Velocity, Acceleration and Distance

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SUMMARY

The discussion focuses on the relationships between position, velocity, and acceleration for an object described by the equation x(t) = A sin(ωt). The instantaneous velocity is derived as v(t) = ωA cos(ωt), while the instantaneous acceleration is calculated as a(t) = -ω²A sin(ωt). Additionally, the acceleration can be expressed in terms of position as a(t) = -ω²x, and the velocity can be represented as v = ±(ω√(A² - x²)). The presence of both positive and negative signs in the velocity equation indicates that the object can move in either direction at a given position.

PREREQUISITES
  • Understanding of calculus, specifically differentiation
  • Familiarity with trigonometric functions and identities
  • Knowledge of angular frequency and its application in harmonic motion
  • Basic concepts of kinematics in physics
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  • Study the derivation of kinematic equations in harmonic motion
  • Explore the application of trigonometric identities in physics problems
  • Learn about the implications of positive and negative velocity in motion analysis
  • Investigate the relationship between angular frequency and oscillatory motion
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Students of physics, particularly those studying mechanics and oscillatory motion, as well as educators looking for clear explanations of kinematic relationships in harmonic systems.

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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{}
 
Last edited:
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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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