About 3.9 feet. In a published set of 50 braking tests, every extra mile per hour added 3.932 feet to the stopping distance on average, and the 95% confidence interval for that figure runs from 3.097 to 4.768 feet. That is the short answer, and it comes from a single straight line fitted by least squares.
The longer answer is more useful, because the same line also tells you how far a single car can stray from it, where it stops making sense, and which one car in the data does not fit. So this walkthrough starts with the headline number and then pushes on it.
The data are the cars data set that ships with every copy of R: speed in mph and stopping distance in feet for 50 cars, recorded about a century ago. Old brakes, so the distances are not a guide to a modern car. But the numbers are real, public and easy to check, which is exactly what a worked example needs. Speeds run from 4 to 25 mph, the mean speed is 15.4 mph and the mean stopping distance 42.98 feet.
The fit behind the 3.9, piece by piece
Paste the 50 speeds as X and the 50 distances as Y into the linear regression calculator and it fits the line that makes the squared vertical gaps as small as possible. The result is y = -17.58 + 3.932x, the same line R prints for lm(dist ~ speed). The table below breaks the fit into its parts; each row is a figure from the calculator.
| Piece of the fit | Value | What it means here |
|---|---|---|
| Slope b | 3.932 | feet added per extra mph |
| Standard error of the slope | 0.4155 | how much b would wobble between samples of 50 cars |
| 95% interval for the slope | 3.097 to 4.768 | b plus or minus 2.0106 standard errors (t on 48 df) |
| t and p for the slope | 9.464, below 0.0001 | the slope is far from zero |
| Intercept a | -17.58 | distance "at 0 mph", a number with no physical meaning |
| R squared | 0.6511 | share of the variation in distance that speed explains |
| Standard error of the estimate | 15.38 | typical miss of the line, in feet |
Two of those rows deserve a second look. The 2.0106 in the interval is the t critical value for 48 degrees of freedom; with a very large sample it would shrink to the familiar 1.96, which is the z-score for the 97.5th percentile. And 15.38 feet of typical miss is a lot next to a mean of 42.98. Speed explains about two thirds of the variation. The other third comes from everything the data did not record: tires, road, driver, brakes.
How much of the spread does the line explain?
R squared is 1 - SSE / SST. For these cars the sum of squared errors left around the line is 11,353.52, and the total sum of squares around the mean distance is 32,538.98. The R squared calculator takes those two numbers and returns R squared 0.6511, adjusted R squared 0.6438, an explained sum of squares of 21,185 and an F statistic of 89.57 on 1 and 48 degrees of freedom.
Is 0.65 good? For 50 observations and one predictor, anything above 0.0777 would already be statistically significant at 0.05, so the relationship is not in doubt. Whether 0.65 is good enough depends on the job. For saying "faster cars need more room" it is plenty. For telling one driver how much room they need, it leaves a third of the variation unexplained, and the next section shows what that looks like in feet.
Six ways to push on the number
How much room does one car need at 21 mph?
At 21 mph the line predicts 65.00 feet. The confidence interval around that, 58.60 to 71.41, is about the average car at 21 mph. A single car is a different question, and its 95% prediction interval runs from 33.42 to 96.58 feet. Sixty-three feet wide. The table shows the same two intervals at five speeds.
| Speed (mph) | Line predicts (ft) | Average car, 95% (ft) | One car, 95% (ft) |
|---|---|---|---|
| 10 | 21.74 | 15.46 to 28.03 | -9.810 to 53.30 |
| 15 | 41.41 | 37.02 to 45.79 | 10.17 to 72.64 |
| 20 | 61.07 | 55.25 to 66.89 | 29.60 to 92.54 |
| 21 | 65.00 | 58.60 to 71.41 | 33.42 to 96.58 |
| 25 | 80.73 | 71.60 to 89.87 | 48.49 to 113.0 |
Notice the first row. At 10 mph the one-car interval starts at -9.810 feet, which is impossible. The interval formula assumes the scatter around the line is the same everywhere, and in these data it is not: slow cars cluster tightly, fast cars spread widely. A constant 15.38 feet of scatter is too much at 10 mph and too little at 25.
Why does the line predict a negative distance at 4 mph?
The line gives -1.849 feet. It crosses zero at about 4.47 mph, and below that it predicts that a car stops before it starts braking. Two cars in the data were doing exactly 4 mph and needed 2 and 10 feet. The line is not wrong about the data it was fitted to on average; it is simply the wrong shape near zero, where the true relationship has to pass through the origin.
Is stopping distance proportional to speed squared?
Physics says it should. The kinetic energy a brake has to absorb grows with speed squared, so at constant deceleration the braking part of the distance grows with v squared too. Fitting y = a + b x + c x squared to the same 50 cars gives an R squared of 0.6673 (0.66733 before rounding) instead of 0.6511.
