Complete question is;
Use a variation model to solve for the unknown value.
The stopping distance of a car is directly proportional to the square of the speed of the car.
a. If a car traveling 50 mph has a stopping distance of 170 ft, find the stopping distance of a car that is traveling 70 mph.
b. If it takes 244.8 ft for a car to stop, how fast was it traveling before the brakes were applied?
Answer:
A) d = 333.2 ft
B) 60 mph
Explanation:
Let the stopping distance be d
Let the speed of the car be v
We are told that the stopping distance is directly proportional to the square of the speed of the car. Thus;
d ∝ v²
Therefore, d = kv²
Where k is constant of variation.
A) Speed is 50 mph and stopping distance of 170 ft.
v = 50 mph
d = 170 ft = 0.032197 miles
Thus,from d = kv², we have;
0.032197 = k(50²)
0.032197 = 2500k
k = 0.032197/2500
k = 0.0000128788
If the car is now travelling at 70 mph, then;
d = 0.0000128788 × 70²
d = 0.06310612 miles
Converting to ft gives;
d = 333.2 ft
B) stopping distance is now 244.8 ft
Converting to miles = 0.046363636 miles
Thus from d = kv², we have;
0.046363636 = 0.0000128788(v²)
v² = 0.046363636/0.0000128788
v² = 3599.99658
v = √3599.99658
v ≈ 60 mph
