Car stopping times (not distances) (Apologies if this is not the correct Stack Exchange site; I was unsure where I should post this question)
I was prompted to write this after reading on Reddit the stories told by drivers who had attended speed awareness courses in the UK (after being caught speeding themselves).
One "fact" apparently passed on by an instructor sounded unbelievable to me but I'm struggling to check the physics of it. The fact was:

In ideal conditions, car doing 31mph that has to do an emergency stop will still be doing 8mph at the point where it would have stopped had it been doing 30mph.

This doesn't sound right to me off the bat, and using this rule-of-thumb formula for calculating stopping distances:

In a non-metric country the stopping distance in feet given a velocity in MPH can be approximated as follows:

*

*take the first digit of the velocity, and square it. Add a zero to the result, then divide by 2.

*sum the previous result to the double of the velocity.


or, in Excel form:

FLOOR([@Speed] / 10, 1)^2 * 5 + (2 * [@Speed])

I have calculated the stopping distances to be 105 and 107 feet, respectively. Given that the second car's stopping distance is only an extra two feet I don't believe that it can still be travelling at 8mph at that point but I'm unable to get any further with my maths to check this.
How can I get stopping times for the above initial speeds? I was unable to find any resources on stopping times.
 A: In the England, Wales and Scotland there is a publication named the Highway Code which is essential reading for everyone. The aim of The Highway Code is to promote safety on the road, whilst also supporting a healthy, sustainable and efficient transport system.
In it there is a graphic which shows typical stopping distances which are the sum of a thinking distance (the distance traveled before the driver applies the brakes) and the braking distance (the distance traveled under braking).

The data in the graphic above can be converted into a table.

You will note that there is an assumption of a constant thinking (or reaction) time of $0.67\,\rm s$ and an approximately constant deceleration of around $6.6\,\rm m/s^2$.
The deceleration can be found by using one of the constant acceleration kinematic equations.  For example at $30\,\rm mph = 13.41\, m/s$
$ v_{\rm final}^2 = v_{\rm initial}^2 + 2 \,a \,s \Rightarrow 0 = 13.41 ^2+2\,a\,14 \Rightarrow a=-6.42\,\rm m/s^2.$
The distance traveled under braking is the area under a speed against time graph which at $30\rm \, mph$ looks like this.

If the starting speed is $31\,\rm mph$ then the graph (red) is a shown below.

Note the increase in thinking distance.
Now remember the statement which we are considering, In ideal conditions, car doing 31mph that has to do an emergency stop will still be doing 8mph at the point where it would have stopped had it been doing 30mph.
So at some time $t$ the car which started out moving at $31\,\rm mph$ will have travelled $9+14=23\,\rm m$ and be travelling at a speed of $v_{31}$.
$v_{31} = 31-(-6.42)\,t$ and $9+14 = 9+0.39 = (31+v_{31})\,t/2 $
Solving these equations gives $t=1.51\,\rm s$ and $v_{31} = 4.16 \rm \, m/s = \bf{9.3}\rm \,mph$.
If one ignores the extra distance $(0.39\,\rm m)$ covered when thinking then $t=1.61\,\rm s$ and $v_{31} =3.52\,\rm m/s \approx \bf{8}\rm \,mph$.
A: Setup
In the simplified "physics 101" model the velocity at time $t$ is:
$v(t) = v_0 - \mu g t$
where $v_o$ is the initial velocity, $\mu$ is the coefficient of friction, and $g$ is the acceleration due to gravity.
Stopping time is the time where $v(t) = 0$, and this is $t=\frac{v_0}{\mu g}$.
Distance is the integral of velocity (with respect to time):
$d(t) = v_0 t - \frac{1}{2} \mu g t^2$
Note some values:
31 mph is 13.9 m/s
30 mph is 13.4 m/s
$g$ is roughly 10 m/s/s
A reasonable value for $\mu$ is 0.8
The example
Therefore, the stopping time at 30 mph (13.4 m/s) is:
$t_1 = \frac{13.4}{0.8 \times 10} = 1.675 \mathrm{s}$.
The car will have traveled a distance of
$d_1(t) = 13.4 \times 1.675 - \frac{1}{2}\times 0.8 \times 10 \times 1.675^2 = 11.2 \mathrm{m}$
A car initially traveling at 31mph would travel that distance in
$t = 1.27$ s (solving the quadratic equation for $d(t)$).
and its speed at that distance would be:
$v(1.27) = 13.9 - 0.8 \times 10 \times 1.27 = 3.74 \mathrm{m/s}$
3.74 m/s is about 8 mph.
Real life is highly likely to be more complicated (intuitively I don't think I could stop a car from 30 mph in 11.2 m).  Perhaps a good way to check would be to use real cars on a racetrack.
A: That sounds about right. Kinetic energy is proportional to speed squared. Kinetic energy eaten up by braking for one meter is some constant. 31^2 = 961, 30^2 = 900, the difference is 61 ≈ 8^2.
When the slower car stops, it has eaten up 900 units of kinetic energy (doesn't matter how much a unit is). The faster car that started with 961 units has 61 units left after the same braking distance, which is just slightly less than the kinetic energy of a car at 8mph.
Interestingly, the result is the same, no matter how good or bad your brakes are, or how heavy the car is. The result is also the same if you have cars driving km/h or metres/second: By the time a car going 30 metres per second stops, a car going 31 metres per second is down to almost 8 metres a second.
A: If a car initially at speed $u$ decelerates at a constant rate $a$ then after $t$ seconds its speed will be
$v(t) = u - at$
so it takes $\frac u a $ seconds to come to rest. In this time it travels a distance of
$\displaystyle s = \frac {u^2} {2a}$
so braking distance increases as the square of the initial speed. This means that the braking distance of a car initially travelling at $31$ mph will be about $6.7 \%$ greater than the braking distance of a car travelling at $30$ mph. Note that the approximate formula that you give seems to underestimate the difference between the two braking distances.
The relationship between $v(t)$, $u$, $a$ and distance travelled at time $t$, $s(t)$, is
$\displaystyle v(t) = \sqrt{u^2 - 2as(t)}$
So if $u=31$ mph and $s(t)$ is the stopping distance for $30$ mph, which is $\frac {30^2} {2a}$, then
$\displaystyle v(t) = \sqrt{31^2 - 30^2} = \sqrt{61} \approx 7.8$ mph
So the car that is initially travelling at $31$ mph will be travelling at just under $8$ mph at the point where the car travelling at $30$ mph would have stopped.
Note that this calculation only takes account of braking distances i.e. the distance travelled after the car starts to decelerate. In reality, you have to add a thinking time/distance onto this to take account of the driver's reaction time. If the thinking time (i.e. reaction time) is the same, then the thinking distance will be about $3.3 \%$ greater at $31$ mph than at $30$ mph, which makes the difference in overall stopping distances even greater.
