Interperting displacement vs time graph

[Purple_Monkey]

I'm having a conceptual problem and having a hard time putting it into words. I think I'm assuming a lot of things have confused the sh*t out of myself. Sorry if this is really confusing and flawed in logic.
So, let's say we have a graph and it's displacement versus time. The slope of the curve would be velocity as velocity is =d/t.
If we pick a single point on the graph, the line tangent to that would be the velocity vector and we can draw the x & y components for the vector. If what I've said so far is right then why is it that when we want to calculate for time using the equations for uniform acceleration we use vsinθ, the y-component. According to the graph y= displacement and x=time...

 

[nasu]

I'm having a conceptual problem and having a hard time putting it into words. I think I'm assuming a lot of things have confused the sh*t out of myself. Sorry if this is really confusing and flawed in logic.
So, let's say we have a graph and it's displacement versus time. The slope of the curve would be velocity as velocity is =d/t.

This is right, in essence. In general "displacement" is a vector but you may be thinking about a one-dimensional motion.
Velocity is the derivative of displacement in respect to time (it is d/t only for the case of constant velocity)

If we pick a single point on the graph, the line tangent to that would be the velocity vector and we can draw the x & y components for the vector.

No, this is not true. The velocity is a vector in space whereas your "vector" will have d-t components.
If you plot, let's say, the x coordinate versus time, the slope of the tangent will give the component of the velocity along the x-axis (Vx).
Similar for y versus t and Vy.