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Physics guide

Kinematics equations: motion with constant acceleration

Displacement, velocity and acceleration, the four constant-acceleration equations, how to pick the right one, and free-fall problems worked step by step.

Kinematics describes motion without worrying about what causes it. In most intro physics classes you will use four equations that work whenever acceleration is constant. The hard part is not the algebra; it is choosing the right equation. This guide gives you a system for that.

The quantities

  • Displacement (Δx): change in position, with a direction. Walking 5 m forward and 5 m back gives a displacement of 0, even though you traveled 10 m.
  • Velocity (v): rate of change of displacement, in m/s, with direction. Speed is velocity without direction.
  • Acceleration (a): rate of change of velocity, in m/s². Slowing down while moving forward means acceleration points backward.
  • Time (t): in seconds.

We usually write v₀ for initial velocity and v for final velocity.

The four equations

  1. v = v₀ + at (no Δx)
  2. Δx = v₀t + ½at² (no v)
  3. v² = v₀² + 2aΔx (no t)
  4. Δx = ½(v₀ + v)t (no a)

Each equation leaves out exactly one variable. That is the key to choosing.

How to pick the right equation

  1. List the five variables: Δx, v₀, v, a, t.
  2. Write down the three you know and the one you want.
  3. The fifth variable is the one you do not know and do not need. Pick the equation that leaves it out.

Worked example 1: a car accelerating

A car starts from rest and accelerates at 3.0 m/s² for 5.0 s. How far does it travel?

  • Known: v₀ = 0, a = 3.0 m/s², t = 5.0 s. Want: Δx. Not needed: v.
  • Use equation 2: Δx = (0)(5.0) + ½(3.0)(5.0)² = ½ × 3.0 × 25 = 37.5 m.

Worked example 2: braking distance

A cyclist moving at 12 m/s brakes with an acceleration of −4.0 m/s². How far does she travel before stopping?

  • Known: v₀ = 12 m/s, v = 0, a = −4.0 m/s². Want: Δx. Not needed: t.
  • Use equation 3: 0² = 12² + 2(−4.0)Δx, so 0 = 144 − 8.0Δx and Δx = 18 m.

Free fall

Near Earth's surface, ignoring air resistance, every falling object accelerates downward at g ≈ 9.8 m/s², no matter its mass. Choose a sign convention and stick with it. If up is positive, then a = −9.8 m/s².

Worked example 3: a ball thrown upward

A ball is thrown straight up at 14.7 m/s. How high does it go, and how long until it returns to the thrower's hand?

  1. At the top, the velocity is momentarily 0. Using v² = v₀² + 2aΔx: 0 = 14.7² + 2(−9.8)Δx, so Δx = 216.09 / 19.6 ≈ 11 m.
  2. Time to the top: v = v₀ + at, so 0 = 14.7 − 9.8t and t = 1.5 s.
  3. By symmetry, the trip down takes another 1.5 s, so the total is 3.0 s.

Reading motion graphs

  • On a position-time graph, the slope is velocity. A straight line means constant velocity; a curve means acceleration.
  • On a velocity-time graph, the slope is acceleration and the area under the line is displacement.

Common mistakes

  • Mixing signs, such as treating g as positive while calling up positive.
  • Using these equations when acceleration is not constant.
  • Confusing distance and displacement.
  • Forgetting that "starts from rest" means v₀ = 0 and "comes to a stop" means v = 0.

Practice questions

  1. A train speeds up from 10 m/s to 30 m/s in 8.0 s. What is its acceleration?
  2. A rock is dropped from a 45 m cliff. How long does it take to hit the ground?
  3. A runner accelerates from rest at 2.0 m/s² over 25 m. What is her final speed?

Answers: (1) a = (30 − 10) / 8.0 = 2.5 m/s². (2) 45 = ½(9.8)t², so t ≈ 3.0 s. (3) v² = 0 + 2(2.0)(25) = 100, so v = 10 m/s.

Last reviewed 2026-10-23 by the BrainBoost team.

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