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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
    Commented 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
    Commented 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
    Commented 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. $\endgroup$ Commented Apr 18, 2023 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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Does a positive acceleration mean speeding up?

Not necessarily. For example, every particle with a negative velocity and positive acceleration, like this one, is slowing down. This is due to the net force on the particle acting opposite in direction to its motion, which is because its acceleration and velocity have different signs.

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

A particle that is slowing down with a negative velocity has a positive acceleration; so, saying that it is "decelerating" is potentially misleading.

A particle that is speeding up with a negative velocity has a negative acceleration; so, saying that it is "accelerating" is potentially misleading.

A particle in uniform circular motion has a constant speed and nonzero acceleration; so, saying that it is "not accelerating" is potentially misleading.

Hence, echoing Lubin's comment: best to say "slowing down" (or "retarding", noting that this refers to decreasing speed rather than decreasing velocity) and "speeding up" instead of "decelerating" and "accelerating".

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  • $\begingroup$ Since words may be open to interpretation then many words could be confusing. That is why I suggest to regard velocity and acceleration as vectors. When they are parallel or antiparallel the result is obvious, isn't it. BTW, no change in speed (or $|velocity|$) but change in direction is also an accelerated locomotion. $\endgroup$
    – m-stgt
    Commented May 25 at 7:45
  • $\begingroup$ @m-stgt No one is suggesting to regard velocity/acceleration as a scalar. (The OP's question is in the context of one-dimensional motion, in which case the direction in space of the vector position/displacement/velocity/acceleration is indicated by its plus/minus sign.) $\endgroup$
    – ryang
    Commented May 26 at 2:14

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