Can You Calculate the Stopping Distance at a Traffic Light?

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The discussion focuses on calculating the stopping distance for a motorist approaching a traffic light that turns amber. It involves determining the minimum distance to stop before the intersection, factoring in reaction time and maximum braking deceleration. Additionally, it addresses the maximum distance from which the driver can safely proceed through the intersection without running a red light. A critical point is established that if the initial speed exceeds a certain threshold, there exists a range where the driver cannot stop in time or safely continue through the intersection. The conversation highlights the need for proper application of kinematic equations to solve these problems effectively.
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Homework Statement



(3) A motorist is approaching a green traffic light with speed v_0 when the light turns
to amber.
(a) If his reaction time is \tau, during which he makes his decision to stop and applies his foot to the
brake, and if the maximum braking deceleration is a, what is the minimum distance s_{min} from the
intersection at the moment the light turns to amber in which he can bring his car to a stop?
(b) If the amber light remains on for a time t before turning red, what is the maximum distance
s_{max} from the intersection at the moment the light turns to amber such that he can continue into the
intersection at speed v_0 without running the red light?
(c) Show that if his initial speed v_0 is greater than

v_{0_{max}} = 2a(t- \tau )

there will be a range of distance from the intersection such that he can neither stop in time nor
continue through the intersection without running the red light.

Homework Equations



kinematics equations

The Attempt at a Solution



A using vf^2=v_0^2 +2(-a)d is s_{min}=\frac{v_0^2}{2a} +v_0\tau

B is simply v_0 t

C I'm almost clueless. i can sub the given expression into both of my derived equations for s_{min} and s_{max} and all i get is that they both equal 2at(t-\tau) which i can't see how to use to prove that s_{max}<s<s_{min}.

This is actually an intermediate mechanics question but seems simple enough.
 
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Substitute your values into the inequality for smax & smin.
Solve for v.
 
the implication needs to go the other way, if p is the statement about velocity and q is the statement about distance i need to prove p -> q, that would prove q->p

even if i did do as you said, how would i "solve" for v?
 
Last edited:
ice109 said:
(c) Show that if his initial speed v_0 is greater than

v_{0_{max}} = 2a(t- \tau )

there will be a range of distance from the intersection such that he can neither stop in time nor
continue through the intersection without running the red light.

s > s_max or s < s_min
v*t > s_max or v*t < s_min

Since we are looking for the maximum initial speed, then
vo_max*t = vo_max^2/(2a) + vo_max*r
t = vo_max/(2a) + r
t - r = vo_max/(2a)
vo_max = 2a(t-r)
 
i don't understand what you've done

why did you equate the two?
 
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