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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{Higher Derivatives}

\begin{example}
Egbert tosses a baseball straight up, and measures its position at each time
\( t \). He finds that the position \( s(t) \) (in feet above his hand) at
time \( t \) seconds is \[
s(t)=-16t^{2}+24t+7.\]
Find the position and velocity at time \( t=0 \), when he let go of the ball. 
\end{example}

\subsubsection*{Solution}

The velocity is just the derivative, \[
v(t)=s'(t)=-32t+24,\]
and so the velocity at time \( t=0 \) is \[
v(0)=24.\]
The acceleration is the derivative of the velocity, \[
a(t)=v'(t)=s''(t)=-32,\]
so no matter what \( t \) is, the acceleration is a constant \( 32\, ft/sec^{2} \).
The reason the acceleration is negative is that the positive direction is up.
The acceleration is downward, and so negative. You should also notice what happened
with the units. Since \( s(t) \) is measured in feet, and time \( t \) in
seconds, the derivative, which is the limit of the change in distance divided
by the change in time, is measured in \( feet/sec \). The derivative of that,
the acceleration, is measured in \[
\left( feet/sec\right) /sec=feet/sec^{2},\]
again because the difference quotient divides the velocity measurements (in
\( feet/sec) \) by time.
\end{document}
