---
title: "An object of mass \\(m\\) is released from rest at time \\(t = 0\\) in a uniform gravitational field of magnitude \\(g\\). As the object falls vertically downward, it experiences a resistive drag force of magnitude \\(F_{\\text{drag}} = bv\\), where \\(b\\) is a positive constant and \\(v\\) is the speed of the object. Which of the following expressions correctly gives the speed \\(v(t)\\) of the object as a function of time \\(t\\)?"
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url: "https://nerd-notes.com/ubq/120549/"
date_modified: "2026-08-23T04:41:33+00:00"
---

# An object of mass \(m\) is released from rest at time \(t = 0\) in a uniform gravitational field of magnitude \(g\). As the object falls vertically downward, it experiences a resistive drag force of magnitude \(F_{\text{drag}} = bv\), where \(b\) is a positive constant and \(v\) is the speed of the object. Which of the following expressions correctly gives the speed \(v(t)\) of the object as a function of time \(t\)?

An object of mass \(m\) is released from rest at time \(t = 0\) in a uniform gravitational field of magnitude \(g\). As the object falls vertically downward, it experiences a resistive drag force of magnitude \(F_{\text{drag}} = bv\), where \(b\) is a positive constant and \(v\) is the speed of the object. Which of the following expressions correctly gives the speed \(v(t)\) of the object as a function of time \(t\)?

- **A.** \(v(t) = \dfrac{mg}{b} e^{-bt/m}\)
- **B.** \(v(t) = \dfrac{mg}{b}\left(1 - e^{-bt/m}\right)\)
- **C.** \(v(t) = \dfrac{mg}{b}\left(e^{bt/m} - 1\right)\)
- **D.** \(v(t) = \dfrac{mg}{b}\left(1 + e^{-bt/m}\right)\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120549/*