The distance required for an emergency stop for a car varies directly as the square of the car’s speed. A car traveling at 50 mph requires 140 feet to stop. What is the stopping distance for a car traveling at 30 mph? What is the stopping distance for a car traveling at 70 mph?
You are told that
d(v) = kv^2
140 = k*2500
k = 0.056
so,
d(30) = .056*900
d(70) = .056*4900
THANK YOU!
350
To answer this question, we need to use the given information to establish a relationship between the speed of the car and the distance required for an emergency stop.
We are told that the distance required varies directly as the square of the car's speed. This means that we can set up a proportion using the given information.
Let's denote the distance required for an emergency stop as d, and the car's speed as s. The proportion can be written as:
d ∝ s^2
Using the first given condition, where the car is traveling at 50 mph and requires 140 feet to stop, we can substitute these values into the proportion:
140 ∝ 50^2
Simplifying, we get:
140 ∝ 2500
To determine the constant of proportionality, we divide both sides of the proportion by 140, resulting in:
1 ∝ 2500/140
Simplifying further, we find:
1 ∝ 25/14
Now that we have the constant of proportionality, we can use it to find the stopping distance for a car traveling at 30 mph and 70 mph.
For a car traveling at 30 mph, we can set up the equation:
d = (25/14) * 30^2
Simplifying, we find:
d = (25/14) * 900
d = 1625 feet
Therefore, the stopping distance for a car traveling at 30 mph is 1625 feet.
For a car traveling at 70 mph, we can use the same equation:
d = (25/14) * 70^2
Simplifying, we find:
d = (25/14) * 4900
d = 8750 feet
Therefore, the stopping distance for a car traveling at 70 mph is 8750 feet.