We know that if velocity is zero for an instant, then acceleration need not be zero (a simple example of which is a ball thrown upwards.)

But if velocity is zero for an interval, will acceleration also be zero throughout?

I thought it should be true, as the instant $v$ comes to $0$, $a$ should become $0$ to allow the velocity to remain zero throughout that interval

Q. Is this true? How can we show this?

I tried the following:

let $a=kt$ (a is dependent on t) $$\implies \dfrac{dv}{dt} = kt$$

$$\int^v_{v_0} dv = \int^t_{t_0} kt~dt$$

$$v-v_0 = \dfrac{k(t^2 - t^2_0)}{2}$$

Now if $v_0$ to $v$ be interval when velocity is zero, $k = 0$ implying acceleration is zero ($kt=0$).

Q. Is this correct way to show this?

  • $\begingroup$ Hi Tim. Do you mean the velocity is zero at all times during the interval, or just that the average velocity is zero over the interval? $\endgroup$ Jun 14, 2015 at 6:00
  • $\begingroup$ @JohnRennie Yes its zero in the whole interval $\endgroup$
    – Max Payne
    Jun 14, 2015 at 6:01

2 Answers 2


The acceleration is the time derivative of the velocity:

$$ a = \frac{dv}{dt} $$

so if the velocity does not change with time the acceleration is necessarily zero. Since in your example the velocity is constant during the interval that means $dv/dt = 0$ and therefore that $a = 0$ during the interval.

The velocity doesn't have to be zero. Any constant non-zero value will also give $a = 0$.

  • $\begingroup$ Yeah if $v = v_0$, then $k = 0$ in my example isnt it? $\endgroup$
    – Max Payne
    Jun 14, 2015 at 6:09
  • $\begingroup$ Yes, though that seems an unnecessarily complicated way of proving that $a = 0$ when it's zero by definition. $\endgroup$ Jun 14, 2015 at 6:16
  • $\begingroup$ @JohnRennie Tim's method only covered the constant acceleration case, so it showed that if the acceleration is constant then it must be zero, whereas your method actually shows that acceleration is zero. $\endgroup$
    – Timaeus
    Jun 14, 2015 at 16:58

Here average velocity is zero let time be infinitesimally small then the same average velocity now become instantaneous velocity for the same interval of time. Let x and x+ dx be the displacement point in interval t to t+dt and hence let v and v+dv be the velocities at those points now instant acceleration = {v+dv} - {v} / dt which is zero.

  • $\begingroup$ Hi aryen, note that we do have an equation editor built into the site; search notation in help center for details. $\endgroup$
    – Kyle Kanos
    May 25, 2019 at 22:09

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