That extra 0.016 is not free: the curve uses a second predictor. The fair comparison is adjusted R squared. Enter R squared 0.66733 with n = 50 and k = 2 in the R squared calculator's third mode and it returns 0.6532, against 0.6438 for the straight line. A small win, but a win. In the middle of the data the two models barely disagree: at 10, 20 and 21 mph they are within a foot of each other (65.73 against 65.00 feet at 21 mph), and the largest gap between 10 and 21 mph is 2.75 feet, at 15. At the edges they part: 7.05 feet at 25 mph, and at 4 mph the curve gives a sensible 7.72 feet where the line gives a negative number.
Is a 120-foot stop an outlier?
Car 49 was doing 24 mph and took 120 feet. The line expects 76.80 for that speed, so the residual is 43.20 feet, and the regression calculator puts its standardized residual at 2.92, the largest of the 50.
Is it an outlier? Paste the 50 distances into the z-score calculator and 120 gets an ordinary z of 2.99. Just short of the popular cut-off of 3, partly because the 120 itself inflates the standard deviation it is measured with (25.77 feet). The modified z-score, built on the median of 36 feet and a median absolute deviation of 16, is not moved by one extreme value. It gives 3.54, above the 3.5 threshold Iglewicz and Hoaglin recommend, and flags the car.
Flagged is not the same as wrong. Drop car 49 and refit, and the slope falls from 3.932 to 3.640 feet per mph, while R squared moves from 0.6511 to 0.6474. One car shifts the headline number by about 0.29 feet per mph. Worth reporting both, and worth finding out whether 120 was a typing slip or a car with bad brakes.
Can this data predict stopping distance at 30 mph?
Nobody in the data went faster than 25. The line says 100.39 feet at 30 mph; the curve says 119.83. In the middle of the data the two models were within a few feet of each other; five mph past its edge they are 19.44 feet apart. The regression calculator prints an extrapolation warning for exactly this reason. Past the data, the model you choose matters more than the data you have.
What is the slope in meters per km/h?
Convert the slope: 3.932 feet per mph is 0.745 meters per km/h. Or convert the raw data and refit, which gives the same line in metric units, because least squares does not care about scale. R squared, adjusted R squared, t, p and every z-score stay exactly the same; they have no units.
Which of these numbers can you lean on?
One row per scenario from the previous section: the figure it produces, the fit or interval behind it, and whether it holds up.
| Scenario | Answer | Fit or interval | Rely on it? |
|---|---|---|---|
| Straight line, all 50 cars | 3.932 ft per mph | R squared 0.6511, adjusted 0.6438 | yes, inside 4 to 25 mph |
| One car at 21 mph | 65.00 ft, range 33.42 to 96.58 | prediction interval | only as a range |
| A car at 4 mph | -1.849 ft | line below zero | no |
| Curve with speed squared | 65.73 ft at 21 mph | adjusted 0.6532 | yes, and safer past the data |
| Without the 120 ft car | 3.640 ft per mph | R squared 0.6474 | as a sensitivity check |
| A car at 30 mph | 100.39 ft line, 119.83 ft curve | extrapolation | no |
The verdict on 3.9 feet per mph
For these 50 cars, between 4 and 25 mph, yes: 3.932 feet per mph, somewhere between 3.1 and 4.8 with 95% confidence, and a third of the variation left to things the data never measured. The number to hand a single driver is the prediction interval, not the line. And the moment the question leaves the range of the data, whether toward 0 mph or past 25, the straight line should give way to the curve that physics expects, which the adjusted R squared already slightly prefers.
Tools discussed in this article
- Linear regression calculator: the least-squares line with t, p and intervals for both coefficients, R squared and adjusted R squared, a prediction with both intervals, standardized residuals and a note when one pair steers the slope.
- R squared calculator: R squared from actual and predicted values, from SSE and SST, or adjusted and tested from a value you already have, with the F test and the smallest R squared that would be significant.
- Z-score calculator: one value against a mean and SD, every value in a list with the modified z-score for outliers, and conversions between z-scores and percentiles.
More statistics tools
Pearson correlation (r with a p-value and Spearman beside it) · Skewness (how lopsided the residuals are) · Kurtosis (how heavy their tails are) · Interquartile range (Tukey fences for one column) · Coefficient of variation (spread as a share of the mean) · Percentile (where one value sits in a list) · Mode (the most frequent value) · Wilcoxon test (rank tests when the tails are heavy) · Fisher's exact test (small 2x2 tables of counts) · Standard deviation (the spread behind every z-score) · Linear function (zero and intercepts of a line you already have) · Quadratic equation (roots of the curved model)