## Related questions with answers

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

VerifiedThe variation equation is

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

where $k$ represents the constant of variation, $F$ is the force, $w$ is weight of the car, $s$ is the speed of the car and $r$ is the radius of the curve.

First, find the constant of variation.

To find the constant of variation, substitute in the variation equation $290$ for $F$, $2,200$ for $w$, $35$ for $s$ and $520$ for $r$.

$\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 $0.056$ for $k$, $2,800$ for $w$, $50$ for $s$ and $415$ for $r$ to find $F$.

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

Required force is $944.58$ lb.

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