How to calculate the Overall Stopping / Braking Distance Watch

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Andylol
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#1
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Ok So I'm doing the theory test questions.

On the safety margins part of it, it asks for the "typical braking distance" or "typical stopping distance" at such and such mph.

But my question is, how the hell am I supposed to know XD?

Is there a way to guess this?

Apparently :-
Typical stopping distance @ 70mph is 96m
Typical braking distance @ 50mph is 55m.

I have absolutely no idea how they got those numbers.

Someone enlighten me? :confused:

//Andy
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Hellier
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F = m*a
F = mu*m*g
therefore a = mu*g
so v^2 = 2*mu*g*s
so s = v^2/(2*mu*g)
easy
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Andylol
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Oh dear not the suvat (or whatever those equations are called) equations!

So is that the stopping distance or braking distance?
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Hellier
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#4
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braking distance
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Glutamic Acid
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Thinking distance is roughly proportional to speed, braking distance is roughly proportional to speed squared.
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Amanda1801
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There is a really easy way to remember it for the speeds and distances in feet (they give you both) - if someone has the distances in feet I may be able to remember it, although was a while since I sat my theory!

(EDIT) I remember now...

Starting at 20mph x 2 - for the total stopping distance (i.e. thinking + braking)... increase the multiplication by 0.5 increments

20mph x 2 = 40ft
30mph x 2.5 = 75ft
40mph x 3 = 120ft
50mph x 3.5 = 175ft
60mph x 4 = 240ft
70mph x 4.5 = 315ft
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Future pilot
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Speed + Speed/2 = Number X 2 and thats the braking distance.

70 MPH + 35 MPH = 105ft*3 = 315ft stopping distance ( includes thinking + reaction time ).

another example :

40 MPH + 20 MPH = 60ft*3 = 180ft stopping distance
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Andylol
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(Original post by Future pilot)
Speed + Speed/2 = Number X 2 and thats the braking distance.

70 MPH + 35 MPH = 105ft*3 = 315ft stopping distance ( includes thinking + reaction time ).

another example :

40 MPH + 20 MPH = 60ft*3 = 180ft stopping distance
(The bold)

What??
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Future pilot
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lool , I'm tipsy I think I made that up . Its wrong it is ? . ? ...
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Talon
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I know how long/far it takes for my car to decelerate...
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Elementric
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Well, anyone who is trying to form some ridiculous mathematical equation must have something wrong with them because 1. Everything I've seen so far is total ********, 2. Stopping distance is hugely dependant on the car, and 3. How the hell are you going to work out the values you need?

Anyway, just learn it. It's a load of nonsense, but for the sake of the theory test you have to just learn the numbers.
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Elementric
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(Original post by Hellier)
F = m*a
F = mu*m*g
therefore a = mu*g
so v^2 = 2*mu*g*s
so s = v^2/(2*mu*g)
easy
I hope this is a joke.
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Laura373
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(Original post by Elementric)
Well, anyone who is trying to form some ridiculous mathematical equation must have something wrong with them because 1. Everything I've seen so far is total ********, 2. Stopping distance is hugely dependant on the car, and 3. How the hell are you going to work out the values you need?

Anyway, just learn it. It's a load of nonsense, but for the sake of the theory test you have to just learn the numbers.
precisely. :yep:
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ashy
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Does anyone who has passed their test actually know these?

I sure as hell don't.
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abcdefghijklmnopqrstuvwxyz
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why not just remember.......
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gbduo
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#16
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IRL you just leave 2 second gap between you and the car in front or more if you have inclement weather...

Sorted.

SUVAT wouldn't work because you are neglecting to take into account brake efficiency and hydraulic pressure asserted so you cannot find out out deceleration of the car over s.

If you could work out the deceleration rate (a) for your car from key speeds then interpolate you could work it out using SUVAT, but every car is different so meh.

Thinking about it, you would also need to consider tyres, road surface type and condition (new, potholes, etc), weather conditions, age of the car and brakes, metal fatigue affecting the pressure asserted onto the pads and metal fatigue/warping of the discs.

In other words, you cannot work it out by using SUVAT, there are many equations that have to be done first.

In saying this, SUVAT would give you a ball park figure...
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gbduo
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(Original post by Hellier)
F = m*a
F = mu*m*g
therefore a = mu*g
so v^2 = 2*mu*g*s
so s = v^2/(2*mu*g)
easy
where did that m go?

Wouldn't it be a = u(2m)*g?
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Hellier
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#18
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m is the mass n00b, so it cancels with the other mass when you equate the forces.
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Amanda1801
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#19
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(Original post by ashy)
Does anyone who has passed their test actually know these?

I sure as hell don't.
Yes - see above post - remembering 6 numbers isn't all that hard...
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ashy
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(Original post by Amanda1801)
Yes - see above post - remembering 6 numbers isn't all that hard...
Why should I remember things that I would never actually think about when actually driving? You judge based on experience, not maths.
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