Random Lamp-post?

We first determine the time taken to travel from the $1^{st}$ to the $n^{th}$ lamp post. Let $t_0$ be the time taken to travel from the $1^{st}$ to the $2^{nd}$ lamp post

Total distance = $(n-1)x$ Using the kinematics 2D formula $x-x_0 = v_0t + 0.5at^2$ , and since the first lamp post is the initial reference, $v_0 = 0, x_0 = 0$ . Hence, Distance travelled $= 0.5at^2$

For $n = 2, x = 0.5at_0^2$ $=> t_0 = \sqrt{\frac{2x}{a}}$

At $n$ , $(n-1)x = 0.5at^2$ $=> t = \sqrt{\frac{2(n-1)x}{a}}$

$=> t = (\sqrt{n-1})\sqrt{\frac{2x}{a}}$

$=> t = \sqrt{n-1}t_0$

Hence, time taken to taken to travel from 486th to 488th lamp post = Time taken to travel from 1st to 488th lamp post - Time taken to travel from 1st to 486th lamp post

$= \sqrt{487}t_0 - \sqrt{485}t_0 = \boxed{0.907219}$

Displacement $s$ of an object at time $t$ moving with constant acceleration $a$ and initial velocity $u$ is given by $s = ut + \frac 12 at^2$ .

Then, for moving from the first to second lamppost, we have:

$\begin{aligned} x & = \frac 12 a (20^2) \\ \implies a & = \frac x{200} \end{aligned}$

Let the time taken to arrived at 486th and 488th lampost be $t_1$ and $t_2$ respectively. Then we have:

$\begin{aligned} 485x & = \frac 12 \cdot \frac x{200} t_1^2 \\ \implies t_1 & = \sqrt{485\cdot 400} \approx 440.4543109 \end{aligned}$

Similarly,

$\begin{aligned} \implies t_2 & = \sqrt{487\cdot 400} \approx 441.3615298 \end{aligned}$

The time taken to travel from 486th and 488th lamppost is $t_2-t_1 \approx \boxed{0.907} \ s$ .