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Stopping distance and reaction time

Free notes · JEE Main Physics

Formulas

s=u22as = \frac{u^2}{2a}

Stopping distance from speed u

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