The first time derivative of velocity is acceleration. Can we then conclude that the first time derivative of speed is the magnitude of acceleration? In the following example I will consider a one dimensional case, for the sake of argument. Suppose the velocity v of a particle as a function of time t is given by v(t) = t^2. The acceleration, a, as a function of t is therefore given by a(t) = 2t. And so the magnitude of acceleration (the absolute value, since we are dealing with a one dimensional case as I have previously stated) is a "piece-wise defined function of t", namely 2|t|. That's observation A.
Now, let's go back to velocity. Since |v(t)| = v(t), we can then conclude that the derivative of speed as a function of time is given by d/dt(|v(t)|) = 2t; which, technically speaking, is not the same as |a(t)|. So am I right in my conclusion that differentiating speed does not yield the magnitude of acceleration?
Now, let's go back to velocity. Since |v(t)| = v(t), we can then conclude that the derivative of speed as a function of time is given by d/dt(|v(t)|) = 2t; which, technically speaking, is not the same as |a(t)|. So am I right in my conclusion that differentiating speed does not yield the magnitude of acceleration?
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