Smooth trajectory on a smooth manifold

Physicists talk about a smooth trajectory of a particle on a smooth manifold and they label it as q(t) where q_1(t)....q_n(t) are component functions coming from the homeomorphism. I don't see how we can meaningfully talk about the whole trajectory this way as it might happen that the particle moves from one point to another point on the manifold and it might happen that the coordinate system at this point is different from that of previous point. How do we ensure that these two patch up and we can meaningfully talk about q(t) independent of the location of particle?

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You essentially ask how we can state a parametrized curve, a one-dimensional submanifold, a smooth function from $t$ to $\mathcal M$. Clearly, if you use an atlas without global coordinates, but a set of coordinates on their respective patches, then you have to state the curve in all the coordinates. If the expressions $q_1(t),\dots,q_n(t)$ denote specific functional dependencies, then they are of course only valid in one coordinate system. In the spaces where the patches intersect, you know how to translate from one to the other by the definition of a manifold. If that system is not covering the whole space, then this is not the full information, but the full information should be available by solving the equations for the mechanical system. If the curve is given by a integral curve of some vector field and you have to compute it by say the Hamiltonian equations of motion, then the tangent vector is determined by the equations at every point and so in every patch. Say you want to compute where the trajectory has to go step by step. Then if you go along and the patch comes to an end, then you'll have to transform to the next patch, transform the equations with you and follow the same procedure againt.
@LakshyaBhardwaj: There is no global coordinate system in general, so there can be no such guaranee. However, the equations of motions of classical systems for $\pi(t)=(q_1(t),\dots,q_n(t),p_1(t),\dots,p_n(t))$ are of the form $\pi'(t)=X(\pi(t),t)$ and so you have a solution in every patch. Yes to the second question - you just have to know how to get from one to the other. – NikolajK Jul 16 '12 at 8:10