Stopping | | DJS


For all of these questions but the last we will assume that the old formula in the Highway Code is the correct model. Using feet and miles per hour, then a speed S mph has a stopping distance in feet given by D = S²/20 + S, where the linear part is the ‘thinking distance and the quadratic part is the braking distance. I’m assuming you know the formulae for constant acceleration (“suvat”).
Because the old formula works in old units, you might prefer to convert everyhting to work in different units, but your answers need to finish up still using mph for speed. Assume that 2 s.f. answers will be acceptable, so work to 3 or 4 s.f. in your intermediate calculations. 
This is mostly just arithmetic, so anyone from Y8 onwards can do them; in general, older students will go a good deal faster. Some of this is surprisingly complicated to sort out. It is the language mostly, even after I’ve tried to make it a little easier to follow….

1.   Using 5280 ft per mile and 3600 seconds per hour, what is the thinking time at 30mph?
      Is the thinking time different at differnt speeds?

2.   What is the assumed deceleration at 30 mph?
      Is the deceleration different at differnt speeds?

3.    When learning to drive we are told that the preferred gap to the car in front is two seconds.
       Show that at 30mph this gap is smaller than the stopping distance and
       Find the time taken to stop from 30mph.

4.   Find the speed at which the two second gap is equal to the stopping distance.
      Relate this to the speed limits in residential areas.

5.   Imagine you’re driving at a regulation 30mph and and a two second gap – and the car in front has an accident you don’t expect, effectively putting a barrier up at your two second limit. How fast will you be going at the moment of impact?
  If, instead, the car in front takes one car’s length, 13feet or 4 metres, to stop, what happens?

6.    At 40mph, what is the minimum stopping time?
       When your two second gap is used up, how much more time and distance do you need in which to stop?

7.   As Q6 but at 70mph.

8.   Jeremy likes to drive on the motorway at 85mph. He thinks of himself as a superior driver. In support of this thinking, what time gap would be sensible to preserve the same levels of safety as the two second gap provides? In particular, what time gap provides the same results as for Q7? 

9. The problems of driving fast on the motorway is not stopping as much as it is that the other drivers cannot see a car going significantly faster; drivers assume that the differences in speed are small. Let us assume you’re driving in Germany where the speeds can be a lot higher.  Assume you are going at 70mph and that you wish to overtake a lorry doing 60mph. Allowing a two second safety margin, show that you need around ten seconds in the overtaking lane.

10. Most motorway driving uses very small acclerations and decelerations; even a tenth of the model’s values are considered extreme. Consider 1 ft/sec/sec as ‘reasonable’, 5% of the braking model. Someone doing 100mph wants (Q8) 4 seconds empty space in front of them. If you are to not cause such a driver to slow, how far behind you do you need to see to complete your overtake and cause a fast driver no inconvenience?

I suggest that to do this last question you might need some research from direct observation. You only need to know what values form ‘acceptable behaviour’ on the motorway. You need to observe what intervals (time and distance) cause a driver to react defensively, and what values cause derogatory comment – or, for that matter, admiring comment.

DJS 20171213

1.  30mph = 44 ft/sec, so 30 feet represents 0.68 seconds.                  No.

2.   a = v²/2s = 44²/90 = 21.511 ft/sec/sec = 6.5566 m/s/s = 0.67g (2s.f.)    No, it is constant.

3.   Gap is 88 feet. Stopping distance is 75 feet. 88 > 75. Done, 
      Thinking time is 0.68 secs. Braking time is v/a = 44/21.5 = 2.05 seconds

4.    v-u=at =>v=at = 21.511 x (2-0.68) = 28.4 ft/sec = 19.4 mph.
       This is why we have 20mph zones in residential areas.

5.  Q3 says you have 13 feet to go at 21.511 ft/sec/sec. Q3 also says you needed 2.72 secsonds to stop, so you have 0.72 seconds of stopping left at 21.511 ft/sec/sec. This says you will hit at 21.5 x 0.72 = 15.5 ft/sec = 10.6 mph.
The extra 13feet means you may just stop in time - a gentle bump, perhaps. No damage in this secondary impact.

6.     40mph = 58.667 feet/sec => 0.68 + 2.727 secs to stop, 1.407 secs more than two secs.
Speed at two secs is therefore 30.27 ft/sec, 20.6 mph. 21.3 feet required, 6.5 metres.

7.     70mph = 102.667 ft/sec  =>0.68 + 4.773 secs to stop, 3.45 sec more than two.
Speed at two seconds gone is 74.3 ft/sec, 50.6 mph, 128 feet required to stop without impact, around nine more car lengths.

8. We want the extra time so that at a two second gap one could be doing 70mph. On stopping distances this is an extra 131.25 feet, which at 85 mph (125ft/sec) is an extra 1.05 seconds. At a hundred, it is one more second. The bad news is that any slight deviation in the bad direction has powerful results on impact.

9   10mph difference on passing a space 20m long is around 5 seconds, plus the 2 seconds each side, plus 1 for margins. You could argue this up by four seconds quite easily. You need something like 8-14 seconds in the other lane. Say up to ten.                 Discuss.

10    Someone doing 1000mph, 44.7m/s, is 13.4 m/s faster than you and in your 10 seconds will catch up 135m; they need 14 seconds or more, which is 625m, more than half a kilometre and possibnly further ythan you can see behind you. A 500m space between you and the fast car will be zero in 37 seconds; it will be four seconds (125m) in 28 seconds. Slowing from 100 to 70 takes 80m and 2 secs at the same hard braking that the model predicts. You really don’t want to have the fast car as close as four seconds and not slowing down.
Using a 1 ft/sec/sec model, reducing from 100 to 70 (a difference of 44 ft/sec) takes 44 seconds, which is around 600m up the road. This is not an answer, but fuel for discussion. suggests (P149) that, for speeds over 12m/s, 1.5m/s/s is a reasonable deceleration value, 10% of the model used in Q1-7, but the table on P150 gives greater detail, suggesting that an ordinary car might use a value of 0.6 m/s/s as a maximum acceleration. More research called for.

The normal decel

value is m sfor sp eeds up to m s m sfor sp eeds within the range

dep ends on the sp eed of the vehicle and was adopted from ITE The

to m s and m sfor sp eeds greater than The acceleration in the car following regime

m s 

The normal decel

value is m sfor sp eeds up to m s m sfor sp eeds within the range

dep ends on the sp eed of the vehicle and was adopted from ITE The

to m s and m sfor sp eeds greater than The acceleration in the car following regime

m s 

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