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title: "A high-speed research probe moves along a straight line through a resisting medium that slows it down according to the velocity function \\(v(t) = v_0 e^{-kt}\\), where \\(v_0\\) is the speed at time \\(t = 0\\) and \\(k\\) is a positive constant. The probe passes position \\(x = 0\\) at \\(t = 0\\). Which of the following statements correctly describes the limiting distance traveled by the probe as \\(t \\to \\infty\\) and the time \\(t_{0.9}\\) required for the probe to cover \\(90\\%\\) of that total limiting distance?"
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url: "https://nerd-notes.com/ubq/123965/"
date_modified: "2026-09-28T13:28:18+00:00"
---

# A high-speed research probe moves along a straight line through a resisting medium that slows it down according to the velocity function \(v(t) = v_0 e^{-kt}\), where \(v_0\) is the speed at time \(t = 0\) and \(k\) is a positive constant. The probe passes position \(x = 0\) at \(t = 0\). Which of the following statements correctly describes the limiting distance traveled by the probe as \(t \to \infty\) and the time \(t_{0.9}\) required for the probe to cover \(90\%\) of that total limiting distance?

A high-speed research probe moves along a straight line through a resisting medium that slows it down according to the velocity function \(v(t) = v_0 e^{-kt}\), where \(v_0\) is the speed at time \(t = 0\) and \(k\) is a positive constant. The probe passes position \(x = 0\) at \(t = 0\). Which of the following statements correctly describes the limiting distance traveled by the probe as \(t \to \infty\) and the time \(t_{0.9}\) required for the probe to cover \(90\%\) of that total limiting distance?

- **A.** The limiting distance diverges to infinity as \(t \to \infty\), and the time \(t_{0.9}\) is infinite.
- **B.** The limiting distance approaches \(\dfrac{v_0}{k}\) as \(t \to \infty\), and \(t_{0.9} = \dfrac{\ln 10}{k}\).
- **C.** The limiting distance approaches \(\dfrac{v_0}{k}\) as \(t \to \infty\), and \(t_{0.9} = \dfrac{9}{10k}\).
- **D.** The limiting distance approaches \(\dfrac{v_0}{k}\) as \(t \to \infty\), and \(t_{0.9} = \dfrac{\ln(10/9)}{k}\).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/123965/*