Expanding my comment on the threeway intersections, we get an auxiliary result:

Lemma 1. If $v>5$, then starting from a threeway intersection where the branches have $0$, $a$ and $b$ units of weed, and enough free space outside that, the mower can clear the intersection in time $2(a+b) \, / \, (v-5)$.

(By "enough space" we mean that any other weed segments will not touch these branches during the operation.)

Proof. Keep doing round-trips to the three branches and back alternatively, always choosing the longest branch. Let us first consider the first round-trip and assume without loss of generality that it is $b$.

• The outgoing leg takes time $t=b/(v-1)$ because the mower starts $b$ units behind the tip, and his relative speed is $v-1$ as the tip is growing. During this time, the other two branches grow $t$ units each.
• The incoming leg also takes time $t$. During this time the mower completely cleans this branch (including any weed that was growing back to the branch from the intersection). The other branches grow another $t$ units each.

Overall in one round-trip, in $2t$ time the net weed decrease is $b - 4t = t(v-5)$, so weed is cleared at rate $(v-5)/2$. It is easy to see that the same average rate holds for each round trip. Having started with $(a+b)$ units of weed, it is all cleared in time $2(a+b) \, / \, (v-5)$.

If $v$ is large — let's take $v=100$ for concreteness — this suggests the following strategy. Let $L,R$ be the left and right intersections.

• Start from $L$. Make three round-trips to $R$: first along the north route ($3$ units and back), then along the middle route ($1$ unit and back), and then along the south route ($3$ units and back). In total this takes time $0.14$.
• Now $L$ has weed lengths north=$0.08$, middle=$0.06$ and south=$0$.
• And $R$ has weed lengths north=$0.11$, middle=$0.07$ and south=$0.03$.
• By Lemma 1, we can clear around $L$ in time $2\cdot 0.14 / 95 < 0.003$. During this time the weed around $R$ grows less than $0.01$ in each direction.
• Move east to $R$ in time $0.01$. During this time the weed grows another $0.01$ units.
• Now $R$ has weed lengths north $< 0.13$ and south $< 0.05$.
• By Lemma 1, we can clear around $R$ in less than $2 \cdot 0.18 / 95 < 0.004$ time.

Clearly $v=100$ suffices. It should now be relatively straightforward to find the minimum $v$ that suffices with this strategy. Of course this says nothing about possible other strategies.

Expanding my comment on the threeway intersections, we get an auxiliary result:

Lemma 1. If $v>5$, then starting from a threeway intersection where the branches have $0$, $a$ and $b$ units of weed, and enough free space outside that, the mower can clear the intersection in time $2(a+b) \, / \, (v-5)$.

(By "enough space" we mean that any other weed segments will not touch these branches during the operation.)

Proof. Keep doing round-trips to the three branches and back alternatively, always choosing the longest branch. Let us first consider the first round-trip and assume without loss of generality that it is $b$.

• The outgoing leg takes time $t=b/(v-1)$ because the mower starts $b$ units behind the tip, and his relative speed is $v-1$ as the tip is growing. During this time, the other two branches grow $t$ units each.
• The incoming leg also takes time $t$. During this time the mower completely cleans this branch (including any weed that was growing back to the branch from the intersection). The other branches grow another $t$ units each.

Overall in one round-trip, in $2t$ time the net weed decrease is $b - 4t = t(v-5)$, so weed is cleared at rate $(v-5)/2$. It is easy to see that the same average rate holds for each round trip. Having started with $(a+b)$ units of weed, it is all cleared in time $2(a+b) \, / \, (v-5)$.

If $v$ is large — let's take $v=100$ for concreteness — this suggests the following strategy. Let $L,R$ be the left and right intersections.

• Start from $L$. Make three round-trips to $R$: first along the north route ($3$ units and back), then along the middle route ($1$ unit and back), and then along the south route ($3$ units and back). In total this takes time $0.14$.
• Now $L$ has weed lengths north=$0.08$, middle=$0.06$ and south=$0$.
• And $R$ has weed lengths north=$0.11$, middle=$0.07$ and south=$0.03$.
• By Lemma 1, we can clear around $L$ in time $2\cdot 0.14 / 95 < 0.003$. During this time the weed around $R$ grows less than $0.01$ in each direction.
• Move east to $R$ in time $0.01$. During this time the weed grows another $0.01$ units.
• Now $R$ has weed lengths north $< 0.13$ and south $< 0.05$. (Nothing on the middle branch because we cleared it on the way.)
• By Lemma 1, we can clear around $R$ in less than $2 \cdot 0.18 / 95 < 0.004$ time.

Clearly $v=100$ suffices. It should now be relatively straightforward to find the minimum $v$ that suffices with this strategy. Of course this says nothing about possible other strategies.