The braking distance for a car

The braking distance for a car

Postby Guest » Wed Nov 18, 2020 6:12 am

Given the formula:
D=[tex]\frac{v^2}{64.4(F+G)}[/tex]

where:
D = breaking distance in feet
v = speed of car (feet per second) when breaks are applied
F = coefficient of friction (no units, number between 0 and 1)
G = grade of the road (no units expressed as decimal)

A car takes 63 feet to complete to a stop after its driver applies the brakes. The grade of the road is 0.01 and the coefficient of friction is 0.5. How fast was the car moving, in miles per hour, when the brakes were applied?

How would the answer be 31 miles an hour? I've tried endless combinations of numbers but I keep getting the wrong answer and I don't know if there's some hidden formula for this that I am missing as well to be able to solve the problem correctly...
Guest
 

Re: The braking distance for a car

Postby Guest » Wed Nov 25, 2020 6:33 pm

If you have correctly quoted the problem then the answer is NOT "31"!

The formula is [tex]D= \frac{v^2}{64.4(F+ G)}[/tex]
[tex]63= \frac{v^2}{64.4(0.5+ 0.01)}= \frac{v^2}{64.4(0.51)}= \frac{v^2}{32.844}[/tex]
so
[tex]v^2= 63(32.844)= 2069.172[/tex]
[tex]v= 45.48815230364935[/tex]

To two decimal places that is 45.49 mph.
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Re: The braking distance for a car

Postby HallsofIvy » Mon Dec 21, 2020 6:38 pm

I am puzzled by the formula itself!

[tex]D=\displaystyle \frac{v^2}{64.4(F+G)}[/tex]
D is in feet and v is in feet per second. You say that G and F have no units. What about the consant,"64.4"? [tex]v^2[/tex] would have units or "feet squared per second square". In order that the right hand side also have units of "feet" the "64.4" would have to have "meters per second squared". Is that "2g", twice the acceleration due to gravity?

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Re: The braking distance for a car

Postby Baltuilhe » Tue Dec 22, 2020 12:02 am

Good night!

The formula:
[tex]D=\frac{v^2}{64(F+G)}[/tex]
D = feet
v = feet per second

With:
D = 63ft
F=0.5
G=0.01

As were calculated:
[tex]v\approx 45,488ft/s[/tex]

Converting to miles per hour:
[tex]v\approx \frac{15}{22}\cdot 45,488\approx 31,015 mph[/tex]

Hope to have helped!

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Re: The braking distance for a car

Postby Guest » Mon Mar 01, 2021 3:46 am

i got it, thank you for the explanation.
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