Answer
a) $11.9\;\rm s$
b) $41.7^\circ $ north of west.
Work Step by Step
We take the center of the circular ice rink to be our origin $(0,0)$. We chose east to be the positive $x$-direction and north to be the positive $y$-direction.
By this coordinate, the initial velocities of the two skaters are given by
The one who is heading north:
$$v_1=(0\;\hat i+2.5\;\hat j)\;\rm m/s\tag 1$$
The one who is heading west:
$$v_2=(-3.5\;\hat i+0\;\hat j)\;\rm m/s\tag 2$$
The two skaters will collide at the origin and stick together while moving to the edge of the rink as one unit in a straight line.
This means that they will travel a distance of $R=25\;\rm m$ where $R$ is the radius of the rink.
Thus, the time of the trip is given by
$$v_f=\dfrac{R}{t}$$
$$t=\dfrac{R}{v_f}\tag 3$$
We assume that the ice is frictionless, so the momentum of the system (the two skaters) is conserved.
Thus,
$$m_1v_1+m_2v_2=(m_1+m_2)v_f$$
Plugging the known;
$$75v_1+60v_2=(75+60)v_f$$
Plugging from (1) and (2);
$$75(0\;\hat i+2.5\;\hat j)+60(-3.5\;\hat i+0\;\hat j)=(75+60)v_f$$
$$(75\times 2.5)\;\hat j +(60\times -3.5)\;\hat i =(75+60)v_f$$
Therefore,
$$v_f=\dfrac{(75\times 2.5)\;\hat j +(60\times -3.5)\;\hat i }{75+60}$$
$$v_f=\dfrac{(75\times 2.5)\;\hat j }{75+60}+\dfrac{(60\times -3.5)\;\hat i }{75+60}$$
$$v_f=({\bf -1.556}\;\hat i +{\bf 1.389}\;\hat j )\;\rm m/s$$
Now we need to find its magnitude which is given by applying the Pythagorean theorem.
$$|v_f|=\sqrt{v_{fx}^2+v_{fy}^2}=\sqrt{( -1.556)^2+(1.389)^2}$$
$$v_f=\bf 2.1\;\rm m/s$$
Plugging into (3);
a) $$t=\dfrac{R}{v_f}=\dfrac{25}{2.1}=\color{red}{\bf 11.9}\;\rm s$$
b) And the direction of their velocity, relative to the west, is given by
$$\tan\theta=\dfrac{v_{fy}}{v_{fx}}$$
$$\theta=\tan^{-1}\left(\dfrac{v_{fy}}{v_{fx}}\right)$$
$$\theta=\tan^{-1}\left(\dfrac{ 1.389}{ 1.556}\right)=\color{red}{\bf41.7^\circ}$$
which is $41.7^\circ $ north of west.
