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Physics: Motion to AP Physics 1 · Module 1 of 8

Motion in One Dimension

Displacement, velocity and acceleration, the kinematic equations, motion graphs, and free fall.

In this module
  1. Displacement, Velocity & Acceleration
  2. The Kinematic Equations
  3. Motion Graphs & Free Fall
  4. Check yourself

Displacement, Velocity & Acceleration

  • Displacement is change in position, with a direction. Distance is the total path length, with no direction.
  • Velocity = displacement ÷ time (a vector). Speed is its magnitude. Acceleration = change in velocity ÷ time.
  • An object slows down when its acceleration points opposite its velocity, whatever the sign convention.
Worked example

You walk 30 m east, then 10 m west, in 20 s. Find your distance, displacement, and average velocity.

Solution. Distance = 40 m. Displacement = 20 m east. Average velocity = 20 m ÷ 20 s = 1.0 m/s east.

The Kinematic Equations

  • For constant acceleration: v = v₀ + at; Δx = v₀t + ½at²; v² = v₀² + 2aΔx.
  • List the knowns (v₀, v, a, t, Δx), identify the unknown, and pick the equation that leaves out the quantity you don't know and don't need.
Worked example

A car starts from rest and accelerates at 3.0 m/s² for 4.0 s. Find its final speed and the distance covered.

Solution. v = 0 + (3.0)(4.0) = 12 m/s. Δx = 0 + ½(3.0)(4.0)² = 24 m.

Motion Graphs & Free Fall

  • Position–time graph: the slope is velocity. Velocity–time graph: the slope is acceleration, and the area under the curve is displacement.
  • In free fall (ignoring air resistance) every object accelerates downward at g ≈ 9.8 m/s², regardless of its mass.
  • At the top of a vertical throw the velocity is momentarily zero, but the acceleration is still g downward.
Reading a velocity–time graph
t (s)v (m/s)slope = a = 3 m/s²area = Δx = 24 m12 m/s
A car speeds up from rest at 3.0 m/s² for 4.0 s. The slope of the v–t graph is the acceleration (3 m/s²), and the shaded area under it is the displacement: ½ × 4.0 s × 12 m/s = 24 m.
Worked example

A ball is dropped from rest. What are its speed and the distance it has fallen after 2.0 s?

Solution. v = gt = (9.8)(2.0) = 19.6 m/s. d = ½gt² = ½(9.8)(2.0)² = 19.6 m.

Common mistake: Zero velocity does NOT mean zero acceleration. At the peak of its flight, a thrown ball is still accelerating at 9.8 m/s² downward.

Check yourself

2 of this module's 5 quiz questions. Pick your answers, then check them in the app, with an explanation for every question.

1. On a velocity–time graph, what does the area under the curve represent?

2. A cyclist speeds up from 2 m/s to 10 m/s in 4 s. What is her acceleration?

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