mechanics ,,

s=ut+1/2at^2
if the deceleration of the car is halved,
a becomes 1/2a,
s=ut+1/4at^2

By the use of v^2=u^2+2as, if the final speed is 0, then we have v=0, a is negetive (braking), then, u^2=-2as, we can see if the deceleration of the car is halved, the stopping distance will be doubled, if the initial speed keeps the same.

But if there is the reaction time taken in, since the reaction time is no depent on anything, except you, so a fixed distance s' are needed before braking is applied. Then the total distance sb=s'+s=ut(reaction)+abs(u^2/2a)^(0.5).

是否用linear motion的equation計算過發覺distance是double呢?
這句子正確在"stopping distance"
stopping distance=braking distance + thinking distance
thinking distance在車的deceleration減半的情況下是不變的
while braking distance is halved
let braking distance=2a, thinking distance=b
stopping distance=2a+b
當deceleration is halved, stopping distance=a+b
所以the statement is correct