# Thread: Solve Real-World Variation Problems

1. ## Solve Real-World Variation Problems

I tried this one myself, then I checked the back of the book and the answer didn't match. I am not sure what I am doing wrong here. If someone could help me set this up, that would be great.
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"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. It requires 226 lb of force to keep a 2200-lb car, traveling at 30 mph, from skidding on a curve of radius 500 ft. How much force is required to keep a 3000-lb car, traveling at 45 mph, from skidding on a curve of radius 400 ft?"

This is what I did:

I tried to find K so that I could solve the second half of the equation.

f = $\frac{Kws^2}{r}$

226 = $\frac{K(2200)(30)^2}{500}$

K = $\frac{226}{500}$

K = ${.452}$
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So now that I had K, I tried to solve the equation "How much force is required to keep a 3000-lb car, traveling at 45 mph, from skidding on a curve of radius 400 ft?"

I did this:

f = $\frac{Kws^2}{r}$

f = $\frac{(.452)(3000)(45)^2}{400}$

f = $\frac{2745900}{400}$

f = ${6864.75}$
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Now obviously I did something wrong because the answer in the back says "about 1020.2 lb of force" -- I'm not understanding how to set it up? Thanks a lot.

2. Originally Posted by CzWJF41
I tried this one myself, then I checked the back of the book and the answer didn't match. I am not sure what I am doing wrong here. If someone could help me set this up, that would be great.
------------------------------------------------------

"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. It requires 226 lb of force to keep a 2200-lb car, traveling at 30 mph, from skidding on a curve of radius 500 ft. How much force is required to keep a 3000-lb car, traveling at 45 mph, from skidding on a curve of radius 400 ft?"

This is what I did:

I tried to find K so that I could solve the second half of the equation.

f = $\frac{Kws^2}{r}$

226 = $\frac{K(2200)(30)^2}{500}~\implies~\bold{\color{bl ue}226=3960 \cdot K ~\implies~K\approx 0.0571}$

K = $\frac{226}{500}$

K = ${.452}$
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So now that I had K, I tried to solve the equation "How much force is required to keep a 3000-lb car, traveling at 45 mph, from skidding on a curve of radius 400 ft?"

...

Now obviously I did something wrong because the answer in the back says "about 1020.2 lb of force" -- I'm not understanding how to set it up? Thanks a lot.
Using the K-value of 0.0571 I get $f \approx 866.8\ lb$