Calc Q: Find Acceleration Using Velocity Function

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The discussion focuses on calculating acceleration from a given velocity function for a falling body, specifically v=8√(s-t)+1 feet per second. The derivative of the velocity function, which represents acceleration, is derived as v'=4(v-1)(s-t)^(-1/2). By substituting the velocity function into the derivative, the acceleration can be confirmed to equal 32 ft/sec², demonstrating the relationship between velocity and acceleration in this context.

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Calculus Question. Using velocity function to find acceleration?

The velocity of a falling body is v=8sqrt(s-t)+1 feet per second at the instant t(sec) the body has fallen s feet from its starting point. Show that the body's acceleration is 32 ft/sec^2.

Well, the derivative of the velocity function is acceleration, so it's v'=4(v-1)(s-t)^(-1/2)

How do I go from there?
 
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I don't know. The RHS of the given equation doesn't have a proper dimension.
 
Well, you know that:
[tex]v=8\sqrt{s-t}+1[/tex]
so you could substitute that in for the [itex]v[/itex] in the second equation:
[tex]v'=4\frac{v-1}{\sqrt{s-t}}[/tex]
It should work out nicely from there.
 

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