How does an airbag reduce injury in a crash?

It increases the time taken to stop, so the deceleration and force are smaller

Typical speeds, reaction times and dangerous decelerations

Physics questions often expect you to estimate. Learn some typical speeds, how to measure reaction time, and why large decelerations are dangerous.

GCSE · 1PH0 · 2.12, 2.13, 2.27, 2.31
About 15 minutes

Practise this lesson free (6 questions)

Typical speeds and accelerations

Values are typical; real speeds vary.
Moving thing	Typical speed
Walking	about 1.5 m/s
Running	about 3 m/s
Cycling	about 6 m/s
Car on a motorway (70 mph)	about 31 m/s
Sound in air	about 330 m/s

Free fall and reaction time

An object in free fall near the Earth accelerates at about g = 10 m/s².
Measuring reaction time: a partner drops a ruler between your finger and thumb, and you catch it. The distance it falls, d, gives the time from d = ½ × g × t², so t = √(2d ÷ g).
Example: the ruler falls 0.20 m. t = √(2 × 0.20 ÷ 10) = √0.04 = 0.2 s.
Typical reaction times are about 0.2 to 0.9 s. Tiredness, alcohol, drugs and distractions make them longer.

Large decelerations

Stopping suddenly means a large deceleration, so a large force acts on the people inside (F = m × a).
Example: a 1000 kg car stops from 20 m/s in 0.1 s in a crash. a = 20 ÷ 0.1 = 200 m/s², so F = 1000 × 200 = 200 000 N.
Seat belts, airbags and crumple zones make the stop take longer, which lowers the deceleration and so the force.

Worked example

A ruler drops 0.45 m before it is caught. Calculate the reaction time. (g = 10 m/s²)

t = √(2d ÷ g)
= √(2 × 0.45 ÷ 10) = √0.09
= 0.3 s

Answer: 0.3 s

Key idea

Walking about 1.5 m/s, running 3 m/s, cycling 6 m/s, sound 330 m/s. g = 10 m/s². Ruler-drop reaction time: t = √(2d ÷ g). Large decelerations mean large forces; safety features increase stopping time to reduce the force.

Quick check

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Extra practice

2 more questions

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A dropped ruler falls 0.05 m before it is caught. Calculate the reaction time in seconds. (g = 10 m/s²)
Show the answer
Answer: 0.1 s
d = ½gt², so t² = 2 × 0.05 ÷ 10 = 0.01 and t = 0.1 s
A 1000 kg car stops from 20 m/s in 0.1 s in a crash. Calculate the force on the car in newtons.
Show the answer
Answer: 200000 N
Change in momentum = 1000 × 20 = 20 000 kg m/s. F = 20 000 ÷ 0.1 = 200 000 N

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Sources and references

Checked by the GCSE-ready editorial team, 17 October 2026. Last updated 2 October 2026. Spotted a mistake? Tell us and we will fix it.

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