In the answer, the person showed a graph of position vs time (for constant acceleration), his graph looks like $$x\propto t$$ but in reality, $$x\propto t^2$$ under constant acceleration. Isn't his answer wrong? Or didn't he mean to graph position vs time squared? $$x=\frac{1}{2}gt^2$$
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He didn't show $x \propto t$, he showed $v \propto t$ ! Have a look at the vertical axis:
As $v = \frac{x}{t}$, you have no contradiction and $v \propto t \Longleftrightarrow \frac{x}{t} \propto t \Longleftrightarrow x \propto t^2$.
Emile Couzin
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The graph is accurate. Cause they plotted velocity vs time not position vs time. And $$v\propto t$$ using the function they found the graph.
Sayma
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