On this page
A child runs onto the road. The driver sees her, decides, moves a foot to the brake, and only then does the car start to slow. Exam questions on stopping distance test whether you remember that first stretch, travelled at full speed. This post builds the full formula from the Class 11 kinematic equations, works three problems with numbers and shows the mistakes that cost marks.
What does NCERT say about stopping distance?
The NCERT Class 11 chapter on motion in a straight line has two short examples on this. Example 2.6 uses v² = v₀² + 2ax with v = 0 to get the braking distance d = v₀²/2a, where a is the size of the deceleration, and notes that doubling the initial speed makes the braking distance four times larger. It quotes test figures for one car: braking distances of 10, 20, 34 and 50 m at 11, 15, 20 and 25 m/s.
The JEE (Main) 2026 syllabus lists "relations for uniformly accelerated motion" under Unit 2, Kinematics, which is all this topic needs. Check the syllabus for your own exam year.
NCERT's test figures are worth checking yourself. Using a = v²/2d gives 6.05, 5.63, 5.88 and 6.25 m/s². So the car's deceleration was roughly constant, near 6 m/s², which is why the v² rule fits.
Example 2.7 then defines reaction time as the time a person takes to observe, think and act, and measures it by catching a falling ruler. A ruler that falls 21.0 cm before being caught gives t = √(2d/g) = √(0.42/9.8) ≈ 0.21 s.
How long does a driver take to react?
Catching a ruler you are waiting for is a simpler task than braking in traffic. Marc Green's review of driver studies, "How long does it take to stop?" (2000), found that expectation is the biggest factor, changing reaction times by a factor of about two. Drivers who know when and where the signal will come take about 0.70 to 0.75 s to detect it and move the foot to the brake. For an unexpected but common signal, such as the brake lights of the car ahead, it is about 1.25 s. For a surprise, such as an object suddenly moving into the road, it is roughly 1.5 s.
What is the full formula?
During the reaction time t_r the car still moves at v, so:
Reaction distance = v × t_r
Braking distance = v²/2a
Total stopping distance = v t_r + v²/2a
On a velocity-time graph, the reaction distance is the rectangle under the flat part and the braking distance is the triangle under the sloping part. Our guide to reading physics graphs covers areas under v-t graphs.
Worked example 1: what changes when you double the speed?
Take a deceleration of 6 m/s², close to NCERT's figures, and a reaction time of 0.75 s. These are sample values for practice.
At 54 km/h:
Convert: 54 × 1000/3600 = 15 m/s.
Reaction distance = 15 × 0.75 = 11.25 m.
Braking distance = 15²/(2 × 6) = 225/12 = 18.75 m.
Total = 30 m.
At 108 km/h (30 m/s):
Reaction distance = 30 × 0.75 = 22.5 m.
Braking distance = 900/12 = 75 m.
Total = 97.5 m.
Doubling the speed doubled the reaction distance, quadrupled the braking distance, and made the total 3.25 times larger.
Worked example 2: a surprise instead of a signal
Same car at 15 m/s, but the driver takes 1.5 s, the surprise figure from Green's review. Reaction distance = 15 × 1.5 = 22.5 m, braking distance still 18.75 m, total 41.25 m. The brakes did not change; the extra 11.25 m came from the reaction alone.
Worked example 3: what is the safe speed?
An obstacle appears 40 m ahead. The driver's reaction time is 1.0 s and the brakes give 5 m/s². What is the highest speed at which the car can still stop in time? (Sample values.)
Set total distance = 40: v(1.0) + v²/(2 × 5) = 40.
Multiply by 10: v² + 10v − 400 = 0.
Quadratic formula: v = [−10 + √(100 + 1600)]/2 = (−10 + 41.23)/2 ≈ 15.6 m/s.
In km/h: 15.6 × 3.6 ≈ 56 km/h.
Reject the negative root, since speed here is a magnitude. Check: 15.6 × 1 + 15.6²/10 ≈ 15.6 + 24.4 = 40 m.
What mistakes cost marks?
Forgetting the reaction distance. If the question gives a reaction time, it wants v t_r included.
Applying braking during the reaction time. The car does not slow until the brakes act.
Leaving speed in km/h. Convert to m/s first; divide by 3.6.
Using a negative a inside v²/2a twice. Write the formula with the size of the deceleration, or keep the signs in v² = u² + 2as, but not both.
Scaling the total by 4 when speed doubles. Only the braking part scales by 4.
A practice routine
Day 1: Measure your own ruler-catch time three times with a friend and average it, as in NCERT Example 2.7.
Day 2: Make a table of total stopping distance at 36, 54, 72 and 90 km/h for reaction times of 0.75 s and 1.5 s, with a = 6 m/s². Notice which column grows faster.
Day 3: Three "safe speed" problems solved with the quadratic formula, each checked by substitution.
Day 4: Two v-t graph versions, finding stopping distance from areas.
When a variation stumps you, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can compare its set-up line by line with yours. For more kinematics practice, see our post on motion under gravity.
Today, redo Worked example 1 at 72 km/h and check that the total comes to 48.3 m.
