Why does doubling your speed make the stopping distance four times longer?

Getting the formula in one line

Start from v² = v₀² + 2ax, the equation that has no time in it. When the car stops, v = 0. So 0 = v₀² + 2a·d_s, which gives

d_s = −v₀² / 2a

With forward as positive, braking makes a negative, and the minus sign turns the distance positive. Using the size of the deceleration, d_s = v₀² / 2|a|. Same thing.

What "four times" actually means

Speed sits inside a square. Go from 10 m s⁻¹ to 20 m s⁻¹ and the stopping distance doesn't double; it goes up four times. NCERT adds a condition in brackets, "for the same deceleration", and it matters. Wet roads or worn brakes change a, and then the comparison is off.

A quick check in your head: a car at 10 m s⁻¹ stops in 8 m. At 15 m s⁻¹ the speed is 1.5 times larger, so the distance is 1.5² = 2.25 times larger: 2.25 × 8 = 18 m.

The book's real data

Example 2.6 also quotes measured braking distances for one make of car:

From 11 to 25 m s⁻¹ the speed goes up a little more than two times. The distance goes up five times. Divide each distance by the square of its speed and you get a number between 0.080 and 0.089 every time, which is why NCERT calls the data "nearly consistent" with the formula. NCERT adds: stopping distance is an important factor in setting speed limits, for example in school zones.

Careful: the clock starts when the brakes go on

In Example 2.6, d_s is measured from the moment the brakes are applied. Before that, the driver has to notice the danger and move a foot, and the car keeps going at full speed.

Example 2.7 measures reaction time with a ruler. A friend holds a ruler between your thumb and forefinger and lets go without warning. You catch it. The ruler fell freely from rest, so d = ½gt², which gives t_r = √(2d/g). In the book the ruler drops 21.0 cm:

t_r = √(2 × 0.21 / 9.8) ≈ 0.207 s, about 0.2 s.

At 20 m s⁻¹, a 0.2 s reaction uses about 4 m of road before the brakes do anything. That part grows only in step with speed; the braking part grows with its square. So the "four times" rule belongs to the braking distance alone.

The unit trap. Put d = 21 into the formula as if it were metres and you get about 2.07 s. Nobody takes two seconds to catch a ruler. Convert 21.0 cm to 0.21 m first.

Two more examples from the same lesson

A ball thrown up from a building (Example 2.3). A ball leaves the top of a 25.0 m building at 20 m s⁻¹ upward, with g taken as 10 m s⁻². It rises v₀²/2g = 20 m, to 45 m above the ground: 2 s up, 3 s down, 5 s in all. The book puts the origin at the ground with up as positive. Its second method solves 0 = 25 + 20t − 5t² in one go and gets the same 5 s.

Galileo's odd numbers (Example 2.5). Drop something from rest and split its fall into equal time intervals. The distances covered in successive intervals are in the ratio 1 : 3 : 5 : 7 … NCERT credits this to Galileo Galilei (1564–1642). It follows from y = ½gt²: total distances go 1, 4, 9, 16 … and the gaps are the odd numbers.

Check yourself

A car stops in 12 m from 10 m s⁻¹. With the same deceleration, how far does it need from 30 m s⁻¹? (Answer at the bottom.)

Questions students ask

Answer: three times the speed gives 3² = 9 times the distance, so 9 × 12 = 108 m.