If I have a $t^2$ vs position graph and a $t$ vs average speed graph (where $t$ is time) for the same data set, shouldn't both their slopes equal and be the acceleration? However, my $t$ vs average speed graph yielded in a slope almost twice as big as the slope from $t^2$ vs position graph.

To clarify, here's the experiment that I did. enter image description here

Please tell me where I went wrong...

  • $\begingroup$ How did you calculate average speed? $\endgroup$ – sammy gerbil Aug 27 '17 at 7:44

The velocity time equation is:

$$ v = u + at $$

so if you graph $v$ against $t$ we get a straight line with gradient $a$. However the distance time equation is:

$$ s = ut + \tfrac{1}{2}at^2 $$

In your experiment the initial velocity $u=0$ so the equation simplifies to:

$$ s = \tfrac{1}{2}at^2 $$

That means when we graph $s$ against $t^2$ we get a straight line with gradient $\tfrac{1}{2}a$.

Response to comment:

The basic equations of motion for a constant acceleration are commonly known as the SUVAT equations because the relate the variables $s$ (distance), $u$ (initial velocity), $v$ (final velocity), $a$ (acceleration) and $t$ (time). Googling for suvat equations will find you lots of info, or see the Wikipedia article Equations of motion.

  • $\begingroup$ Oh... I thought that since in the first graph the time is squared vs position (x), and acceleration is $\delta x / \delta t^2$ that it would actually result in acceleration. Put aside your proof (that makes sense to me, thank you), could you please explain why my reasoning above may be wrong? $\endgroup$ – jjhh Aug 27 '17 at 7:06
  • $\begingroup$ Also, I'm new to physics, so are the equations you used above like standard equations applicable to every situation? $\endgroup$ – jjhh Aug 27 '17 at 7:07
  • $\begingroup$ @Helena: I've extended my answer to respond to your comment $\endgroup$ – John Rennie Aug 27 '17 at 9:05

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