The initial velocity of the object is $\vec{v}_0 = v_1 \hat{i}$. The velocity of belt 2 is $\vec{v}_{belt} = v_2 \hat{j}$. The object stops sliding when its velocity becomes equal to that of belt 2, so the final velocity is $\vec{v}_f = v_2 \hat{j}$.

The initial relative velocity of the object with respect to belt 2 is:

$$\vec{v}_{rel} = \vec{v}_0 - \vec{v}_{belt} = v_1 \hat{i} - v_2 \hat{j}$$

The kinetic friction force $\vec{f}$ opposes this relative motion. Its magnitude is $f = \mu mg$. The acceleration of the object is constant, with magnitude $a = f/m = \mu g$, and its direction is opposite to $\vec{v}_{rel}$.

The change in the object's velocity is $\Delta \vec{v} = \vec{v}_f - \vec{v}_0 = v_2 \hat{j} - v_1 \hat{i}$. The time $t$ taken for this change is given by $\vec{a}t = \Delta \vec{v}$. Taking the magnitude of both sides:

$$|\vec{a}|t = |\Delta \vec{v}|$$ $$(\mu g)t = \sqrt{(-v_1)^2 + (v_2)^2}$$ $$t = \frac{\sqrt{v_1^2 + v_2^2}}{\mu g}$$

For motion with constant acceleration, the displacement can be found using the average velocity. The final coordinates $(x,y)$ are the components of the displacement vector $\Delta\vec{r}$.

$$\Delta\vec{r} = \vec{v}_{avg} t = \frac{\vec{v}_0 + \vec{v}_f}{2} t$$ $$x\hat{i} + y\hat{j} = \frac{(v_1 \hat{i}) + (v_2 \hat{j})}{2} t$$

Substituting the expression for $t$:

$$x = \frac{v_1}{2} t = \frac{v_1 \sqrt{v_1^2 + v_2^2}}{2 \mu g}$$ $$y = \frac{v_2}{2} t = \frac{v_2 \sqrt{v_1^2 + v_2^2}}{2 \mu g}$$