Position x vs Time t x_{o}(m) v_{o}
(m/s) a_{o}(m/s^{2})
da/dt(m/s^{3})

In a position versus time graph, the vertical axis marks the position of
the object (x) and the horizontal axis marks the time (t). x_{o} is the position
when t=0s. That is the intercept with the vertical axis. Position the cursor on the
intercept and click the left mouse button. The coordinates of the position of the cursor
appear on the lower left corner of the graph. In the example shown, with the cursor
highlighted in red, the position for t=0 is approximately equal to -10, thus x_{o}
is equal to -10.

A straight line curve in a position versus time graph means that the change in position
is the same for equal time intervals. In other words, this means that the (change in
position)/(time) is constant. Since the velocity is defined as the change in
position over time, we can deduce that the velocity is constant.
The same conclusion infers that the acceleration is zero. We can then write that a_{o}=0
and da/dt=0. To find the velocity, we need to find the slope of the line. We already have
the coordinates of one point on the line (0,-10). Position the cursor on the line and read
its coordinate by pressing the left mouse key.

(please note the coordinates are approximated to the nearest integer value)

In the example shown, the point selected has the coordinates (4,10). The velocity is
then given by:

v_{o}= v = [10-(-10)]/[4-0] = 20/4 = 5

Note that when you press the SimulateIt button, you are provided with a
simulation of the displacement of the object. Note that in this case the ghost images of the
puck are equidistant, proving that the displacement is the same of equal intervals of time.