Question

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An object moves at a constant speed along a circular path in a horizontal XY plane. With the center at the origin. When the object is at x=−2m, its velocity is −(4m/s) j^.What is the object's acceleration when it is y=2m?
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Answer 1

Answer:

Velocity, v = −(4m/s) j^.

at y = 2, centripetal acceleration magnitude,

$a = \frac{v {}^{2} }{r} \\ = \frac{(4) {}^{2} }{2} \\ = \frac{16}{2} = 8m {}^{2} .$

So, ar = −(8m/s 2 )j^.