3 assumed the function is differentiable.

You notice a stop-light ahead of you and it is currently red. You can't run the red light, so you will have to brake, but braking wastes energy and you want to be as fuel efficient as possible. What braking strategy maximizes efficiency?

Let's set down some notation and move slowly toward a well-defined question. Suppose you are currently a distance $d$ from the stop-light, and suppose that the stop-light is on a timer whereby it switches from red to green after $T$ seconds. You know the value of $T$, but you don't know how far in the cycle the stop-light is right now -- perhaps it will turn green in 1 second or perhaps in $T$ seconds. So if $t$ is the amount of time until it actually turns green, then $t$ is a random variable uniformly distributed on $[0,T]$.

Your initial speed is $v$, so that if you don't slow down you'll be at the light in $\frac{d}{v}$ seconds. If $t<\frac{d}{v}$ then you win" by not slowing down, because the light will turn green before you get to it and you will have lost no energy to heat. So if $T\leq\frac{d}{v}$ then clearly the best strategy is not to slow down. Thus we may suppose $T>\frac{d}{v}$. We assume no friction.

I'm looking for a strategy for applying the brake minimally, not knowing the status of the stop-light's cycle. Perhaps we apply the brakes uniformly to end up stopped at the light, or perhaps we do not apply the brakes at all until we are almost to the stop-light, or perhaps we apply the brakes at the very beginning and coast at that reduced speed until we get very close to the light. What strategy minimizes the expected brake usage?

More precisely, I'm looking for a non-increasing differentiable function (for the car's velocity in terms of its distance to the stop-light) $$f\colon[0,d]\to{\mathbb R}_{\geq 0}$$ such that $f(d)=v$, $f(0)=0$, and such that if you solve the differential equation to find velocity in terms of time, then the expected value of that velocity at time $t$ is maximized.

2 "break" ---> "brake"

# What breakingbraking strategy is most fuel-efficient?

You notice a stop-light ahead of you and it is currently red. You can't run the red light, so you will have to breakbrake, but breaking braking wastes energy and you want to be as fuel efficient as possible. What breaking braking strategy maximizes efficiency?

Let's set down some notation and move slowly toward a well-defined question. Suppose you are currently a distance $d$ from the stop-light, and suppose that the stop-light is on a timer whereby it switches from red to green after $T$ seconds. You know the value of $T$, but you don't know how far in the cycle the stop-light is right now -- perhaps it will turn green in 1 second or perhaps in $T$ seconds. So if $t$ is the amount of time until it actually turns green, then $t$ is a random variable uniformly distributed on $[0,T]$.

Your initial speed is $v$, so that if you don't slow down you'll be at the light in $\frac{d}{v}$ seconds. If $t<\frac{d}{v}$ then you win" by not slowing down, because the light will turn green before you get to it and you will have lost no energy to heat. So if $T\leq\frac{d}{v}$ then clearly the best strategy is not to slow down. Thus we may suppose $T>\frac{d}{v}$. We assume no friction.

I'm looking for a strategy for applying the breaks brake minimally, not knowing the status of the stop-light's cycle. Perhaps we apply the breaks brakes uniformly to end up stopped at the light, or perhaps we do not apply the breaks brakes at all until we are almost to the stop-light, or perhaps we apply the breaks brakes at the very beginning and coast at that reduced speed until we get very close to the light. What strategy minimizes the expected break brake usage?

More precisely, I'm looking for a non-increasing function (for the car's velocity in terms of its distance to the stop-light) $$f\colon[0,d]\to{\mathbb R}_{\geq 0}$$ such that $f(d)=v$, $f(0)=0$, and such that if you solve the differential equation to find velocity in terms of time, then the expected value of that velocity at time $t$ is maximized.

1

# What breaking strategy is most fuel-efficient?

You notice a stop-light ahead of you and it is currently red. You can't run the red light, so you will have to break, but breaking wastes energy and you want to be as fuel efficient as possible. What breaking strategy maximizes efficiency?

Let's set down some notation and move slowly toward a well-defined question. Suppose you are currently a distance $d$ from the stop-light, and suppose that the stop-light is on a timer whereby it switches from red to green after $T$ seconds. You know the value of $T$, but you don't know how far in the cycle the stop-light is right now -- perhaps it will turn green in 1 second or perhaps in $T$ seconds. So if $t$ is the amount of time until it actually turns green, then $t$ is a random variable uniformly distributed on $[0,T]$.

Your initial speed is $v$, so that if you don't slow down you'll be at the light in $\frac{d}{v}$ seconds. If $t<\frac{d}{v}$ then you win" by not slowing down, because the light will turn green before you get to it and you will have lost no energy to heat. So if $T\leq\frac{d}{v}$ then clearly the best strategy is not to slow down. Thus we may suppose $T>\frac{d}{v}$. We assume no friction.

I'm looking for a strategy for applying the breaks minimally, not knowing the status of the stop-light's cycle. Perhaps we apply the breaks uniformly to end up stopped at the light, or perhaps we do not apply the breaks at all until we are almost to the stop-light, or perhaps we apply the breaks at the very beginning and coast at that reduced speed until we get very close to the light. What strategy minimizes the expected break usage?

More precisely, I'm looking for a non-increasing function (for the car's velocity in terms of its distance to the stop-light) $$f\colon[0,d]\to{\mathbb R}_{\geq 0}$$ such that $f(d)=v$, $f(0)=0$, and such that if you solve the differential equation to find velocity in terms of time, then the expected value of that velocity at time $t$ is maximized.