Probability to be overtaken on circuit

Question

I'm running on a treadmill in the gym and use the software (provided) which basically enables me to see other people running around me in the virtual race arena (standard 400 m circuit).

I'm running pretty fast (13.2 km/h) and most of the time I close on other users and overtake them. Once in a while I face some other speedy user and the exercise become much more fun (turned out competitive running is less boring).

Suppose, when I start my race other users "lags" behind me with the lag distributed uniformly and independently. If I safely assume other users' speeds are distributed something like $N(10 \text{ km/h}, 4/3 \text{ km/h})$,

what is the probability someone will overtake me during the race?

Let $v$ - be my speed, $d$ - be the distance I run, $l$ be the "lag", then $$l \sim U[0, 400]$$ I basically thought of two approaches:

1. "Uniform lag" approach. Let $t_0 = \frac{d}{v}$ - be the time it takes me to finish the race. Let other user's speed be $u$, then $t_u = \frac{d+l}{u}$ and the lowest speed at which s/he still catches me is $$t_0 = t_u \Rightarrow u_{\text{crit}} = \frac{d+l}{d}v.$$ Hence the probability the user with lag $l$ catches me is $1-\Phi_{\mu,\sigma}(u_{\text{crit}})$, where $\Phi_{\mu,\sigma}$ -is cdf of Normal Distribution with parameters $\mu,\sigma$ (in my case 10 km/h, 4/3 km/h). Integrate for "lags" and get the answer.
2. "Speed-based" approach. Look at other user, if s/he runs slower than me - the probability s/he can catch me is $0$, no matter that the "lag" is. There is some critical speed, for which the user catch me with $400$ meter "lag" and for any speed above "catching" probability is $1$. In between, some "lags" are closable and some are not, depending on the speed. From the previous time equation it follows $$ l = \left( \frac{u}{v} - 1 \right) \cdot d$$ Here, integration is on speed, with probability to catch being the ratio of catchable "lags-to-lap length.

All-in-all, how can I be sure, both methods produce the same probability value? If the values are different, which approach is the right one?

From my MATLAB numerical estimations the values are close (around $0.1$% difference).