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.
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:
\begin
[ t_
= \mbox
][t = \mbox
][x_
= \mbox
] [ v_
= \mbox
] [ a = \mbox
^
] \end
We begin by finding the velocity. The simplest Law of Change appropriate to our givens is:
\begin
[ v = v_
+ a (t-t_
) ]\end
which, after substituting the givens, tells us:
\begin
[ v = (\mbox
^
) t ] \end
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:
\begin
[ x = x_
+ v_
(t-t_
) + \frac
a (t-t_
)^
= \frac
(\mbox
^
)t^
\end
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 ti = 4.0 s, t = 8.0 s, and a = – 2.0 m/s 2. Unfortunately, this is not enough. We do not know xi, x, vi 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:
\begin
[ t_
= \mbox
][t = \mbox
] [ x_
= \mbox
] [ v_
= \mbox
] [ a = \mbox{-2.0 m/s}^
] \end