Chapter 5 - Force and Motion-I - Problems - Page 118: 28c

$4.0$

$v^{2} = v_{0}^{2} - 2 a \Delta x$ $v_{0}$ = initial speed; $v$ = final speed = 0. So, $v_{0}^{2} = 2 a \Delta x$ Force remained the same, the acceleration(deceleration in this case) $a$ remains the same. So, if speed doubles, the stopping distance $\Delta x$ with be 4 times that earlier. So the factor is $4.0$.