The more kinetic energy an object has, the larger velocity it has if the mass holds constant. Therefore, unless the stopping force is larger for a vehicle of higher kinetic energy with the same mass, than under constant acceleration, the stopping distance of a vehicle traveling at higher velocities would be greater than the stopping distance of a vehicle traveling at lower velocities.
Mathematically, if we call the deceleration \(a\) and the time to decelerate for the car traveling at 30mph is \(t\), than \(30\text{mph} - \text{at} = 0\), than it is obvious that the 60mph car requires \(2t\), therefore the deceleration would total \(2at\). Using a kinematic equation, we see that \(30t + \frac{1}{2}at^2 \ \textless \ 120t + 2at^2\).
