constant or non-constant
Constant Acceleration: Velocity	$v = v_{0} + a t$	Only valid for constant $a$
Constant Acceleration: Position	$x = x_{0} + v_{0} t + \frac{1}{2} a t^{2}$	Only valid for constant $a$
Constant Acceleration: Time-Independent	$v^{2} = v_{0}^{2} + 2 a (x - x_{0})$	Only valid for constant $a$
Motion Graph Rules	Slope = derivative, Area = integral	Slope of $x$ - $t$ = $v$ , slope of $v$ - $t$ = $a$ ; area of $a$ - $t$ = $Δ v$ , area of $v$ - $t$ = $Δ x$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Average vs instantaneous velocity
2022 · FRQ
Non-constant acceleration kinematics

Going deeper

syllabusOfficial AP Physics C CEDOfficial course framework

What's Next

One-dimensional kinematics is the fundamental foundation for all subsequent topics in AP Physics C: Mechanics. You will next extend these calculus-based definitions of position, velocity, and acceleration to two dimensions to solve problems involving projectile motion and relative motion. Without mastering the sign conventions, derivative/integral relationships, and constant acceleration rules you learned here, multi-dimensional kinematics and all later topics involving motion will be significantly harder to master. This topic also feeds directly into Newton's second law of motion, where you will connect acceleration to net force, and later into energy and momentum, where you will integrate acceleration and velocity to find work and impulse.