The stopping distance s of a car varies directly as the square of its speed v. If a car traveling at 30 mph requires 81 ft to stop, find the stopping distance for a car traveling at 55 mph.
s = kv^2, so s/v^2 is constant. That means you want s where
s/55^2 = 81/30^2
To solve this problem, we need to use the concept of direct variation. The problem states that the stopping distance (s) of a car varies directly as the square of its speed (v). Mathematically, we can express this relationship as:
s = kv^2
where k is the constant of proportionality.
We are given that when the car is traveling at 30 mph, it requires 81 ft to stop. Let's substitute these values into the equation to solve for k:
81 = k(30)^2
81 = 900k
Divide both sides of the equation by 900 to isolate k:
k = 81/900
k = 0.09
Now that we have found the value of k, we can use it to find the stopping distance for a car traveling at 55 mph. Let's substitute the new speed (v = 55) and the value of k into our equation:
s = 0.09(55)^2
s = 0.09(3025)
s = 272.25
Therefore, a car traveling at 55 mph would require 272.25 ft to stop.