Velocity, Acceleration and Distance

AI Thread Summary
The discussion focuses on deriving instantaneous velocity and acceleration from the position function x(t) = A sin(ωt). The instantaneous velocity is found to be v(t) = ωA cos(ωt), while the instantaneous acceleration is a(t) = -ω²A sin(ωt). The user successfully substitutes x into the acceleration equation, resulting in a(t) = -ω²x. However, they struggle to express velocity in terms of x, ω, and A, ultimately deriving v = ±(ω√(A² - x²)). It is clarified that both positive and negative values for velocity are valid, as they depend on the direction of motion.
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1.11. If position of an object is given by: x(t) = A sin(ωt) where A is a constant and ω is
the angular frequency.
a) What is the instantaneous velocity at time t?
b) What is the instantaneous acceleration at time t?
c) Express the instantaneous velocity and the instantaneous acceleration in terms of
x, ω and A.




2. dx(t)/dt= v(t)
dv(t)/dt=a(t)




3. a) For (a) I just took the derivative of the x function and got wAcos(wt)
b) Same thing here but took the derivative for the answer i got in A and got
-w^2Asin(wt)
c) This is where i had some trouble. I got the instantaneous acceleration part of the
problem by substituting x for Asin(wt) and thus got a(t)=-w^2x. But i had no
clue how to get v(t) in terms of x,w,A.

What i did was set sin^2(wt) + cos^2(wt)=1 and solved for cos^2(wt). then plugged that into v^2(t)=w^2A^2cos^2(wt). I got v^2=w^2(A^2-x^2). After taking the square root i got v = +/-(w * sqrt(A^2-x^2). Is this right? should there be the +/- or is one ruled out?


Thanks for the help in advance!\sqrt{}
 
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Good job! No, you can't pick one of the +/-. You need them both. At a given value of x, v can be either negative or positive depending on whether x is increasing or decreasing. And it could be doing either.
 
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