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

[DrummingAtom]

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

a_rad = 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.

 

[SrEstroncio]

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.

 

[DrummingAtom]

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?