Question

A car purchased for $24,000 is expected to lose value, or depreciate, at a rate of 8% per year. This situation can be modeled by P(t)=24,000(1-0.08)^t
What is the meaning of the output P(t) in context of the problem?
What is the meaning of the input t in context of the problem?

Answer 1

$\bf \qquad \textit{Amount for Exponential Decay} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &24000\\ r=rate\to 8\%\to \frac{8}{100}\dotfill &0.08\\ t=\textit{elapsed time}\dotfill &t\\ \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{current value}}{\stackrel{A}{\stackrel{\downarrow }{P(t)}}}=24000(1-0.08)^{\stackrel{\stackrel{\textit{years}}{\downarrow }}{t}}$