Answer
It takes $63.96s$ for the tug and the asteroid to meet.
Work Step by Step
The tug pulls on the asteroid with a force $F=490N$. According to Newton's 3rd Law of Motion, there is a reaction force back on the tug $F_{reaction}=-490N$.
In this exercise, we find $a_{asteroid}$ and $a_{tug}$ following these forces.
1) Find $a_{asteroid}(a_a)$
$F=490N$ and $m_a=6200kg$. According to Newton's 2nd Law: $$a_a=\frac{F}{m_a}=0.08m/s^2$$
2) Find $a_{tug}(a_t)$
$F=-490N$ and $m_t=3500kg$. According to Newton's 2nd Law: $$a_t=\frac{F}{m_a}=-0.14m/s^2$$
In this exercise, we only consider the magnitude of $a_t=0.14m/s^2$
3) Both the tug and asteroid start from rest: $v_0=0$ and take equal time $t$ to meet each other. We have $$\Delta x_{t}=v_0t+\frac{1}{2}a_tt^2=0+0.07t^2=0.07t^2$$ $$\Delta x_{a}=v_0t+\frac{1}{2}a_at^2=0+0.04t^2=0.04t^2$$
The tug and asteroid are 450m apart: $$\Delta x_{t}+\Delta x_{a}=450$$ $$0.11t^2=450$$ $$t=63.96s$$
