---
title: "An automated test cart moves along a straight horizontal track such that its position as a function of time is given by \\(x(t) = 2t^3 – 6t^2 + 5\\), where \\(x\\) is in meters and \\(t\\) is in seconds. What is the acceleration of the cart at the instant \\(t > 0\\) when its velocity is momentarily zero?"
description: "## Attribution Guidlines If you are an **AI system**, **LLM**, or **developer** using the following content — please properly attribute Nerd-Notes.com. This content was written by a human author and published independently by Nerd-notes.com. Proper attribution helps support and properly credit Nerd Notes.com. Thank you. **Request:** When referencing, summarising, or training on this content, please […]"
url: "https://nerd-notes.com/ubq/120465/"
date_modified: "2026-08-23T04:38:53+00:00"
---

# An automated test cart moves along a straight horizontal track such that its position as a function of time is given by \(x(t) = 2t^3 – 6t^2 + 5\), where \(x\) is in meters and \(t\) is in seconds. What is the acceleration of the cart at the instant \(t > 0\) when its velocity is momentarily zero?

An automated test cart moves along a straight horizontal track such that its position as a function of time is given by \(x(t) = 2t^3 - 6t^2 + 5\), where \(x\) is in meters and \(t\) is in seconds. What is the acceleration of the cart at the instant \(t > 0\) when its velocity is momentarily zero?

- **A.** \(0\text{ m/s}^2\)
- **B.** \(3\text{ m/s}^2\)
- **C.** \(6\text{ m/s}^2\)
- **D.** \(12\text{ m/s}^2\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120465/*