the to approaching this problem is knowing some differentiation. As an example, if you differentiate distance with respect to time, you create a rate of change of distance dependent upon time, more commonly know as velocity (ie miles per hour...). If you differentiate again, you then have distance per unit time squared (or rate of change of velocity) more commonly know as acceleration.
In classical physics, distance can be determined by X(t)=Xo+V(t)+1/2A(t)^2 where Vo is the starting position, V is velocity (the t value cancels out the time variable in the denominator of V, leaving just distance), and the A corresponds to acceleration (with the 1/2 co-efficient as a result of integration, with time square to cancel out the time variable in the denominator
extension of differentiation
If you differentiate the X(t) equation with respect to time you then have a velocity equation in the form of:
V(t)=V+A(t), (all variables have the form of unit distance per unit time)
this should help start you out
Joe