Question

The stopping distance of a car varies directly as the square of its speed. If a car traveling at 20mph requires 25.5ft to stop, find the stopping distance for a car traveling at 50mph.

Answer 1

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 $y=kx$, where y is directly proportional to x and k is constant of proportionality.

We can represent our given information in an equation as:

$d=ks^2$.

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:

$25.5=k(20)^2$

$25.5=400k$

$k=\frac{25.5}{400}$

$k=0.06375$

Now, we will substitute $k=0.06375$ and $s=50$ in our equation and solve for d as:

$d=0.06375(50)^2$

$d=0.06375(2500)$

$d=159.375$

Therefore, a car traveling at 50 mph requires 159.375 feet to stop.