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Kinematic equations in one dimension

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Kinematic Equations in One Dimension

The kinematic equations describe motion in a straight line at constant acceleration, linking initial velocity, final velocity, acceleration, time, and displacement. Learn the four equations, how to choose the right one, how a velocity-time graph shows acceleration and displacement, and a worked example.

What the kinematic equations are

The kinematic equations are a set of formulas that describe motion in a straight line when acceleration is constant. They connect five quantities: initial velocity (v₀), final velocity (v), acceleration (a), time (t), and displacement (s). Knowing any three lets you solve for the rest. They rest on the definitions of position, velocity, acceleration, and time.

The four equations

  • v = v₀ + at
  • s = v₀t + ½at²
  • v² = v₀² + 2as
  • s = ½(v₀ + v)t

Each equation leaves out one of the five quantities, so you pick the equation that skips the variable you neither know nor need.

Reading a velocity–time graph

Because acceleration is constant, a graph of velocity against time is a straight line. Two features of that line carry all the physics:

Velocity–time graph for constant acceleration A straight line rising from an initial velocity, plotted on velocity (vertical) against time (horizontal). The slope of the line equals the acceleration, and the shaded trapezoid area under the line equals the displacement. v t slope = a area = displacement v₀
On a velocity–time graph, the slope is the acceleration and the area beneath the line is the displacement.

The slope of the line equals the acceleration a, and the area under the line equals the displacement s. That area is a trapezoid, which is exactly what the equation s = ½(v₀ + v)t computes.

Worked example

A car starts at v₀ = 2 m/s and accelerates at a = 1.5 m/s² for t = 6 s. Final velocity: v = 2 + 1.5×6 = 11 m/s. Displacement: s = 2×6 + ½×1.5×6² = 12 + 27 = 39 m.

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