SOLUTION: Skidding Car: The force needed to keep a car from skidding on a curve varies inversely as the radius r of the curve and jointly as the weight of the car and the square of the speed

Algebra ->  Customizable Word Problem Solvers  -> Evaluation -> SOLUTION: Skidding Car: The force needed to keep a car from skidding on a curve varies inversely as the radius r of the curve and jointly as the weight of the car and the square of the speed      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 1089390: Skidding Car: The force needed to keep a car from skidding on a curve varies inversely as the radius r of the curve and jointly as the weight of the car and the square of the speed. It takes 3000 lb of force to keep a 200 lb car from skidding on a curve of radius 500 ft at 30 mph. What force will keep the same car from skidding on a curve of radius 800 ft at 60 mph?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
F= force needed to keep a car from skidding, in lb of force
r= radius of the curve, in feet
s= speed of the car, in mph
We could make
w= weight of the car, but that does not change.

For inverse and direct variations there is always a constant,
multiplied times the variables involved in direct variation,
and divided by the variables involved in inverse variation.
In this case the relation of F with the variables is
F=kws%5E2%2Fr
As it is the same car, w is constant, so we can write it simply as
F=Ks%5E2%2Fr (with a new constant K=kw ).
We are told that F=3000 for r=500 and s=30 , so
3000=K30%5E2%2F500
3000=900K%2F500
3000%2A500%2F900=K
K=5000%2F3
So, the relation is
F=5000s%5E2%2F3r .
When the speed is s=60 and the radius is r=800 ,
F=5000%2A60%5E2%2F%283%2A800%29
F=5000%2A3600%2F2400
F=5000%2A3%2F2
highlight%28F=7500%29 .
The force needed to keep a car from skidding when taking a curve
with radius 800 ft at 60 mph is highlight%287500%29 lb of force.