Stopping distance and reaction time

Stopping distance grows as the square of speed — doubling speed quadruples the stopping distance.

Stopping distance and reaction time0 min · Free lecture

In this lesson

  • Stopping distance grows as the square of speed — doubling speed quadruples the stopping distance.

At 2× speed you need 4× the stopping distance. That is why speed limits exist.

From v² = u² + 2as with v = 0: s = u²/2a. Doubling u quadruples s.

With reaction time τ, add the reaction distance u·τ travelled before braking begins.

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

Transcript (0 min)
WEBVTT

1
00:00:00.000 --> 00:00:05.000
Stopping distance and reaction time — FemtoLearn.
2
00:00:05.000 --> 00:00:15.000
Stopping distance grows as the square of speed — doubling speed quadruples the stopping distance.
Printable notes (free)

Practice

Free · 4 questions with full solutions
  1. Q1 · Numerical · difficulty 2/5

    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.

  2. Q2 · MCQ · difficulty 2/5

    A particle moves with constant velocity. Which a–t graph describes it?

    • A horizontal line at a = 0
    • A horizontal line at a = 2 m/s²
    • A straight line through the origin
    • A parabola
  3. Q3 · Numerical · difficulty 2/5

    A body moves with v = 10 m/s for 3 s, then v = 5 m/s for 2 s in the same direction. Find the total distance.

  4. Q4 · MCQ · difficulty 2/5

    The slope of a velocity–time graph gives:

    • Displacement
    • Acceleration
    • Speed
    • Distance