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Answer:
0.12 m/s²
Explanation:
For an object moving with constant acceleration, the average velocity (vavg) is the average of the initial velocity (u) and the final velocity (v), and the acceleration (a) is the change in velocity over time (t).
[tex]\Large \text {$ v_{avg}=\frac{1}{2}(u+v) $}\\\\\Large \text {$ a= $}\huge \text {$ \frac{v-u}{t} $}[/tex]
If the object starts at rest (u = 0), then the final velocity is twice the average velocity, and the acceleration is equal to the final velocity divided by time.
[tex]\Large \text {$ v=2v_{avg} $}\\\Large \text {$ a= $}\huge \text {$ \frac{v}{t} $}[/tex]
Substituting and rearranging:
[tex]\Large \text {$ a= $}\huge \text {$ \frac{2v_{avg}}{t} $}\\\\\Large \text {$ 2v_{avg}=at $}\\\\\Large \text {$ 2v_{avg}=(2a)\frac{1}{2} t $}[/tex]
The slope of the [tex]2v_{avg}[/tex] vs [tex]\frac{1}{2}t[/tex] graph is 2a. Therefore, if the slope is 0.23 m/s², then the acceleration is:
2a = 0.23 m/s²
a = 0.115 m/s²
Rounded to two significant figures, the acceleration is 0.12 m/s².