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title: "A test cart moves along a straight, horizontal track with a velocity given as a function of time by \\(v(t) = 3t^2 – 12t + 9\\), where \\(v\\) is in meters per second and \\(t\\) is in seconds. What is the ratio of the total distance traveled by the cart to the magnitude of its net displacement over the time interval from \\(t = 0\\text{ s}\\) to \\(t = 4\\text{ s}\\)?"
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url: "https://nerd-notes.com/ubq/123962/"
date_modified: "2026-09-28T13:28:11+00:00"
---

# A test cart moves along a straight, horizontal track with a velocity given as a function of time by \(v(t) = 3t^2 – 12t + 9\), where \(v\) is in meters per second and \(t\) is in seconds. What is the ratio of the total distance traveled by the cart to the magnitude of its net displacement over the time interval from \(t = 0\text{ s}\) to \(t = 4\text{ s}\)?

A test cart moves along a straight, horizontal track with a velocity given as a function of time by \(v(t) = 3t^2 - 12t + 9\), where \(v\) is in meters per second and \(t\) is in seconds. What is the ratio of the total distance traveled by the cart to the magnitude of its net displacement over the time interval from \(t = 0\text{ s}\) to \(t = 4\text{ s}\)?

- **A.** \(1\)
- **B.** \(2\)
- **C.** \(3\)
- **D.** \(4\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/123962/*