Object moving in equilateral triangle - magnitude of acceleration

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SUMMARY

An object moving around an equilateral triangle at a constant speed of 22.3 m/s takes 12.69 seconds to complete its path. The magnitude of the average acceleration vector calculated using the appropriate kinematic equations is 3.04 m/s². The solution involves breaking down the motion into x and y components and applying the formula for average acceleration. The final answer confirms the correct application of physics principles in uniform circular motion.

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Homework Statement


An object moves around a equilateral triangle with a constant speed (magnitude of velocity is constant) of 22.3 m/s. If it moves from the starting to ending point (as shown below) in a time of 12.69 seconds, what is the magnitude of the average acceleration vector in m/s2?



Homework Equations



V[tex]\Delta[/tex]V= (rFx-rix/delta(t)) - (rfy-riy\delta (t))


The Attempt at a Solution

 
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Break the vector into x and y components then use the equation:
(delta)y=v0t-(1/2)gt^2
 


Thanks! i figured out the answer.. its 3.04
 

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