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- Thread starter petern
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- #2

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(The acceleration will equal zero at any point where the f(t) graph changes concavity)

- #3

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Please help me with this:

and this:

I don't really understand most of it.

and this:

I don't really understand most of it.

- #4

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Your first assertion, that acceleration occurs only at curves is correct. Velocity is only equal to zero when it (the position function) has zero slope (ie, no motion--- straight lines and relative max/mins) and acceleration equals zero when the velocity is constant (velocity is a straight line) and when the position graph switches concavity (inflection point). It is important to note that, though velocity may =0 at some point, acceleration may not (although it can).

Here's a few rules to help you out.

When the function is concave (up) its derivative (in this case velocity) is increasing, which means that its acceleration is positive

When it's convex (concave down) its derivative is decreasing

which means that its acceleration is negative

Here's a few rules to help you out.

When the function is concave (up) its derivative (in this case velocity) is increasing, which means that its acceleration is positive

When it's convex (concave down) its derivative is decreasing

which means that its acceleration is negative

Last edited:

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