---
title: "A test vehicle moves along a straight track with its position given as a function of time by \\(x(t) = (2.0 \\text{ m/s}^3)t^3 – (6.0 \\text{ m/s}^2)t^2\\), where \\(x\\) is in meters and \\(t\\) is in seconds. What are the vehicle’s instantaneous acceleration at \\(t = 3.0 \\text{ s}\\) and its average acceleration over the time interval from \\(t = 0\\) to \\(t = 3.0 \\text{ s}\\)?"
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url: "https://nerd-notes.com/ubq/117538/"
date_modified: "2026-08-04T07:48:56+00:00"
---

# A test vehicle moves along a straight track with its position given as a function of time by \(x(t) = (2.0 \text{ m/s}^3)t^3 – (6.0 \text{ m/s}^2)t^2\), where \(x\) is in meters and \(t\) is in seconds. What are the vehicle’s instantaneous acceleration at \(t = 3.0 \text{ s}\) and its average acceleration over the time interval from \(t = 0\) to \(t = 3.0 \text{ s}\)?

A test vehicle moves along a straight track with its position given as a function of time by \(x(t) = (2.0 \text{ m/s}^3)t^3 - (6.0 \text{ m/s}^2)t^2\), where \(x\) is in meters and \(t\) is in seconds. What are the vehicle's instantaneous acceleration at \(t = 3.0 \text{ s}\) and its average acceleration over the time interval from \(t = 0\) to \(t = 3.0 \text{ s}\)?

- **A.** Instantaneous acceleration: \(24 \text{ m/s}^2\); Average acceleration: \(6.0 \text{ m/s}^2\)
- **B.** Instantaneous acceleration: \(18 \text{ m/s}^2\); Average acceleration: \(6.0 \text{ m/s}^2\)
- **C.** Instantaneous acceleration: \(24 \text{ m/s}^2\); Average acceleration: \(18 \text{ m/s}^2\)
- **D.** Instantaneous acceleration: \(36 \text{ m/s}^2\); Average acceleration: \(12 \text{ m/s}^2\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/117538/*