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How do i work out the total stopping distance for a car at 80mph given that:

SPEED ... TOTAL STOPPING DISTANCE

20mph ... 12m

30mph ... 23m

40mph ... 36m

50mph ... 53m

60mph ... 73m

70mph ... 96m

Note that mph denotes miles per hour and m metres

The above figures are widely agreed on (as stated in the Highway Code of Great Britain), but it stops at 70mph and I must now work out the total stopping ditance at 80mph. Please give formula(e), as I am totally stumped.

I have tried the following methods, so please tell me if any are correct (d = total stopping distance):

1) Using the APPROXIMATION formula:

d = (v^2 / 20) + v

where v is velocity

2) Using the mathematical forecast formula which I want to avoid:

d (80mph as x in this formula ONLY) = a + bx

where a = AVERAGEy - (b * AVERAGEx)

and b = (SUM((x - AVERAGEx)*(y - AVERAGEy))) / (SUM((x - AVERAGEx)^2))

using the different speeds in mph as x and the known stopping distances as y

by SUM i mean sigma - so each value of x and y have to be subbed in

3) Finding the average increase between each known stopping sitance, and adding it to 70mph

4) Using t = (v - u) / a

where t is the time taken to stop

v is the final velocity (always 0 in this case)

u is the starting velocity

and a is the acceleration

Then using AVERAGEv = (v + u) / 2

where v is the final velocity (always 0 in this case)

and u is the starting velocity

And substituting the value of AVERAGEv and t in d = AVERAGEv * t

5) Using KE = 0.5 * m * v^2

where m is the mass of the vehicle

and v is the starting velocity (i think it should be starting)

And then substituting the value for kinetic energy in d = WD / f

where WD is the work done (equal to kinetic energy i think)

and f is the force applied to the brakes

The best ones i think are 4 and 5 (esp. 4), but the problem is i have so many unknowns, and when i try to work my way back to get acceleration, for example, i get different values for every known stopping distance. So now i am totally stumped.

Please help!

Addendum:

Hmmm... I've just had a flash of lightening but it might just be me. Using KE = 0.5 * m * v^2 it is obvious that doubling the velocity quadruples the kinetic energy. Hence, the braking distance is quadrupled.

This is proven in my figures as the stopping distance at 20mph is 12m, and at 40mph it is 36m (36 / 12 = 4).

It is also kind of disproven in the figures as the stopping distance at 60mph is not quadruple 30mph (73m and 23m, respectively).

Can any one explain why this is the case for the 60/30mph? And is my theory right?

Final Word:

Woohoo! I have now finally solved it. The concept introduced in the addendum is actually correct, but it applies to the BRAKING distance not the total stopping distance - which includes the thinking distance. So by multiplying the braking distance at 40mph by 4, and then finding the trend for the thinking distance (which is simple - t(n) = 3n). By taking the sum of the braking and thinking distances, i have now arrived at the total stopping distance at 80mph!

SPEED ... TOTAL STOPPING DISTANCE

20mph ... 12m

30mph ... 23m

40mph ... 36m

50mph ... 53m

60mph ... 73m

70mph ... 96m

Note that mph denotes miles per hour and m metres

The above figures are widely agreed on (as stated in the Highway Code of Great Britain), but it stops at 70mph and I must now work out the total stopping ditance at 80mph. Please give formula(e), as I am totally stumped.

I have tried the following methods, so please tell me if any are correct (d = total stopping distance):

1) Using the APPROXIMATION formula:

d = (v^2 / 20) + v

where v is velocity

2) Using the mathematical forecast formula which I want to avoid:

d (80mph as x in this formula ONLY) = a + bx

where a = AVERAGEy - (b * AVERAGEx)

and b = (SUM((x - AVERAGEx)*(y - AVERAGEy))) / (SUM((x - AVERAGEx)^2))

using the different speeds in mph as x and the known stopping distances as y

by SUM i mean sigma - so each value of x and y have to be subbed in

3) Finding the average increase between each known stopping sitance, and adding it to 70mph

4) Using t = (v - u) / a

where t is the time taken to stop

v is the final velocity (always 0 in this case)

u is the starting velocity

and a is the acceleration

Then using AVERAGEv = (v + u) / 2

where v is the final velocity (always 0 in this case)

and u is the starting velocity

And substituting the value of AVERAGEv and t in d = AVERAGEv * t

5) Using KE = 0.5 * m * v^2

where m is the mass of the vehicle

and v is the starting velocity (i think it should be starting)

And then substituting the value for kinetic energy in d = WD / f

where WD is the work done (equal to kinetic energy i think)

and f is the force applied to the brakes

The best ones i think are 4 and 5 (esp. 4), but the problem is i have so many unknowns, and when i try to work my way back to get acceleration, for example, i get different values for every known stopping distance. So now i am totally stumped.

Please help!

Addendum:

Hmmm... I've just had a flash of lightening but it might just be me. Using KE = 0.5 * m * v^2 it is obvious that doubling the velocity quadruples the kinetic energy. Hence, the braking distance is quadrupled.

This is proven in my figures as the stopping distance at 20mph is 12m, and at 40mph it is 36m (36 / 12 = 4).

It is also kind of disproven in the figures as the stopping distance at 60mph is not quadruple 30mph (73m and 23m, respectively).

Can any one explain why this is the case for the 60/30mph? And is my theory right?

Final Word:

Woohoo! I have now finally solved it. The concept introduced in the addendum is actually correct, but it applies to the BRAKING distance not the total stopping distance - which includes the thinking distance. So by multiplying the braking distance at 40mph by 4, and then finding the trend for the thinking distance (which is simple - t(n) = 3n). By taking the sum of the braking and thinking distances, i have now arrived at the total stopping distance at 80mph!

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