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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

Aolong Li
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Sullo
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  • If you solved simbolically Smin=Smax where the unknown is v0, you would get two values at best. The lower value represents the slowest he could go to clear the intersection while the higher value represents the fastest it could go to stop ahead. Any speed value between those two he could either stop ahead or pass through safely. – WindSoul Dec 04 '19 at 16:36
  • @WindSoul I did that I evened but gave an equation. So I thought this was proof that didn't give zero, so it should have a space – Sullo Dec 04 '19 at 16:45
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    Not correct. You have the space, Smin=Smax. If the times t and T are right, you will obtain two solutions that gives you a range of speeds for which either passing or stopping is possible. If your speed is outside the range to the right, then you may pass safely but not stop ahead. If the speed is outside the range to the left, you may stop safely ahead but can’t pass before red light. – WindSoul Dec 04 '19 at 16:57

1 Answers1

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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.

Quanto
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  • Very good! Thank you!! – Sullo Dec 04 '19 at 16:50
  • @Sullo - You bet. No problem. – Quanto Dec 04 '19 at 16:52
  • This seems like a potentially useful result! If you know your reaction time $T$, the yellow-light time $t$, and your car's braking deceleration $a$, then you'd better not drive faster than $2a(t-T)$ as you approach the light. A quick online search for typical values gives me $T=1.5$, $t=3.6$, and $a=7.1$ (on the low end) in SI units, giving $30.,\mathrm{m/s}$, which is $67,\mathrm{mi/hr}$, or $107,\mathrm{km/hr}$. But a distracted driver can have $T=3.0$ (double the usual), giving only $19,\mathrm{mi/hr}$, or $31,\mathrm{km/hr}$. Very different! – Toby Bartels Dec 04 '19 at 17:31