A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t_{1} = 5.00 s, it is at point (5.00 m, 6.00 m) with velocity (3π m/s)^jand acceleration in the positive x direction. At time t_{2} = 10.0 s, it has velocity (-3π m/s)^i and acceleration in the positive y direction. What are the (a) x and (b) y coordinates of the center of the circular path if t_{2} - t_{1} is less than one period?

Options:

A . (15 m, 6 m)

B . (-5 m, 6)

C . (15 m, 0)

D . (-5 m, 0)

Answer: Option A : A We know the body is undergoing uniform circular motion, ∴ only possible path is As acceleration should always be directed toward the centre To calculate the position of the centre, we need the radius, to calculate radius,v=ωr we can use the relation,as we can calculate ω Δθ=3π2,Δt=5s =3π2×5=3π10S−1 usingv=ωr r=vω |→v|=3πm/sgiven ∴r=3π3π10=30π3π=10m ∴ifr=10m, coordinates of the centre (15 m,6m)

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