A Puzzle for you............

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MB Enthusiast
Mar 30, 2007
North Wilts
XC60 MY2014 SeLux Nav plus lot and lots of toys...
Two vehicles are travelling along the Motorway, one in lane two and one in lane three.
The vehicle in lane two is travelling at 70 mph
The vehicle in lane three is travelling at 100mph
At a given point the vehicles are alongside one another.
At this point they both have stop because of an obstruction ahead.
The vehicle in lane 2 skids from 70mph to a stop. At what speed is the vehicle in lane three still travelling when the vehicle in lane 2 is stationary ?

It's an interesting answer ... and for the benefit of this assume average person using average car with good well maintained brakes and tyres.....
Assuming speed decrease is linear, then 52.5mph, I reckon
I guesstimate: 60mph.
we dont know its stopping curve but anything between 30 & 55mph..

lots of variables like abs, traction, surface, tyres etc etc
Mark was the closest...

The answer is ................... 71 mph......a scary thought....

100mph = 147ft per sec. On impact it all ends in less than half a second?
If the vehicle in lane 2 had ABS it wouldnt have skidded ;)

Im sure I read all all the stopping distances based on 1960's Vaxuhalls/Fords etc?
Nothing like a "stopping distance" question to get people going..:D

But even if the figures I read weren't 100% accurate its enough to make one think............ well it made me think.....;)
i dont belive them at all, i mean there are just WAY too many variables to be able to provide a universal stopping distance chart.

yet they make spotty 17 year olds learn them off by heart for the theory.
most of these young chaps wouldnt know how far a foot was if someone gave them a tape measure...
Try plotting it on a graph. My logic was correct, even if I forgot to take away the number I first thought of.

If stopping distance at 70 is 315ft, then after 315ft, the car that was doing 100 will be doing 47.5
Well, if the slower car only just managed to stop immediately before the obstacle, the faster car would have reached the obstacle first, crashed into it and already come to a halt. So zero :D
It's a good theory - but what if ......

...the vehicle in lane 3 has brake assist, abs, new, correctly inflated tyres and perfect shock absorbers,

...the vehicle in lane 2 has tyres at the legal minimum, say 3psi underinflated, worn shock absorbers (but still an MOT pass), no brake assist, no abs, and maybe no brake servo (ie Mk1 Golf), and laden to the vehicle's maximum mgw ?

Would the question then be "When the vehicle in Lane 3 has stopped, what speed is the perfectly roadworthy vehicle in lane 2 still doing?"


These "statistics" are pure B.S.
Well, if the slower car only just managed to stop immediately before the obstacle, the faster car would have reached the obstacle first, crashed into it and already come to a halt. So zero :D

And it's only because the faster car has shunted the obstacle 50 yards up the road that the slower car has missed it........
Jay is of course right in saying that you can't give absolute numbers that apply to all cars in all circumstances.

The most important thing to remember is that you can't change the laws of physics. ;) In this case, the laws of physics that govern motion and energy are the most important ones:

F * d = 0.5 x m x v²

In the above d = distance and v = speed.

So, if you double the speed, you need to quadruple the distance.

Imadoofus, that means that your graph needs to be exponential, not linear :D:


But that is what Crockers is really saying: given two identical cars on identical surfaces etc, the stopping distance quadruples with every doubling of the speed. What the actual numbers are depends from car to car and from surface to surface, etc, but the physics of the relationship between speed and distance remains. ;)

Obviously, there are other important factors at play, notably the distance covered during the reaction time of the driver. Assuming the same reaction speed, you'll cover more distance obviously at higher speed than at a lower speed.
now thats an answer..

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