Red Light Dilemma

Let's denote $D_a$ and $D_d$ as the distance you travel when accelerating and decelerating, respectively. Let's also denote $x$ as distance between the car and the stoplight at the moment when the light turns yellow.

In order to find our solution, we simply have to find the values of $x$ that satisfy $D_a < x < D_d$ . All we have to do is find $D_a$ and $D_d$ and we'll have our bounds, therefore our answer will be of the form $D_a+D_d$ .

Let's first find $D_a$ . For this method, Alice's dad starts at a speed of $40m/s$ and accelerates at $8m/s$ for $2$ seconds. Using the formula $d=v_it+\frac{1}{2}at^2$ :

$D_a = (40m/s)(2s)+\frac{1}{2}(8m/s^2)(2s)^2=96m$

Now all that's left is to find $D_d$ . In this case, Alice's dad again starts with an initial velocity of $40m/s$ but ends with a final velocity of $0m/s$ after an unknown period of time. Using the formula $v_f^2=v_i^2+2ad$ and solving for $d$ :

$D_d=\frac{v_i^2}{2a}=\frac{(40m/s)^2}{2(8m/s^2)}=100m$

We know now that Alice's dad would have had to have been between $96m$ and $100m$ away from the stoplight in order for this unfortunate situation to occur. We've found our bounds $D_a$ and $D_d$ , therefore our answer is $D_a+D_d=96+100=\boxed{196}$ .

@Garrett Clarke Check your units for acceleration. I think it should be $8 m/s^2$ and NOT $8 m/s$ .