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

When considering the stopping power required for a vehicle, it's essential to understand the relationship between speed, weight, and stopping distance. The stopping power is influenced by both the mass of the vehicle and the velocity at which it is traveling.

When both the speed and weight of the vehicle are doubled, the stopping power increases significantly. Specifically, if you double the speed of a vehicle, the kinetic energy, which is dependent on the square of the speed, increases by a factor of four (since kinetic energy is proportional to the square of speed: KE = 1/2 mv^2). Therefore, if speed is doubled, the kinetic energy becomes four times greater.

In addition, if you also double the weight of the vehicle, you will have a direct increase in the force required to bring the vehicle to a stop, as the force required for deceleration is directly proportional to the vehicle's mass.

Putting this all together: Doubling both the speed and weight leads to a scenario where the stopping power required is proportional to the increase in kinetic energy due to speed (4 times) multiplied by the increase in mass (2 times). This results in a combined factor of 4 (from speed) multiplied by 2 (from weight), which equals 8