---
title: "A small cart is released from rest at time \\( t = 0 \\) and moves down a straight track with constant acceleration \\( a \\). A motion sensor measures the speed \\( v \\) of the cart at different positions \\( x \\) along the track, where \\( x = 0 \\) is the starting position. The data collected is used to create the graph shown of the square of the speed \\( v^2 \\) as a function of position \\( x \\). Based on the graph, what is the acceleration of the cart?"
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url: "https://nerd-notes.com/ubq/112351/"
date_modified: "2026-04-24T01:12:48+00:00"
---

# A small cart is released from rest at time \( t = 0 \) and moves down a straight track with constant acceleration \( a \). A motion sensor measures the speed \( v \) of the cart at different positions \( x \) along the track, where \( x = 0 \) is the starting position. The data collected is used to create the graph shown of the square of the speed \( v^2 \) as a function of position \( x \). Based on the graph, what is the acceleration of the cart?

A small cart is released from rest at time \( t = 0 \) and moves down a straight track with constant acceleration \( a \). A motion sensor measures the speed \( v \) of the cart at different positions \( x \) along the track, where \( x = 0 \) is the starting position. The data collected is used to create the graph shown of the square of the speed \( v^2 \) as a function of position \( x \). Based on the graph, what is the acceleration of the cart?

![A linear graph with the vertical axis labeled v squared in meters squared per second squared and the horizontal axis labeled x in meters. The vertical axis has grid lines at 0, 4, 8, 12, and 16. The horizontal axis has grid lines at 0, 0.5, 1.0, 1.5, and 2.0. A solid straight line starts at the origin (0,0) and passes exactly through the points (0.5, 4), (1.0, 8), (1.5, 12), and (2.0, 16).](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1776993168-zlVsAv.jpg)

- **A.** \( 2.0 \text{ m/s}^2 \)
- **B.** \( 4.0 \text{ m/s}^2 \)
- **C.** \( 8.0 \text{ m/s}^2 \)
- **D.** \( 16 \text{ m/s}^2 \)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/112351/*