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{excerpt:hidden=true}An introduction to continuously piecing together kinematic solutions for time intervals with different accelerations. {excerpt} 

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.

h4. Solution

{toggle-cloak:id=sys} *System:* {cloak:id=sys} The object will be treated as a [point particle].{cloak}

{toggle-cloak:id=int} *Interactions:* {cloak:id=int}The accelerations as described in the problem.{cloak}

{toggle-cloak:id=mod} *Model:* {cloak:id=mod}[One-Dimensional Motion with Constant Acceleration|1-D Motion (Constant Acceleration)], applied twice to cope with the change in acceleration.{cloak}

{toggle-cloak:id=app} *Approach:*
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{toggle-cloak:id=understand} {color:red}{*}Understand the Limitations of the Model{*}{color}
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We will have to use the model twice. Even though the acceleration has a _size_ of 2.0 m/s{^}2^ throughout the motion, the _direction_ of the acceleration changes at _t_ = 4.0 s. Therefore, the acceleration is _not_ constant over the first 8 seconds of the motion. The acceleration _is_, however, separately constant in the time intervals from 0 to 4 seconds and from 4 to 8 seconds. We will therefore have to 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.
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{toggle-cloak:id=part1} {color:red}{*}The First Interval{*}{color}
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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}
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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:

!vel1.gif!

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:

!pos1.gif!
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{toggle-cloak:id=part2} {color:red}{*}The Second Interval{*}{color}
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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}
This allows us to write an equation for the velocity:
{latex}\begin{large} \[ v = v_{\rm i} + a(t-t_{\rm i}) = \mbox{8.0 m/s} + (-\mbox{2.0 m/s}^{2})(t-\mbox{4.0 s}) \] \end{large}{latex}
and the position:
{latex}\begin{large} \[ x = x_{\rm i} + v_{\rm i}(t-t_{\rm i}) +\frac{1}{2}a(t-t_{\rm i})^{2} = \mbox{16 m} + (\mbox{8.0 m/s})(t-\mbox{4.0 s}) + \frac{1}{2}(-\mbox{2.0 m/s}^{2})(t-\mbox{4.0 s})^{2} \] \end{large}{latex}
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{toggle-cloak:id=together} {color:red}{*}Putting them Together{*}{color}
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These new expressions can be used to finish the plots we started in the first part. Plotting the functions from the first part in the interval _t_ = 0 s to _t_ = 4.0 s and the functions from the second part in the interval _t_ = 4.0 s to _t_ = 8.0 s gives the complete graphs shown below. For completeness, we also show the acceleration graph.

!accelall.gif! !velall.gif! !posall.gif!

Here we can see that although the acceleration is discontinuous (has a break in the graph at _t_ = 4.0 s) the velocity is not discontinuous. The velocity is a connected graph, though it does display a _kink_ at 4.0 s, when the graph suddenly takes a sharp bend. This kink is a sign of a discontinuous 1st derivative (the 1st derivative of velocity with respect to time is acceleration, which is discontinuous). The position, however, is completely smooth. It is connected and displays no kinks. (The first derivative of position is velocity, which is continuous.){note}We _ensured_ that the position graph would be smooth when we set the initial position and velocity of the second portion of the motion equal to the final position and velocity of the first portion\! {note}{info}The fact that smooth position curves are obtained from a motion composed of a sequence of motions with constant acceleration is a major reason that modeling a motion using segments of constant acceleration is a powerful tool. {info}
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