# What is an unsafe acceleration for a road vehicle from the perspective of a pedestrian caught in its path? [closed]

Recently there has been a UK-based petition - which you can view here - that caught my eye. The petition relates to the story of an elderly man's mistaken application of the acceleration pedal as opposed to the break (which killed a pedestrian), and calls for retesting of drivers aged 70yrs and older every 3 yrs.

The petition got me thinking about other likewise unintentional dangers that (particularly elderly) drivers can pose, like unsafe acceleration from standstill. Several times I have witnessed the following:

• Elderly person whom I suspect to be deaf (but likely not to have been born deaf),
• Revving the [insert expletive] out of their car,
• Just prior to departure from a driveway or parking space e.g. at local supermarket.

I've often wondered then about the physics:

• What would happen if for any reason, the clutch were to be suddenly disengaged?
• How quickly might they make up the distance between starting point and say, a pedestrian subsequently caught in their path?

When a car moves slowly enough then most people will easily move out of the way unless they are visually impaired, or not looking at where the vehicle is and are themselves deaf or mentally impaired in some other relevant way, or the vehicle is exceptionally or unusually quiet (I concede for example that electric vehicles are extremely quiet).

I hesitated to walk behind this gentleman's car: I did not trust I would be safe. I could not assess a safe distance.

So what is an unsafe acceleration of a vehicle from standstill, for a pedestrian?

We're probably talking a strike time from stop to final speed of under what, 3-4 seconds (most people will seek to move away in that time). Pedestrian location/distance relative to vehicle assumed to be whatever distance the car would cover in that time-frame, but no greater than say six metres?

I understand if this question is removed for being too subjective. Nevertheless I hope you find it the food for thought if nothing else. I really think it needs research, but if already researched, that research desperately needs greater publicity.

Driving is one of the last bastions of independence that people with deteriorating bodies (age-related or other causes) have. It is hard to argue against that, but perhaps education could mitigate the problem (e.g. coping mechanisms for hearing loss, awareness raising for risks), and when necessary, facilitate the reasoning process applied in handling the vulnerable individuals who stand so much to lose from loss of their driving licences. We cannot educate or raise awareness in any way, without facts.

Maybe you could delete this question, but pose a better one for discussion?

I'm not convinced I've asked the 'right question'. I don't know how fast vehicles can accelerate from standstill, let alone backwards (e.g. parking space), or the difference between damage caused to a person from initial impact, versus being displaced and experiencing a subsequent full stop when their body strikes an obstacle.

## closed as too broad by Yashas, Jon Custer, mmesser314, John Rennie, peterhMay 12 '17 at 16:10

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Off topic because it is more the dangers of elderly drivers than about physics of cars. – mmesser314 May 12 '17 at 13:34

I think this question is probably not appropriate here, but I'll answer a small part of it: what kind of acceleration can a car manage?

Well, we can use the uniform acceleration formula that people learn at school (UK sense: high school in the US?) to answer this: $v = u + at$. If we assume $u=0$ -- acceleration from a standstill -- then we get $v = at$ or $a=v/t$. Well, people quote acceleration times of cars to 60 mph, which is $27\,\mathrm{ms^{-1}}$, and we can use these to work out $a$ for any kind of sustained acceleration.

For a Formula 1 car there are figures which claim $t=2.4$, giving $a = 27/2.4 \approx 11\,\mathrm{ms^{-2}}$. However:

• this is for acceleration over a significant time (a couple of seconds), the momentary value will be larger;
• but this is a Formula 1 car -- a thing made of carbon fibre and frozen magic which will barely fit around a person, with huge wings holding it onto the road (not effective at low speeds of course), special tyres, an engine which needs to be rebuilt every race or something &c &c.

So let's assume that the outer reaches of a domestic car's acceleration are $10\,\mathrm{ms^{-2}}$: almost certainly they can't do anything like that, but let's assume they can. There are good reasons to believe that accelerations much over $g$ are not really plausible for anything which is not using some cleverness to glue itself the road (which a Formula 1 car is using, of course), so this is a good number to have arrived at.

And now we can use the other famous constant-acceleration formula that schoolchildren learn: $s = ut + at^2/2$. Again, assume $u=0$, and we get $s = at^2/2$. So, if you think you can get out of the way in a couple of seconds, then you need $s = 10\times2^2/2\,\mathrm{m} = 20\,\mathrm{m}$

So, if you come across an elderly maniac in a high-performance car (that would be me, except for the high-performance car bit), stay about 20 metres away and you'll be OK.