Question

Proportional Reasoning: Assuming equal rates of acceleration in both cases, how much longer would it take a car to stop if braking from 56 mi/h than from 28 mi/h?

A) 8 times as long
B) 4 times as long
C) 2 times as long
D) 1.4 times as long
E) the same in both cases

Answer 1

It would take twice as long for a car to stop if it was braking from 56 mi/h as opposed to 28 mi/h, assuming a constant deceleration rate, because the stopping time is proportional to the velocity squared.

Step-by-step explanation:

The amount of time it takes for a car to stop from a certain speed is directly related to the initial speed squared when assuming a constant deceleration rate. According to the kinematic equations, the stopping time is proportional to the velocity squared, given by the formula v2 = 2ad, where v is the velocity, a is the acceleration (in this case, deceleration so it's negative), and d is the distance. Given that the deceleration rate is constant, if we double the speed from which we are stopping (28 mi/h to 56 mi/h), then we are effectively quadrupling the required stopping distance because the square of 2 (for doubling the speed) is 4. Since the stopping distance is quadrupled, and if we assume the deceleration rate does not change, it would take twice as long to stop from 56 mi/h as it would from 28 mi/h because distance is proportional to time squared in the uniformly accelerated motion.