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Suppose an object is moving along a one-dimensional position axis.  The object starts its motion at _t_ = 0 at the position _x_ = 0 and with velocity _v_ = 0.  It has an acceleration of +2.0 m/s ^2^.  After 4.0 seconds, the object's acceleration instantaneously changes to - 2.0 m/s ^2^.  Plot velocity and position versus time graphs for the first 8.0 seconds of the object's motion.

System:  The object will be treated as a point particle.

Model:  [One-Dimensional Motion with Constant Acceleration|1-D Motion (Constant Acceleration)].

Approach:  We will use the model twice.  Once to describe the motion from _t_ = 0 seconds to _t_ = 4.0 seconds, and once more to describe the motion from _t_ = 4.0 seconds to _t_ = 8.0 seconds.  

For the first part of the motion, our givens are:

{panel:title=givens}{latex}\begin{large}\[ t_{\rm i} = \mbox{0 s} \]\[t = \mbox{4.0 s} \]\[x_{\rm i} = \mbox{0 m} \] \[ v_{\rm i} = \mbox{0 m/s} \] \[ a = \mbox{2.0 m/s}^{2} \] \end{large}{latex}{panel}

We begin by finding the velocity.  The simplest Law of Change appropriate to our givens is:

{latex}\begin{large}\[ v = v_{\rm i} + a (t-t_{\rm i}) \]\end{large}{latex}

which, after substituting the givens, tells us:

{latex}\begin{large} \[ v = (\mbox{2.0 m/s}^{2}) t \] \end{large}{latex}

This is the equation for a line with slope 2 and intercept 0, giving the graph:

VELPLOT 1

We then find the position.  The most direct Law of Change is:

{latex}\begin{large} \[ x = x_{\rm i} + v_{\rm i} (t-t_{\rm i}) + \frac{1}{2} a (t-t_{\rm i})^{2} = \frac{1}{2}(\mbox{2.0 m/s}^{2})t^{2}\] \end{large}{latex}

which yields the parabolic graph:

POSPLOT1

We now wish to analyze the second part of the motion.  In this part of the motion, we must change our givens.  We now have _t_~i~ = 4.0 s, _t_ = 8.0 s, and _a_ = -- 2.0 m/s ^2^. Unfortunately, this is *not* enough.  We do not know _x_~i~, x, _v_~i~ or _v_.  We have too few givens to proceed.

The answer to this dilemma is simple.  We have just derived expressions that give _x_ and _v_ for any time between 0 s and 4.0 s.  We would like to know _x_ and _v_ at 4.0 s.  Thus, we can use the _final_ time of the first part of the problem to obtain the _initial_ conditions for the second part.  From our graphs or from the equations, we can complete our list of givens for the second part:

{panel:title=givens (second part)}{latex}\begin{large}\[ t_{\rm i} = \mbox{4.0 s}\]\[t = \mbox{8.0 s} \] \[ x_{\rm i} = \mbox{16 m} \] \[ v_{\rm i} = \mbox{8 m/s} \] \[ a = \mbox{-2.0 m/s}^{2}\] \end{large}{latex}{panel}