Calc Q: Find Acceleration Using Velocity Function

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To find the acceleration of a falling body using the given velocity function v=8sqrt(s-t)+1, the derivative of the velocity function is calculated, resulting in v'=4(v-1)(s-t)^(-1/2). Substituting the expression for v into the derivative yields v'=4(v-1)/sqrt(s-t). This approach helps in simplifying the equation to find acceleration. Ultimately, it is shown that the body's acceleration is 32 ft/sec^2.
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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:
v=8\sqrt{s-t}+1
so you could substitute that in for the v in the second equation:
v'=4\frac{v-1}{\sqrt{s-t}}
It should work out nicely from there.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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