Motion in a Straight Line · Lesson 7 of 10

Braking Distance and Reaction Time

“Separate reaction travel from braking travel and use a ruler-drop experiment to estimate response time.”

Learning Objectives

• Derive stopping time and braking distance for uniform deceleration. • Explain how speed affects reaction and braking distances. • Estimate reaction time from a dropped ruler. • Handle units and the physical end of a braking phase correctly.

Stopping is usually a two-stage process

When a driver notices a hazard, braking does not begin instantaneously. First comes a reaction interval: recognising the situation and initiating a response. During a simple model of this interval, the vehicle continues at its initial speed u. Next comes braking, during which the speed decreases. Keep these phases separate because they have different accelerations. Some descriptions use “stopping distance” for braking distance alone; here total stopping distance explicitly includes both phases.

Reaction distanceLaTeX

d_{r} = u t_{r}

u is initial speed and t_r is reaction time. This assumes approximately constant speed before braking begins.

Deriving the braking formulas

Take the initial direction as positive and write acceleration during braking as a = −b, where b > 0 is the magnitude of deceleration. At the end of braking v = 0. Substituting this in v = u − bt gives t_b = u/b. Substituting v = 0 and a = −b in v² = u² + 2ad_b gives 0 = u² − 2bd_b, so d_b = u²/(2b). The distance is positive because b is a positive magnitude.

Uniform brakingLaTeX

t_{b} = \frac{u}{b}, d_{b} = \frac{u ^{2}}{2 b}, d_{total} = u t_{r} + \frac{u ^{2}}{2 b}

If using signed a < 0 instead, d_b = −u²/(2a). Do not insert negative a into a formula that already assumes positive b.

For fixed reaction time, reaction distance is proportional to u. For fixed deceleration, braking distance is proportional to u². Doubling speed therefore doubles reaction distance and quadruples braking distance. Total distance contains both terms and does not generally quadruple. Braking capacity varies with road surface, tyres, vehicle condition and other factors; the constant-b model is an approximation for the interval considered.

Stopping from 126 km/h

Problem
A car stops uniformly within 200 m from 126 km/h. Find signed acceleration and stopping time.

1.Convert speed: 126 × 5/18 = 35 m/s. Take the initial direction positive.
2.Use 0 = 35² + 2a × 200. Thus a = −1225/400 = −3.0625 m/s²; deceleration magnitude is 3.0625 m/s².
3.Use 0 = 35 + at. Then t = 35/3.0625 = 11.43 s.
4.The problem supplies a braking distance without a separate reaction interval, so do not add an invented reaction time.

Why higher speed costs much more distance

Problem
Take reaction time 0.7 s and braking deceleration 5 m/s². Compare initial speeds 20 and 40 m/s.

1.At 20 m/s, reaction distance is 14 m and braking distance is 400/10 = 40 m. Total is 54 m.
2.At 40 m/s, reaction distance is 28 m and braking distance is 1600/10 = 160 m. Total is 188 m.
3.Braking distance quadruples, but total distance increases by 188/54 ≈ 3.48 times because the reaction contribution only doubles.

Reading stopping distance from a graph

Reaction phase followed by uniform braking— One second of reaction at 20 m/s gives 20 m. Four seconds of uniform braking gives a triangle of area 40 m. Total = 60 m.

The reaction portion of this graph forms a rectangle; the braking portion forms a triangle. Their areas give the two distances. Once the vehicle stops, the braking model ends. Continuing v = u − bt to later times would predict backward motion, which ordinary braking on a stationary vehicle does not cause. A valid model must change with the physical phase.

Measuring reaction time with a ruler

Hold your thumb and forefinger near the zero of a vertical ruler without gripping it. Another person releases it without giving a predictable signal. Catch it and read the distance the ruler fell. In the ideal model, the ruler starts from rest and falls freely until caught. Its fall time estimates the response time for this particular task. Convert centimetres to metres before using g in m/s².

Ruler-drop estimateLaTeX

d = \frac{1}{2} g t_{r}^{2} ⟹ t_{r} = \frac{2 d}{g}

d is the positive distance fallen. The estimate depends on the task and measurement conditions; it is not a universal personal constant.

A ruler falls 21.0 cm

Problem
Use g = 9.8 m/s² to estimate reaction time.

1.Convert 21.0 cm to 0.210 m.
2.Compute t_r = √(2 × 0.210/9.8) = √0.042857… s = 0.207 s, approximately 0.21 s.
3.Repeat trials to see the spread. Anticipating release, giving the ruler an initial push, or reading the catch position inconsistently changes the result. The ruler task does not directly establish reaction time in a complex driving situation.

Checking a braking model with data

If b is approximately constant for a given set of tests, the ratios d_b/u² should be approximately constant. For speeds 11, 15, 20 and 25 m/s with braking distances 10, 20, 34 and 50 m, the inferred b values are about 6.05, 5.63, 5.88 and 6.25 m/s². They are similar rather than identical, supporting an approximate square-speed relationship without claiming exact uniform braking in every test.

Quick check

If initial speed doubles while reaction time and braking deceleration stay fixed, what happens?

Key Takeaways

• Distinguish reaction distance from braking distance. • Braking distance depends on the square of initial speed for fixed deceleration. • Use a positive deceleration magnitude or a signed acceleration consistently. • The braking formula applies only until the vehicle stops. • A ruler-drop response time follows from free fall and correct unit conversion.

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Free Fall and Vertical Motion

Next · Lesson 8

Relative Motion and Meeting Problems