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Given an acceleration function and a velocity function, how do I determine whether the particle is decelerating or accelerating?

I understand that if velocity $\times$ acceleration is positive then it is accelerating, and if negative then it is decelerating, but must I only determine this with a graph or interval chart?

Does a positive acceleration mean speeding up?

Also, when calculation at what time a function is changing direction, would you find the zeroes of a position time graph?

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  • $\begingroup$ Because the word “accelerate” has a technical meaning in mathematics, and the acceleration of a thrown object is constant in that technical sense, I for one would be much happier if you said “slowing down” instead of “decelerating”, and “speeding up” instead of “accelerating”. You’re really talking about the derivative of the speed function, $s'(t)$, where $s(t)=|v(t)|$, $v$ being the velocity (first derivative of position). But this $s'$ has no good physical use, since it doesn’t fit into Newton’s Law $F=MA$. $\endgroup$
    – Lubin
    Oct 7, 2013 at 3:19

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Something is accelerating when the acceleration function, $a(t)$, is positive. Something is decelerating when the acceleration function is negative.

Note, for position function $x(t)$ and velocity function $v(t)$:

$x'(t) = v(t)$ and $x''(t) = v'(t) = a(t)$.

Something changes direction when the slope of $x(t)$ changes signs. By this, the slope either goes from positive to negative, or negative to positive. You could also view this as $v(t)$, which is the slope of $x(t)$ crossing the horizontal axis.

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  • $\begingroup$ Ah, yes, my mistake. I corrected it now. I believe it should be correct. $\endgroup$
    – David
    Oct 7, 2013 at 3:10
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    $\begingroup$ Yes, I like it in the new form, so I deleted my critical comment. $\endgroup$
    – Lubin
    Oct 7, 2013 at 3:39
  • $\begingroup$ This implies that the only difference between acceleration and deceleration is your choice of what is the positive direction. I disagree with that, a driver that steps on the gas in a car at rest is not decelerating. $\endgroup$ Apr 18 at 16:56
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Let $f(x)$ be an acceleration function. Now, let $f'(x)$ be the derivative of this function. All points where $f'(x)$ is negative will be where there is deceleration, and positive otherwise.

To find when a function is changing, let $f'(x)$ be the derivative of our function. A critical point is when the derivative is equal to 0 or is undefined. Find such points by looking for wherein there are divergences or infinite/undetermined expressions, or setting the derivative equal to 0.

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I understand that if velocity $\times$ acceleration is positive then it is accelerating, and if negative then it is decelerating

Echoing Lubin's comment:

  • if velocity and acceleration are in the same direction, please say speeding up instead of "accelerating";
  • if velocity and acceleration are in opposite directions, please say slowing down (or retarding; but do note that this refers to decreasing speed, not decreasing velocity) instead of "decelerating" (this is a bad, ambiguous word; avoid it).

Because when the particle's velocity is negative:

  • when the particle is speeding up, its acceleration is negative, so saying that it is "accelerating" is either not meaningful or falsely suggesting that its acceleration is positive;
  • when the particle is slowing down, its acceleration is positive, so saying that it is "decelerating" may give the wrong impression that its acceleration is negative.

Does a positive acceleration mean speeding up?

In the previous bullet point whose particle has a positive acceleration, its acceleration and velocity having different signs means that the net force acting on the particle is opposite in direction to its motion; that is, the particle is slowing down.

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