Answer
$2.0$.
Work Step by Step
$v = v_{0} - a \Delta t$
$v_{0}$ = initial speed; $v$ = final speed = 0.
So, $v_{0} = a \Delta t$.
Force remained the same, the acceleration(deceleration in this case) $a$ remains the same.
$\Delta t$ is directly proportional to $v_{0}$
So, if speed doubles, the stopping distance $\Delta t$ with be 2 times that earlier.
So the factor is $2.0$.
