\begin{problem*}{2.5}
A position-time graph for a particle moving along the $x$ axis is
shown in Figure~P2.5.  \Part{a} Find the average velocity in the time
interval $t=1.50\U{s}$ to $t=4.00\U{s}$.  \Part{b} Determine the
instantaneous velocity at $t=2.00\U{s}$ by measuring the slope of the
tangent to the graph.  \Part{c} At what value of $t$ is the velocity
zero?
\begin{center}
\begin{asy}
import graph;

size(4cm, IgnoreAspect);

/* Parameter computation from the back-of-the-book answers:
 *   2*a*2 + b = -3.8            (1)  (answer to part b)
 *   2*a*4 + b = 0               (2)  (answer to part c)
 *   2*a*2     = 3.8             (3)  ((2)-(1))
 *     a = 3.8/4 = 0.95 ~= 1.0   (4)  (solve (3), rounding to likely value)
 *   b = -8*a                    (5)  (rearrange (2))
 *   b = -8*1 = -8               (6)  (plug (4) into (5))
 *   a*4^2 + b*4 + c = 2         (7)  (location of minimum from graph)
 *   c = 2 - 16a - 4b            (8)  (rearrange (7))
 *   c = 2 - 16*1 + 4*8 = 18     (9)  (plug (4) and (6) into (8))
 */
real a=1, b=-8, c=18;
real tanj_time = 2;

real parab(real t) {
  return a*t*t + b*t + c;
}

real tanj(real t) {
  real slope = 2*a*tanj_time + b;
  return slope * (t - tanj_time) + parab(tanj_time);
}

draw(graph(tanj, 0, 3.5), green);
draw(graph(parab, 1, 6.5), red);

xaxis("$t\U{s}$", BottomTop, LeftTicks(extend=true, ptick=gray+thin()));
yaxis("$x\U{m}$", LeftRight, RightTicks(extend=true, ptick=gray+thin()));
\end{asy}
\end{center}
\end{problem*}

\begin{solution}
\Part{a}
The average velocity is the total displacement over the elapsed time, so
\begin{equation}
  v_\text{avg} = \frac{x(4.00\U{s}) - x(1.50\U{s})}{4.00\U{s}-1.50\U{s}}
    = \frac{2.0\U{m} - 8.0\U{m}}{2.5\U{s}}
    = \ans{-2.4\U{m/s}}
\end{equation}

\Part{b}
Using rise-over-run to determine the tangent slope
\begin{equation}
  v(t=2.00\U{s}) = \frac{0\U{m} - 11.5\U{m}}{3.5\U{s} - 0.5\U{s}}
    = \ans{-3.8\U{m/s}}
\end{equation}

\Part{c}
Looking for the minimum of $x(t)$ (where the tangent curve is flat),
we see that $v(\ans{4.0\U{s}})=0$.
\end{solution}
