A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of 3 m/s and a centripetal acceleration →aof magnitude 2 m/s^{2}. Position vector →rlocates him relative to the rotation axis. What is the magnitude of →r? What is the direction of →rwhen →ais directed due east?

Options:

A . 1.5 m, east

B . 4.5 m, east

C . 1.5 m, west

D . 4.5 m, west

Answer: Option D : D v=3m/s,a=2m/s2 a=v2r⇒r=v2a=322=92=4.5m ∴|→r|=4.5m When →a is due east →r will be due west as r points to the particle's position W.r.t. centre As →a always points toward the centre of motion Please watch the video if you are still doubtful about why a:v2r

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