Answer:
159.375 feet.
Step-by-step explanation:
Let d represent stopping distance of a car and s represent speed of car.
We have been given that the stopping distance of a car varies directly as the square of its speed.
We know that two directly proportional quantities are in form [tex]y=kx[/tex], where y is directly proportional to x and k is constant of proportionality.
We can represent our given information in an equation as:
[tex]d=ks^2[/tex].
We are also told that a car traveling at 20 mph requires 25.5 ft to stop. We can represent this information in our equation as:
[tex]25.5=k(20)^2[/tex]
[tex]25.5=400k[/tex]
[tex]k=\frac{25.5}{400}[/tex]
[tex]k=0.06375[/tex]
Now, we will substitute [tex]k=0.06375[/tex] and [tex]s=50[/tex] in our equation and solve for d as:
[tex]d=0.06375(50)^2[/tex]
[tex]d=0.06375(2500)[/tex]
[tex]d=159.375[/tex]
Therefore, a car traveling at 50 mph requires 159.375 feet to stop.
