Question

A car's stopping distance varies directly with the speed it travels and inversely with the friction value of the road's surface. If a car takes 60 feet to stop at 32 mph on a road whose friction value is 4, what would be the stopping distance of a car traveling at 60 mph on a road with a friction value of 2?

A. 120 feet
B. 90 feet
C. 80 feet
D. 60 feet

Answer 1

The stopping distance of a car is directly proportional to its speed and inversely proportional to the friction value of the road's surface. Using the given information, we can determine the stopping distance of a car traveling at 60 mph on a road with a friction value of 2.

Step-by-step explanation:

The stopping distance of a car varies directly with its speed and inversely with the friction value of the road's surface. This can be represented by the equation:

Stopping Distance = k × Speed / Friction Value

Given that the car takes 60 feet to stop at 32 mph on a road with a friction value of 4, we can plug these values into the equation to find the constant of variation, k:

60 = k × 32 / 4

Simplifying the equation, we find k = 15.

Now, let's use this constant of variation to find the stopping distance of a car traveling at 60 mph on a road with a friction value of 2:

Stopping Distance = 15 × 60 / 2 = 450 / 2 = 225 feet

Therefore, the stopping distance of a car traveling at 60 mph on a road with a friction value of 2 would be 225 feet.