AP PHYSICS 1 — physicsMedium3 min read

Kinematics

Master kinematics for AP Physics 1: motion in one and two dimensions, kinematic equations, projectile motion, and graphical analysis.

Tutor AI Team2026-02-26T00:06:57.429Z

# Kinematics — AP Physics 1

Kinematics is the study of motion without considering the forces that cause it. It is one of the foundational topics in AP Physics 1, forming the basis for nearly every other unit. You'll need to describe motion using displacement, velocity, and acceleration, and apply kinematic equations to solve problems in one and two dimensions.

Key Concepts

Position, Displacement, and Distance

Position ( $x$ or $y$ ): location of an object relative to a reference point.
Displacement ( $\Delta x = x_f - x_i$ ): change in position (vector quantity).
Distance: total path length traveled (scalar quantity).

Velocity and Speed

Average velocity: $\bar{v} = \dfrac{\Delta x}{\Delta t}$
Instantaneous velocity: velocity at a specific moment in time.
Speed: magnitude of velocity (scalar).

Acceleration

Average acceleration: $\bar{a} = \dfrac{\Delta v}{\Delta t}$
An object accelerates when its speed changes, its direction changes, or both.

The Kinematic Equations (Constant Acceleration)

$v = v_0 + at$

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

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

$\Delta x = \frac{v + v_0}{2} \cdot t$

Free Fall

Objects near Earth's surface experience $g \approx 9.8\ \text{m/s}^2$ downward.
In free fall (ignoring air resistance), $a = -g$ (taking upward as positive).

Projectile Motion

Horizontal: $a_x = 0$ , so $v_x = v_{0x}$ and $\Delta x = v_{0x}t$ .
Vertical: $a_y = -g$ , standard kinematic equations apply.
Components are independent of each other.
Launch velocity: $v_{0x} = v_0 \cos\theta$ , $v_{0y} = v_0 \sin\theta$ .

Motion Graphs

Position vs. time: slope = velocity.
Velocity vs. time: slope = acceleration; area under curve = displacement.
Acceleration vs. time: area under curve = change in velocity.

Worked Example

Problem: A ball is thrown vertically upward from ground level with an initial speed of $20\ \text{m/s}$ . Find (a) the maximum height and (b) the total time in the air.

Solution:

(a) At maximum height, $v = 0$ .

$v^2 = v_0^2 + 2a\Delta y$ $0 = (20)^2 + 2(-9.8)\Delta y$ $\Delta y = \frac{400}{19.6} \approx 20.4\ \text{m}$

(b) Time to reach the top: $v = v_0 + at$ $0 = 20 + (-9.8)t$ $t_{\text{up}} = \frac{20}{9.8} \approx 2.04\ \text{s}$

Total time (symmetric trajectory): $t_{\text{total}} = 2 \times 2.04 \approx 4.08\ \text{s}$

Practice Questions

1. A car accelerates uniformly from rest at $3\ \text{m/s}^2$ for $8\ \text{s}$ . How far does it travel?

$\Delta x = 0 + \frac{1}{2}(3)(8)^2 = 96\ \text{m}$

2. A projectile is launched at $30°$ above the horizontal at $40\ \text{m/s}$ . What is the horizontal range? (Assume level ground.)

$v_{0x} = 40\cos 30° = 34.6\ \text{m/s}$ , $v_{0y} = 40\sin 30° = 20\ \text{m/s}$ . Time of flight $= 2v_{0y}/g = 4.08\ \text{s}$ . Range $= 34.6 \times 4.08 \approx 141\ \text{m}$ .

3. From a velocity–time graph, the velocity drops linearly from $10\ \text{m/s}$ to $0$ over $5\ \text{s}$ . What is the displacement?

Area under the curve = $\frac{1}{2}(10)(5) = 25\ \text{m}$ .

4. An object is dropped from a height of $45\ \text{m}$ . How long does it take to hit the ground?

$45 = \frac{1}{2}(9.8)t^2 \Rightarrow t = \sqrt{\frac{90}{9.8}} \approx 3.03\ \text{s}$ .

Want to check your answers and get step-by-step solutions?

Summary

Kinematics describes motion using displacement, velocity, and acceleration.
The four kinematic equations apply only under constant acceleration.
Projectile motion separates into independent horizontal and vertical components.
Graphs of position, velocity, and acceleration are interconnected through slopes and areas.

Tags:

#ap#physics#kinematics

Ready to Ace Your AP PHYSICS 1 physics?

Get instant step-by-step solutions to any problem. Snap a photo and learn with Tutor AI — your personal exam prep companion.