The braking distance for a car

[Guest]

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]

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.

[HallsofIvy]

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?

[Baltuilhe]

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!

[Guest]

i got it, thank you for the explanation.