\begin{problem*}{2.29}
The driver of a car slams on the brakes when he sees a tree blocking
the road.  The car slows uniformly with an acceleration of
$-5.60\U{m/s$^2$}$ for $4.20\U{s}$, making straight skid marks
$62.4\U{m}$ long, all the way to the tree.  With what speed does the
car then strike the tree?
\end{problem*}

\begin{solution}
The speed of the car is given by
\begin{equation}
  v(t) = at + v_0
\end{equation}
and its position is given by
\begin{equation}
  x(t) = \frac{1}{2} a t^2 + v_0 t + x_0
\end{equation}

We know $a$, $t$, and $x-x_0$, so we can use the second equation to
find $v_0$.
\begin{equation}
  v_0 = \frac{x-x_0-\frac{1}{2}at^2}{t}
\end{equation}
We can plug this into the first equation to find the final velocity.
\begin{align}
  v &= at + \frac{x-x_0-\frac{1}{2}at^2}{t} \\
    &= -5.60\U{m/s$^2$} \cdot 4.20\U{s}
       +\frac{62.4\U{m} - \frac{1}{2}\cdot(-5.60\U{m/s$^2$})\cdot(4.20\U{s})^2}
             {4.20\U{s}} \\
    &= \ans{3.10\U{m/s}}
\end{align}
\end{solution}
