SOLUTION: Solve: The force needed to keep a car from skidding on a curve varies jointly as the weight of the car and the square of the car's speed, and inversely as the radius of the curv

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: Solve: The force needed to keep a car from skidding on a curve varies jointly as the weight of the car and the square of the car's speed, and inversely as the radius of the curv      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 145009: Solve:
The force needed to keep a car from skidding on a curve varies jointly as the weight of the car and the square of the car's speed, and inversely as the radius of the curve. If a force of 3600 pounds is needed to keep an 1800 pound car traveling at 20mph from skidding on a curve of radius 600 feet what force would be required to keep the same car from skidding on a curve of radius 570 feet at 50mph? round to the nearest pound of force.
Please help...I do not know where to begin with this one!

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The force needed to keep a car from skidding on a curve varies jointly as the weight of the car and the square of the car's speed, and inversely as the radius of the curve.
F = k[ws^2/r]
-------------------
If a force of 3600 pounds is needed to keep an 1800 pound car traveling at 20 mph from skidding on a curve of radius 600 feet
3600 = k[1800*20^2/600]
k = 3
Equation:
F = 3[ws^2/r]
----------------
What force would be required to keep the same car from skidding on a curve of radius 570 feet at 50 mph? Round to the nearest pound of force?
F = 3[1800*50^2/570]
F = 23684
----------------
Cheers,
Stan H.