What is the combined stopping power necessary when both the vehicle speed and weight are doubled?

To understand the correct answer, it's essential to consider the physics involved in stopping a vehicle, which is primarily governed by the principles of inertia and kinetic energy. When you double both the speed and weight of a vehicle, the stopping power required to bring it to a halt increases significantly.

First, let's analyze the effect of doubling the vehicle's speed. The kinetic energy of a vehicle is proportional to the square of its speed. Therefore, if you double the speed, the kinetic energy increases by a factor of four (since ( KE \propto v^2 )).

Next, consider the effect of doubling the vehicle's weight. The weight of the vehicle directly affects the frictional force available for braking, as more weight typically means more friction. If you double the weight, you also double the force necessary to stop the vehicle.

Now, combining these two effects results in an overall increase in the required stopping power. Since the kinetic energy increases fourfold due to the increased speed and the force required to counteract the weight doubles, the total stopping power needed becomes a multiplication of these two factors:

[

\text{Total required stopping power} = 4 \times 2 = 8

]

Thus, when both the vehicle speed and