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Dimensional sanity checks in kinematics

Free notes · JEE Main Physics

Formulas

[x]=L, [v]=LT−1, [a]=LT−2[x] = L,\ [v] = LT^{-1},\ [a] = LT^{-2}

Dimensions of kinematic quantities

Worked example

The acceleration of a body starting from rest is a = 2 m/s² for 4 s, then zero for 2 s. Find its velocity at t = 6 s.

  1. Area under the a–t graph = change in velocity.
  2. Area = 2 m/s² × 4 s = 8 m/s (the last 2 s add nothing).

Answer: 8