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Can you have a zero velocity and nonzero average acceleration?

I am confused with the word "average" here. If the question would be, "Can you have a zero velocity and nonzero acceleration?" my answer would be yes. An example would be a ball thrown upward. At the highest point, the velocity is zero and instantaneous acceleration is -9.8 m/s$^2$. Since the question states that average acceleration, I can't think of an example that would satisfy the question.

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    $\begingroup$ If the acceleration is constant the average acceleration is equal to that constant. You've answered your own question. Now, are you asking about a constant zero velocity or an instantaneous zero velocity? $\endgroup$
    – Bill N
    Commented Mar 23, 2015 at 16:28
  • $\begingroup$ As you say, it doesn't make sense to compare a time-averaged quantity to an instantaneous quantity. The question could be "in a given time interval, is it possible to have zero velocity at some point in the interval, while the average acceleration over this interval is zero?" or "in a given time interval, can the average velocity over that interval be zero while the average acceleration over the interval is nonzero?" $\endgroup$
    – Kyle Arean-Raines
    Commented Jun 24, 2015 at 14:15

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Can you have a zero velocity and nonzero average acceleration?

  • If you by velocity mean instantaneous velocity, then the question makes no sence. The corresponding acceleration will as well be instantaneous (and the answer would be yes.) $$\lim_{\Delta t \to 0}a_{av}=a_{inst}$$

    • ... unless it is not a requirement that the acceleration is averaged over the same timespan as the velocity is measured. Then the answer is yes, and an upwards thrown stone is an example (it reaches a halt of $v=0$ and starts falling down, but the acceleration is at all times equal to the gravitational acceleration $-g$, so the average acceleration is as well, $a_{av}=-g$.)

  • If you mean average velocity, then the answer is no. Average acceleration doesn't take into account what happen in between; only the end points are interesting: $$a_{av}=\frac{v_2-v_1}{\Delta t}$$ If average acceleration is non-zero, then $v_1 \neq v_2$ and the average of these is surely non-zero as well.

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