A driver approaches a traffic light, which is green with speed $ v_0 $ when it turns yellow.
a) If the driver's reaction occurs within temp $ T $, during which he decides to stop and apply the brake foot, and if the maximum brake deceleration is $ a $, what is the minimum distance $ S_ {min} $ before hitting Does the intersection, the moment the light turns yellow, can it make the car stop without crossing it?
b) If the yellow light stays on for a time $ t $ before turning red, what is the distance $ S_ {\max} $, before the intersection, the instant the yellow light comes on so that it can cross the intersection with speed $ v_0 $ without the red light coming on?
c) Show that in case the initial velocity is greater than$$ v_ {0\max} = 2a (t-T) $$there will be a range of distances before the rally so that the driver does not stop in time and cannot cross it without the red light coming on.
Solution:
''a) The car travels a distance $d_1=v_0T$ while the driver is reacting. It then travels a further distance $d_2=\frac{v_0^2}{2a}$ while decelerating at a rate $a$ before it comes to rest. So the total distance travelled is:
$S_{min}=d_1+d_2 = v_0T + \frac{v_0^2}{2a}$
b) This is simpler than you think. The car is travelling at a constant velocity $v_0$, so it will reach the intersection before the red light comes on as long as its distance from the intersection when the yellow light comes on (t seconds earlier) is less than
$S_{max} = v_0t$
c) The driver can stop before the intersection if his initial distance is greater than $S_{min}$. And he can cross the intersection before the red light comes on if his initial distance is less than $S_{max}$. If $S_{min} \le S_{max}$ then for any initial distance he can either stop before the intersection or cross the intersection before the red light comes on (and for distances between $S_{min}$ and $S_{max}$ he can choose from both actions).
For what value of $v_0$ is $S_{min}$ equal to $S_{max}$ ? If $v_0$ is greater than this value then $S_{min} > S_{max}$. What happens now if the driver's initial distance is between $S_{min}$ and $S_{max}$ ?''
I still don't know how to prove it. I couldn't think of a distance left or missing to play in the equations
Best Answer
Note that, if $v_0=2a(t-T)$, the result from a) $S_{min}= v_0T + \frac{v_0^2}{2a}$ leads to,
$$S_{min}= 2a(t-T)T + \frac1{2a}[2a(t-T)]^2 = 2a(t-T)t = S_{max}$$
Therefore, $v_0=2a(t-T)$ is the critical speed with which the driver can either break to stop before the intersection or cruise cross the intersection before the red light.
In the case where $v_0 > 2a(t-T)$, we have
$$S_{min}>2a(t-T)t=S_{max}$$
That is, if the drive is closer than $S_{min}$ and further than $S_{max}$ from the the intersection, there is no way over this range $(S_{max}, S_{min})$ for the drive to stop in time or cruise cross the intersection without the red light coming on.