Question

1. describe the appearance of an acceleration $(m/s^{2})$ vs. time (s) graph that shows an acceleration of $3m/s^{2}$.

a line through the origin with a slope of $\frac{1}{3}$
a line through the origin with a slope of 3
a horizontal line at $y = 3$
a vertical line at $x = 3$

Explanation:

Step1: Analyze the graph axes

The \( y \)-axis is acceleration (\(m/s^{2}\)) and the \( x \)-axis is time (\(s\)).

Step2: Consider the nature of constant acceleration

Since acceleration \(a = 3m/s^{2}\) is constant. In a \(a - t\) (acceleration - time) graph, for a constant value of \(a\), as time \(t\) ( \(x\)-variable) changes, the value of \(a\) ( \(y\)-variable) remains \(3\). A horizontal line \(y = c\) (where \(c\) is a constant) has the property that for all \(x\) values, \(y\) is fixed. Here \(c = 3\).

Step3: Eliminate other options

• For a line through the origin \(y=mx\) ( \(m\) is slope). If \(y\) (acceleration) were a function of \(x\) (time) like \(y=\frac{1}{3}x\) or \(y = 3x\), acceleration would be changing with time (not constant).
• A vertical line \(x = k\) ( \(k\) is a constant) is not a function (fails the vertical line test for a function \(y=f(x)\) as there are infinitely many \(y\) - values for a single \(x\) - value). In the context of \(a - t\) graph (where \(a\) is a physical quantity that should be a function of \(t\)), a vertical line is not meaningful for representing acceleration as a function of time.

Answer:

C. a horizontal line at \( y = 3 \)