Answer
$0.0425\;\rm m/s$
Work Step by Step
We know that on air-track, the gliders move without friction.
So that we can use the conservation of momentum principle.
The 3 gliders, as we see in the figure below, were initially, were at rest and then they were released together at the same moment. This means that the initial momentum of the system is zero.
Thus,
$$p_{f,system}=p_{i,sys}=0$$
$$m_1v_{fx,1}+m_2v_{fx,2}+m_3v_{fx,3}=m_1v_{ix,1}+m_2v_{ix,2}+m_3v_{ix,3}$$
$$m_1v_{fx,1}+m_2v_{fx,2}+m_3v_{fx,3}=0+0+0$$
$$m_1v_{fx,1}+m_2v_{fx,2}+m_3v_{fx,3}=0 $$
Plugging the known;
$$0.3 v_{fx,1}+0.4 v_{fx,2}+0.2v_{fx,3}=0 $$
Solving for $v_{fx,2}$
$$ v_{fx,2}=\dfrac{-(0.2v_{fx,3} +0.3 v_{fx,1}) }{0.4} \tag 1$$
Now we need to find the velocity of the two gliders, gliders 1 and 3, from the given graph.
The equation of glider 1 is
$$y_1=-0.41x+0.53$$
Differentiating relative to $t$;
$$\dfrac{dy_1}{dt}=-0.41 +0$$
where we know that $dy/dt=v_{fx}$;
Thus,
$$v_{fx,1}=\bf -0.41 \;\rm m/s\tag 2$$
The equation of glider 3 is
$$y_3=0.53x+0.42$$
Differentiating relative to $t$;
$$\dfrac{dy_3}{dt}=0.53 +0$$
where we know that $dy/dt=v_{fx}$;
Thus,
$$v_{fx,3}=\bf 0.53\;\rm m/s\tag 3$$
Plugging (2) and (3) into (1);
$$ v_{fx,2}=\dfrac{-\left[(0.2 \times 0.53) +(0.3\times -0.41)\right] }{0.4} $$
$$ v_{fx,2}= \color{red}{\bf 0.0425}\;\rm m/s $$
And since the result is positive, its direction is to the right.
