What is the Acceleration Vector of a Passenger on a Ferris Wheel?

AI Thread Summary
The discussion revolves around calculating the acceleration vector of a passenger on a Ferris wheel with a radius of 42.0 m, moving at a speed of 3.25 m/s and experiencing an acceleration of 0.550 m/s². Two types of acceleration are identified: centripetal acceleration, calculated using the formula arad = v²/R, and tangential acceleration due to the increase in speed. The centripetal acceleration points towards the center of the circular motion, while the tangential acceleration is directed along the path of motion. The total acceleration vector combines these two components, with the centripetal acceleration pointing in the y-axis and the tangential acceleration along the x-axis. Understanding these components is crucial for solving the problem effectively.
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Homework Statement


A Ferris wheel of radius 42.0m is just starting up. At a given instant, a passenger on the edge of the wheel and passing through the lowest point of his circular motion is moving 3.25m/s and is gaining speed at a rate of 0.550m/s2. Find the magnitude and direction of the passenger's acceleration vector at this instant.


Homework Equations


arad = v2/R


The Attempt at a Solution


I'll say that I have no clue what this question is actually asking. Considering, it gave the acceleration and 3.252/42 \neq 0.550, which is the given acceleration. Thanks for any help.
 
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You've got two different accelerations in this problem, what you got to do is compute the value and direction of the total acceleration experienced by the passenger.

Hint: where's the 3.25^2/42 acel. pointing to? and the 0.555 one?
I hope this helps.
 
Yes, that makes a lot of sense. So, the given acceleration would be pointing along the x-axis and the v2/R acceleration would be pointing in the y axis. Right?
 
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