Final answer:
A vehicle that is twice the weight and traveling at twice the speed of a car requires eight times the stopping power to stop in the same distance, since it has eight times the kinetic energy.
Step-by-step explanation:
The question necessitates an understanding of basic principles of physics, specifically kinetic energy and stopping power, since it asks about the required braking power of a vehicle based on its weight and speed. The kinetic energy (KE) of a moving object is proportional to its mass (m) and the square of its velocity (v), described by the formula KE = 1/2 mv2. As such, if the weight of a vehicle is doubled (2m) and its speed is doubled (2v), its kinetic energy becomes four times the mass times the velocity squared, or KE = 1/2 × 2m × (2v)2 = 8 × 1/2 mv2, meaning eight times the kinetic energy of the original car.
If a vehicle with eight times the kinetic energy of another needs to stop in the same distance, it will require eight times the stopping power. Therefore, the correct answer is d. 8 times the stopping power.