Just in case you are not taking a calculus level class, the position can be deterimined by the area under the curve of the velocity graph up to the time in question. If the graph drops below the x axis, count this as negative area.
Since you have the velocity as a function of time all you have to do is integrate it and add the initial position i.e
Let dx/dt = f(t) then x(t) = ∫f(t)dt + x0 , x(t) is the position as a function of time and x0 is the position at say t=0.
For example if the velocity is dx/dt = kt then the position x(t), is 1/2kt^2 assuming the particle was located at x=0 at t=0. The graph of velocity would be a straight line and the graph of the position would be a parabola.
Velocity times time equals position. There you go!
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The prior two responders to this question are correct. What you will see happen to the graph is it will transform into a graph resembling a power of t one greater than the power that represents the velocity. So, if the velocity graph displays a linear relationship, for example v(t) = 4t, then your graph of position will have an x to the second power, so in our example, x(t) = 2t^2 + x(0). This is obtained by solving the differential equation for x. Though this is a calculus-based argument, perhaps it makes sense that the position graph would represent one higher power of t. If your velocity is of the first order, i.e. v(t) = kt, this means that after each increment of time, velocity increases by k. This is no doubt linear. However, position would increase faster than velocity, because the object is moving faster and faster as time goes on. So, if we saw a line for velocity, position would be parabolic.