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I'm having trouble understanding velocity and acceleration.

Is it possible for velocity to be increasing while acceleration decreases and vice versa?

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    $\begingroup$ Let $v(t)=-1/t$; then $a(t)=1/t^2$. $v$ is increasing and $a$ is decreasing. Change sign of $v$ and...? $\endgroup$ – Michael Galuza Aug 4 '15 at 10:41
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Sure, as long as acceleration is positive, velocity increases, even if acceleration is decreasing (as long as it doesn't reach zero).

Likewise, as long as acceleration is negative, velocity decreases even if acceleration is increasing.

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  • $\begingroup$ how is this possible that acceleration is negative and acceleration is increasing. $\endgroup$ – user258250 Aug 4 '15 at 11:21
  • $\begingroup$ Well, if acceleration is negative at some point and assuming it is continuous, it must increase before it becomes positive. $\endgroup$ – fkraiem Aug 4 '15 at 11:22
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Since acceleration $a$ is deferential of velocity $v$, so as Michael pointed out it is possible to have a situation that $v >0$ while $a = dv/dt < 0$.

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Acceleration is the rate of change of velocity defined by $a=dv/dt$, so even if acceleration is decreasing as long as it is positive it velocity would increase. So if for example at time $t=0$ we have that $a=2m/sec^2$ which means that every second the velocity increases by 2 m/sec , and then say at time $t=1$ we have $a=1 m/sec^2$ i.e $da/dt<0$ we still have that the velocity would increase by $1m/sec$ You can argue for the second part of your question similarly. I hope this helps.

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As long as a is not zero v is increasing because the body has certain instantaneous a. But if a is increasing v does not decrease. If the body has decalaration only then v decreases.

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