If the speed of the vehicle doubles, how much more braking force is required to stop over the same distance?

When the speed of a vehicle doubles, the kinetic energy of the vehicle increases by a factor of four. This relationship is based on the formula for kinetic energy, which is expressed as ( KE = \frac{1}{2} mv^2 ), where ( m ) is the mass of the vehicle and ( v ) is its velocity.

If you double the velocity (speed), the new kinetic energy becomes ( KE' = \frac{1}{2} m (2v)^2 = \frac{1}{2} m (4v^2) = 4 \cdot \left(\frac{1}{2} mv^2\right) ). This shows that the kinetic energy becomes four times greater when the speed increases from ( v ) to ( 2v ).

To bring the vehicle to a stop over the same distance, the braking force must overcome this increased kinetic energy. The work done by the brakes (which is force times distance) must equal the kinetic energy of the vehicle. Since work equals force times distance, to stop the vehicle over the same distance, the braking force must be increased to compensate for the fourfold increase in kinetic energy. Hence, four times the braking force