Kinematics: AP Physics 1 Unit 1

Why this unit matters

Clear motion models are the base for every later unit, including forces, energy, and momentum.

What you will learn

Interpret and connect position-time, velocity-time, and acceleration-time graphs.
Solve constant-acceleration problems with correct signs, units, and known assumptions.
Model projectile motion by separating horizontal and vertical components.

Understand the core ideas

Kinematics describes motion without attributing causes. Start by defining an origin, axis directions, and symbols before writing equations. Displacement is the change in position and can be positive or negative. Average velocity is displacement divided by elapsed time, while instantaneous velocity is represented by the slope of a position-time graph at a point. Acceleration is the rate of change of velocity and controls how the velocity vector evolves over time.

For constant acceleration in one dimension, AP Physics 1 uses connected equations such as v = v0 + at and $x = x0 + v0 t + 0.5 a t^2$ . Graphs provide the same information in another form: slope of x-t gives velocity, slope of v-t gives acceleration, and area under v-t gives displacement. In projectile motion, horizontal and vertical components are analyzed independently under the standard model of negligible air resistance, with horizontal acceleration zero and vertical acceleration equal to -g when up is positive.

A strong solution routine is: choose signs first, list knowns with units, select the equation matching the unknown, then check whether your result has a reasonable magnitude and direction. This prevents common sign errors and keeps units consistent from setup through final answer.

Key terms

displacement: Change in position from initial to final point, including direction.
velocity: Rate of change of position; direction matters and sign follows axis choice.
acceleration: Rate of change of velocity, measured in meters per second squared.
projectile motion: Two-dimensional motion modeled with independent horizontal and vertical components.

Horizontal launch from a table

A ball leaves a 1.20 m high table horizontally at 4.0 m/s. Neglect air resistance. Use +x to the right and +y upward.

Write the vertical displacement equation: delta $y = v0y t + 0.5 a_y t^2$ , with delta y = -1.20 m, v0y = 0, and $a_y = -9.8 m/s^2$ .
Solve for time: $-1.20 = 0.5(-9.8)t^2$ gives $t^2 = 1.20/4.9 = 0.245$ , so t = 0.495 s.
Use horizontal motion with zero horizontal acceleration: x = v_x t = (4.0 m/s)(0.495 s) = 1.98 m.
Check units and sign: meters per second times seconds gives meters, and positive x indicates landing to the right of the launch point.

Result: The ball is in flight for about 0.50 s and lands about 2.0 m from the table edge.

A common misconception

Claim: Gravity reduces the horizontal speed of a projectile in this model.

Correction: With negligible air resistance, gravity changes only vertical velocity. Horizontal velocity remains constant during flight.

Lessons in this unit

Position, displacement, and velocityDistinguish vector and scalar ideas and interpret motion direction from signs and graphs.
Acceleration and motion graphsRelate slope and area on motion graphs to physical quantities.
Constant-acceleration equationsChoose and apply kinematic equations from known and unknown variables.
Projectile motionRepresent 2D motion with independent horizontal and vertical models.

Study task

A ball is launched horizontally from a table. Create a diagram of known quantities, write separate horizontal and vertical equations, and estimate time of flight and horizontal range for a chosen table height and launch speed.

Unit checkpoint

A car starts from rest and accelerates at

2.0 m/s^2

for 5.0 s. What is its final velocity?

Use v = v0 + at. With v0 = 0, v = 0 + (2.0)(5.0) = 10 m/s.

Kinematics