---
title: "An object moves in the \\(xy\\)-plane with an instantaneous velocity \\(\\vec{v} = (3.0\\hat{\\imath} – 4.0\\hat{\\jmath})\\text{ m/s}\\) and an instantaneous acceleration \\(\\vec{a} = (-2.0\\hat{\\imath} + 1.0\\hat{\\jmath})\\text{ m/s}^2\\). What is the instantaneous rate of change of the object’s speed, \\(\\dfrac{dv}{dt}\\)?"
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url: "https://nerd-notes.com/ubq/120469/"
date_modified: "2026-08-23T04:38:58+00:00"
---

# An object moves in the \(xy\)-plane with an instantaneous velocity \(\vec{v} = (3.0\hat{\imath} – 4.0\hat{\jmath})\text{ m/s}\) and an instantaneous acceleration \(\vec{a} = (-2.0\hat{\imath} + 1.0\hat{\jmath})\text{ m/s}^2\). What is the instantaneous rate of change of the object’s speed, \(\dfrac{dv}{dt}\)?

An object moves in the \(xy\)-plane with an instantaneous velocity \(\vec{v} = (3.0\hat{\imath} - 4.0\hat{\jmath})\text{ m/s}\) and an instantaneous acceleration \(\vec{a} = (-2.0\hat{\imath} + 1.0\hat{\jmath})\text{ m/s}^2\). What is the instantaneous rate of change of the object's speed, \(\dfrac{dv}{dt}\)?

- **A.** \(-2.0\text{ m/s}^2\)
- **B.** \(-1.0\text{ m/s}^2\)
- **C.** \(+1.0\text{ m/s}^2\)
- **D.** \(+2.0\text{ m/s}^2\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120469/*