Direct and Inverse Variation problem

Question

The force needed to keep a car from skidding on a curve varies directly as the weight of the car and the square of the speed and inversely as the radius of the curve. Suppose a 3,960 lb. force is required to keep a 2,200 lb. car traveling at 30 mph from skidding on a curve of radius 500 ft. How much force is required to keep a 3,000 lb. car traveling at 45 mph from skidding on a curve of radius 400 ft.?

Answer 1

The problem is a direct variation problem that requires you to translate into a formula.

1. Force to Hold = (weight of car)(Speed)²

(radius)

This formula does not need to be given as it is given in the explanation. Varies Directly (in the numerator) AND the square of the speed (also in numerator) and Inversely (in the denominator) with the radius

You can check the formula given. 3960 = K (30)²

500

Solving this gives you 500* 3960. = 900K...

1980000=900K, so k = 2200

Now use the same formula and plug in 3000 for K , 45 for speed and 400 for radius to get the new Force

F = (3000) (45)²

⁴⁰⁰

So F = 15,187.5 lbs of force needed to hold the car on the road

Answer 2

Force, F=kWs²/r

k=rF/(Ws²) = 500(3960)/(2200(30²)) = 1

F= (1)(3000)(45²)/400