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Question

The force required to prevent a car from skidding on a flat curve varies directly as the weight of the car and the square of its speed, and inversely as the radius of the curve. It requires 290 lb of force to prevent a 2,200 lb car traveling at 35 mph from skidding on a curve of radius 520 ft. How much force is required to keep a 2,800 lb car traveling at 50 mph from skidding on a curve of radius 415 ft ?

Solution

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The variation equation is

F=kws2rF=\dfrac{kws^2}{r}

where kk represents the constant of variation, FF is the force, ww is weight of the car, ss is the speed of the car and rr is the radius of the curve.

First, find the constant of variation.

To find the constant of variation, substitute in the variation equation 290290 for FF, 2,2002,200 for ww, 3535 for ss and 520520 for rr.

290=k2,200352520290=k2,2001225520290=k2,695,000520Multiply both sides by 520.290520=2,695,000k150,800=2,695,000kDivide both sides by 2,695,000.150,8002,695,000=kk=0.056\begin{align*} 290&=\dfrac{k\cdot2,200\cdot 35^2}{520}\\ 290&=\dfrac{k\cdot2,200\cdot 1225}{520}\\ 290&=\dfrac{k\cdot2,695,000}{520}&&\text{Multiply both sides by $520$.}\\ 290\cdot520&=2,695,000k\\ 150,800&=2,695,000k&&\text{Divide both sides by $2,695,000$.}\\ \dfrac{150,800}{2,695,000}&=k\\ k&=0.056 \end{align*}

Now, substitute in the variation eqaution 0.0560.056 for kk, 2,8002,800 for ww, 5050 for ss and 415415 for rr to find FF.

F=0.0562,800502415F=0.0562,8002500415F=392,000415F=944.58\begin{align*} F&=\dfrac{0.056\cdot2,800\cdot50^2}{415}\\ F&=\dfrac{0.056\cdot2,800\cdot2500}{415}\\ F&=\dfrac{392,000}{415}\\ F&=944.58 \end{align*}

Required force is 944.58944.58 lb.

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