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enter image description here

Please explain this graph to me as why acceleration is positive

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  • $\begingroup$ There is a rule that tells us the function is convex if its second derivative is greater than zero. $a=\dfrac{d^2x}{dt^2}$ and it is greater than zero, so $x$ is convex. Another explanation: the acceleration is positive, then the velocity $v(t)$ is an ascending function; we could consider an elementary case, when $V=ax$, i.e. $a$ is constant. Then $x\sim Ca^2, C>0$, so the path would represent a part of parabola with "branches up". $\endgroup$ – nicael Apr 13 '18 at 9:40
  • $\begingroup$ @nicael I think your answer is fine and if I were you I wouldn't have deleted it. If you would like to undelete it I think it would make a useful contribution. This is an elementary question but I still think it's a valid question. We were all beginners at some point. $\endgroup$ – John Rennie Apr 13 '18 at 10:41
  • $\begingroup$ @John well, ok! $\endgroup$ – nicael Apr 13 '18 at 13:38
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The velocity is the gradient of the line on your position:time graph. If I roughly estimate the gradient by eye I get something like this:

Gradients

And the acceleration is the gradient of the velocity:time graph

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    $\begingroup$ Alternatively, the velocity is increasing (becoming less negative) with time, so the acceleration must be positive. $\endgroup$ – jim Apr 13 '18 at 9:46
  • $\begingroup$ Please tell me why velocity is increasing? $\endgroup$ – brahamdeep singh Apr 13 '18 at 11:19
  • $\begingroup$ @brahamdeepsingh The velocity starts out as a negative number and it becomes less negative. That means the change in the velocity is a positive number i.e. the velocity is increasing. $\endgroup$ – John Rennie Apr 13 '18 at 11:21
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There is a rule that tells us the function is convex if its second derivative is greater than zero.

$a=\dfrac{d^2x}{dt^2}$ and it is greater than zero, so $x$ is convex.

As another example, you could consider that:

  • the acceleration is positive
  • then the velocity $v(t)$ is an ascending function
  • we could consider an elementary case, when $V=ax$, i.e. $a$ is constant.
  • then $x\sim Ca^2, C>0$, so the path would represent a part parabola with "branches up"

enter image description here

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