---
title: "A small cart of mass \\(m\\) moves along a horizontal track in the \\(+x\\)-direction.  At time \\(t = 0\\), the cart passes the origin \\(x = 0\\) with speed \\(v_0\\) and experiences a net drag force directed along the track given by \\(F_x = -b\\sqrt{v}\\), where \\(b\\) is a positive constant and \\(v\\) is the instantaneous speed.  Which of the following integral equations correctly relates the position \\(x\\) of the cart to its speed \\(v\\)?"
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url: "https://nerd-notes.com/ubq/120605/"
date_modified: "2026-08-23T04:41:56+00:00"
---

# A small cart of mass \(m\) moves along a horizontal track in the \(+x\)-direction.

At time \(t = 0\), the cart passes the origin \(x = 0\) with speed \(v_0\) and experiences a net drag force directed along the track given by \(F_x = -b\sqrt{v}\), where \(b\) is a positive constant and \(v\) is the instantaneous speed.

Which of the following integral equations correctly relates the position \(x\) of the cart to its speed \(v\)?

A small cart of mass \(m\) moves along a horizontal track in the \(+x\)-direction.

At time \(t = 0\), the cart passes the origin \(x = 0\) with speed \(v_0\) and experiences a net drag force directed along the track given by \(F_x = -b\sqrt{v}\), where \(b\) is a positive constant and \(v\) is the instantaneous speed.

Which of the following integral equations correctly relates the position \(x\) of the cart to its speed \(v\)?

![A horizontal line represents the x-axis with a tick mark at the origin labeled 0. A small rectangular cart labeled m is positioned at the origin, with a horizontal arrow extending to the right from its front edge labeled v_0. A horizontal arrow pointing to the left from the cart center is labeled F_x = -b\sqrt{v}. A coordinate arrow below the track points to the right and is labeled +x. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/setup-fig-1-1787460115-hXXcJS.jpg)

- **A.** \(\int_0^x dx' = -\dfrac{m}{b} \int_{v_0}^v (v')^{1/2}\,dv'\)
- **B.** \(\int_0^x dx' = -\dfrac{m}{b} \int_{v_0}^v (v')^{-1/2}\,dv'\)
- **C.** \(\int_0^x dx' = -\dfrac{b}{m} \int_{v_0}^v (v')^{1/2}\,dv'\)
- **D.** \(\int_0^x dx' = -\dfrac{m}{b} \int_0^v (v')^{1/2}\,dv'\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120605/*