Question

A car's stopping distance in feet is modeled by the equation d(v) = 2.15v² / 58.4f, where v is the initial velocity of the car in miles per hour and f is a constant related to friction. If the initial velocity of the car is 47 mph and f = 0.34, what is the approximate stopping distance in feet?

Answer 1

Using the stopping distance equation d(v) = 2.15v² / 58.4f, the approximate stopping distance for a car with an initial velocity of 47 mph and a friction constant of 0.34 is 239.22 feet.

Step-by-step explanation:

To calculate the approximate stopping distance of a car using the provided model, we plug in the initial velocity (v) of 47 mph and the friction constant (f) of 0.34 into the equation d(v) = 2.15v² / 58.4f. Performing the calculation:

d(47) = 2.15 * 47² / 58.4 * 0.34

d(47) = 2.15 * 2209 / 19.856

d(47) ~ 2.15 * 111.265

Therefore, the approximate stopping distance is:

d(47) ~ 239.22 feet

This is the distance required for the car to come to a complete stop given an initial velocity of 47 mph and a friction factor of 0.34, under the assumptions of the model.