Good Driver, Bad Decisions

Our first step should be to find formulas to calculate the distance each person travels as a function of time. We'll denote $D(t)$ and $P(t)$ as Devin and the policeman's distance formulas, respectively.

Finding $P(t)$ is easy. Using the formula $d=v_it+\frac{1}{2}at^2$ :

$P(t) = \frac{1}{2}(3m/s^2)t^2=\frac{3}{2}t^2$

Finding $D(t)$ is a little bit harder. First, let's find how far away Devin has to be from the stop light in order for him to come to a complete stop at the stop sign. Using the formula $v_f^2=v_i^2+2ad$ :

$(0m/s)^2=(50m/s)^2+2(25m/s^2)d \Longrightarrow d = \frac{50^2}{2(25)}=50m$

$t=\frac{v}{a}=\frac{50m/s}{25m/s^2}=2s$ , therefore it takes $2s$ to travel the $50m$ required to slow down. This will be the same time and distance required to speed up, so a total $100m$ out of the $300m$ is spent accelerating and decelerating. The other $200m$ , Devin is traveling at a constant speed of $50m/s$ , which takes $t=\frac{d}{v}=\frac{200m}{50m/s}=4s$ . Therefore, it takes $2s+4s+2s=8s$ to travel the $300m$ in between each stop sign; $2s$ to accelerate, $4s$ to travel $200m$ , then $2s$ more to decelerate to a stop. If we follow this general idea that it takes Devin $8s$ to travel $300m$ , then we can have a simplified formula of $D(t)=300\bigg(\frac{t}{8}\bigg)$ . Please note that this formula is only valid when $t=8n$ , AKA he is at a stop sign.

Let's just start by seeing how many stop signs Devin can get to and still be ahead of the cop.

$P(t) < D(t)$

$\frac{3}{2}t^2 < 300\bigg(\frac{t}{8}\bigg)$

$t^2 < 25t$

$0 < t < 25$

Again, Devin's formula is only valid when $t=8n$ , so we can only trust the formula up until $t=24$ , AKA he has passed three stop signs and traveled $900m$ . How far has the policeman traveled in this time? $P(24)=\frac{3}{2}(24)^2=864m$ .

At this point the cop is moving pretty fast at a speed of $v=at=(3m/s^2)(24s)=72m/s$ , so let's check to see if he catches Devin while he's accelerating after the $3$ rd stop sign.

$864+(72m/s)t+\frac{1}{2}(3m/s^2)t^2 = 900+\frac{1}{2}(25m/s^2)t^2$

$11t^2-72t+36=0$

$(11t-6)(t-6)=0$

$t=\{\frac{6}{11},6\}$

After excluding the $6s$ solution, we can conclude that our policeman does catch Devin $\frac{6}{11}$ seconds after he leaves the $3$ rd stop sign. Add that to our $24s$ that it took to get up to that point and plug it into our formula for $P(t)$ and we have our answer!

$t=24+\frac{6}{11}=\frac{270}{11}s$

$P\big(\frac{270}{11}\big)=\frac{3}{2}\big(\frac{270}{11}\big)^2\approx \boxed{903.719...}$

You need to specify that Delvin does not [instantly achieve after starting] $50 m/sec$ , and that he accelerates from the stop light at the same rate as he accelerates from all the stop signs. This might seem obvious, but for the sake of removing ambiguity of this problem, include this proviso.

Time for Devin (D) to acc. from 0 m/s to 50 m/s, = 50/25= 2 s. He moves (1/2) * 25 * (2 * 2)=50 m during these 2 s. Same for decc. So acc. and decc. for 2 + 2= 4 s , moves 50+50=100 m. remaining 300-100=200 m with 50 m/s takes 200/50=4 s. From stop to stop D takes 8s to move 300 m at an average speed of
$300/8=\color{#3D99F6}{37.5 m/s}$ . THIS IS TRUE ONLY AT STOPS, that is for 8,16,24 ...seconds.

To achieve 37.5 m/s Police (P) takes $T_t s ~given ~by \dfrac{0*T_t+3*T_t} 2 =37.5~ m/s~~\therefore ~~T_t=25 s$

But this is true only at 24 s. That means D moves 37.5 * 24 =900 m and P moves (1/2) * 3 * 24 * 24 = 864 m in 24 s; P's speed then is 3 * 24= 72 m/s .
D is still 900-864= 36 m ahead. P still needs t < 1 second to catch up, for the speed of D is now less than 37.5 average.

Distance moved by P= 36+distance moved by D in t s (in acc. period for t is not more than 2.) $\implies 72*t + \dfrac 1 2 *3*t^2= 36 + \dfrac 1 2 *25*t^2~\implies11t^2-72t+ 36=0.\\\therefore~t=\dfrac 6 {11}~ or~~ t = 6.~~ t=6 ~does~ not~ hold~ since~ acc~ of~ D ~is ~only~ for~ ~2~ s.\\For~t=\dfrac 6 {11} ,\text{ distance moved by both is} \\ 0*(24+\frac 6 {11})+\dfrac 1 2 *3*(24+\frac 6 {11} )^2=~~~~~\color{#D61F06}{903.719}.\\Or~~~~~~~37.5*24+\dfrac 1 2 *25*(\frac 6 {11})^2=903.719$