Simple classical physics inquiry

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The discussion focuses on deriving acceleration from a velocity function expressed in terms of position rather than time. It emphasizes the need to apply the chain rule to relate acceleration to velocity and position. The formula presented is a = (dv/dx)(dx/dt), highlighting how to correctly derive acceleration when velocity is a function of position. Clarification is sought on how an additional velocity term arises during this derivation. Understanding this relationship is crucial for solving classical physics problems involving motion.
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[Mentor's note -- this post does not use the homework template because it was moved here from a non-homework forum.]

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Not sure how the extra velocity quantity appears after deriving both side of the velocity function to get acceleration. Please help.
 
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You're given v as a function of x, not as a function of t. Therefore you have to use the chain rule: $$a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt}$$
 
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