The stopping distance d of a car after the brakes are applied...

eddy2017

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this is another exercise of the kind i just posted. it is interesting for me because the use of this formula is advised. i do not understand it and the formula is not explained.​

The stopping distance d of a car after the brakes are applied varies directly as the square of the speed r, if a car traveling 40 mph can stop in 70 ft, how many feet will it take the same car to stop when its traveling 80 mph.
use the following formula
d = kr^2
and you can get k from the given information.
thanks in advance for any explanation or help of any kind.
eddy
 

bigjohn 2

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This is a proportion problem...typically written "range variable = k (expression in a domain variable) "....for this query ' d ' is the range value and 'r² ' is the expression ...you are give that d = 70 when r = 40...use this to find a value for ' k '.........then use this k value with r = 80 to find the correct value for distance d .
 

skeeter

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doubling the speed increases the stopping distance by a factor of four … see if you can figure out why
 

eddy2017

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braking distance will quadruple when your speed doubles because the work that it takes to stop your car means removing all of the kinetic energy. That means that for a fixed maximum braking force, your braking distance will be proportional to the square root of the velocity.
 

Dr.Peterson

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this is another exercise of the kind i just posted. it is interesting for me because the use of this formula is advised. i do not understand it and the formula is not explained.​

The stopping distance d of a car after the brakes are applied varies directly as the square of the speed r, if a car traveling 40 mph can stop in 70 ft, how many feet will it take the same car to stop when its traveling 80 mph.
use the following formula
d = kr^2
and you can get k from the given information.
thanks in advance for any explanation or help of any kind.
eddy
This formula is more realistic than the other problem, though it ignores reaction time. See here.

I presume you got this from an algebra source that just provides the formula in terms of variation, rather than from a physics source that explains the reason, or a calculus source that derives the formula.

Do you see how you can use "a car traveling 40 mph can stop in 70 ft" to find k, and then use that to answer "how many feet will it take the same car to stop when its traveling 80 mph?"
 

eddy2017

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Right from an Algebra source. I'll give it my best.
Only one question,
What does k stand for?
Perhaps, constant?. If so what is the constant?.
I know in d=k*r^2, I can solve for k
k = d/r2,
But need to what it is. Some kind of constant?

r is the square of the speed r, that I see.
 
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skeeter

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braking distance will quadruple when your speed doubles because the work that it takes to stop your car means removing all of the kinetic energy. That means that for a fixed maximum braking force, your braking distance will be proportional to the square root of the velocity.

correction … braking distance will be proportional to the square of velocity, not the square root, which is what the equation [imath] d = kr^2[/imath] states.
 

eddy2017

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correction … braking distance will be proportional to the square of velocity, not the square root, which is what the equation [imath] d = kr^2[/imath] states.
What does k stand for, skeeter?
 

eddy2017

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I know that 'cause the distance d varies directly with the square of the speed r, i can come up with the equation d=kr^2 where k is some constant.
What kind of constant?
I'm not getting that
 

skeeter

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k is a constant of proportionality

given your equation, d = kr^2, the ratio d/r^2 is a constant value

to find k, you were given d = 70 ft when r = 40 mph = 176/3 ft/sec

k = (70 ft)/(176/3 ft/sec)^2 = 315/15488 sec^2/ft, but you really don’t need to calculate it …

[math]\frac{d_1}{r_1^2} = \frac{d_2}{r_2^2} \implies d_2 = \dfrac{d_1 \cdot r_2^2}{r_1^2} = \dfrac{70 \cdot 80^2}{40^2} = 70 \cdot \left(\dfrac{80}{40}\right)^2 = 70 \cdot 2^2 = 280 \, ft[/math]
 
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Otis

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… What does k stand for? …
The factor k is what we call a parameter. This particular parameter is a Real number (i.e., a constant). The parameter is needed in the formula because not all vehicle types, under different braking and road conditions, have the same braking distance. By altering the value of k, the formula adjusts for different scenarios.

You find the particular value of k for the given exercise by first substituting the values provided for variables d and r in the formula, followed by solving the resulting equation for k.

😎
 

Otis

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… very interesting all that
Anytime you have an equation containing three symbolic numbers, and you have values for two of them, you can find the value of the third symbol by substituting the known values and solving the resulting equation. That's more than interesting, Eddy.

;)
 

eddy2017

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Anytime you have an equation containing three symbolic numbers, and you have the values for two of them, you can find the value of the third symbol by substituting the known values and solving the resulting equation. That's more than interesting, Eddy.

;)
Wht do you mean by symbolic numbers?
Variables?
 
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Otis

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Wht do you mean by symbolic numbers? …
I mean numbers that are represented by symbols. They can be variables or constants (and symbols for constants in a formula might represent values that change from scenario to scenario; if they do, then we call those constants 'parameters').

EGs:

y = ax^2 + bx + c

Five symbolic numbers (x,y are variables and a,b,c are parameters).

A = pi * r^2

Three symbolic numbers (A,r are variables and symbol pi is a fixed constant, not a parameter).

😎
 

eddy2017

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Very kind of you taking the time to explain that. It clears up a lot in my mind!. Thanks.
 

eddy2017

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k is a constant of proportionality

given your equation, d = kr^2, the ratio d/r^2 is a constant value

to find k, you were given d = 70 ft when r = 40 mph = 176/3 ft/sec

k = (70 ft)/(176/3 ft/sec)^2 = 315/15488 sec^2/ft, but you really don’t need to calculate it …

[math]\frac{d_1}{r_1^2} = \frac{d_2}{r_2^2} \implies d_2 = \dfrac{d_1 \cdot r_2^2}{r_1^2} = \dfrac{70 \cdot 80^2}{40^2} = 70 \cdot \left(\dfrac{80}{40}\right)^2 = 70 \cdot 2^2 = 280 \, ft[/math]
hi, skeeter, i want to ask you if possible when you find some time, plase, can you explain this to me?
i do not understand :
how you got 176/3 ft/sec. how you got that result and why?
and why do you have r^2 subscripted with a 1?
can you walk me thru that, please.

i understood the problem. i really did.
d = kr2
Get k from the given information.
k = d/r2 = (70 ft) / (40 mph)2
k= 0.04375 ft/mph2
constant of proportionality (k) = d(distance you can stop) / (r)the square of the speed
now, when r = 80 mph
d = kr2 = (0.04375 ft/mph^2)(80 mph)^2
d=280 ft
but i am interested in your approach, which gives the same result but it is totally different and a little bit more involved. maybe not, the problem is i do not understand it.
 
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skeeter

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converting miles per hour to feet per second …

[math]\frac{40 \, miles}{1 \, hr} \cdot \frac{5280 \, ft}{1 \, mile} \cdot \frac{1 \, hr}{3600 \, sec} = \dfrac{176}{3} \, ft/sec[/math]
in solving for the stopping distance, [imath]d_2[/imath], for the faster initial speed, I set …

[imath]r_1 = 40 \, mph[/imath], [imath]d_1 = 70 \, ft[/imath], and [imath]r_2 = 80 \, mph[/imath]
 

eddy2017

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wow, that is so good!. thanks. got it now. it takes care of the units right then and there and makes it much more clean.
wow, i like that.
 
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