[Physics] The instant an accelerating object has zero speed, is it speeding up, slowing down, or neither

accelerationkinematicsspeedterminologyvelocity

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This problem is from Khan Academy. Specifically for the blue point circled in red, the answer is that at this blue point, the object is neither speeding up nor slowing down. When I think about the rule about the signs of velocity and acceleration and what this means for change in speed, this makes sense: if velocity and acceleration and have the same sign, the object is speeding up, and if velocity and acceleration have opposite signs, the object is slowing down. At the blue point, the instantaneous velocity is zero and because zero is neither positive nor negative, the object is neither speeding up nor slowing down.

However, intuitively, this doesn't make sense. At the blue point circled in red, the velocity is zero so the speed must be zero. Also, the slope of the function at this blue point is negative so acceleration is negative, meaning that velocity is decreasing every second. If velocity is decreasing every second, then right after 6 seconds, the velocity will turn negative but the speed will have increased. So at the blue point, the object, even though it has zero speed, is in the process of speeding up. Why is this thinking wrong?

Best Answer

The tricky part of this question is that you are given a graph of velocity but asked a question about speed.

Several others have said essentially the same thing, but what really makes this clear for me is a graph of speed:

y = |4 - ((x - 2) / 2) ^ 2|

The above is the graph of $$ y = \left \lvert 4 - \left ( \frac{x - 2}{2} \right ) ^2 \right \rvert \text{,}$$ which is just the absolute value of the velocity graph in your screenshot.

This represents the fact that speed is the absolute value of velocity.

We understand "slowing down" to mean that the slope of the speed is negative, and "speeding up" to mean that the slope of the speed is positive. What is the slope of point $(6, 0)$ on the graph (which corresponds to your circled dot)?

This point is a cusp. The notion of "slope" only exists for differentiable points, and as Wikipedia says,

a function with a bend, cusp, or vertical tangent may be continuous, but fails to be differentiable at the location of the anomaly.

Thus the slope of speed does not exist at this point, and so the object is neither speeding up nor slowing down in this instant.

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