Motion of a particle in one dimension when given a velocity or acceleration function Suppose we know that the velocity of a particle moving back and forth on the xaxis is given by v(t) = 2t – 2 and the initial position x(0) = –3. Notice that this is the same particle as in Part I above. Some basic facts:

v(t) = 2t – 2 is the derivative of the position function
x(t) = t^{2} – 2t – 3
.

a(t) = 2 is the derivative of v(t) and is the uniform (or constant) acceleration of the particle.

v(t) = 0 when t = 1 which invites us to investigate the sign of our velocity function
 v(t) < 0 when t < 1 and v(t) > 0 when t > 1.
To determine an object's position given its velocity function and an initial position we will use a basic fact from calculus: Given the rate of change of a quantity and a time interval, the “net change” is the integral of the rate of change over the given time period. Let's use these fact to look at three final questions. 