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Velocity, Acceleration and Distance

by clipperdude21
Tags: acceleration, distance, velocity
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clipperdude21
#1
Nov1-07, 02:21 PM
P: 49
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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Dick
#2
Nov1-07, 03:34 PM
Sci Advisor
HW Helper
Thanks
P: 25,246
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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