A particle moves along a circular path over a horizontal xy coordinate system,

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Question

A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t₁ = 5.00 s, it is at point (5.00 m, 6.00 m) with velocity (3π m/s)

^j

and acceleration in the positive x direction. At time t₂ = 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 t₂ - t₁ 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,

A Particle Moves Along A Circular Path Over A Horizontal X...

∴

only possible path is

A Particle Moves Along A Circular Path Over A Horizontal X...

As acceleration should always be directed toward the centre
To calculate the position of the centre, we need the radius, to calculate radius,

v = ω r

we can use the relation,as we can calculate

ω

Δ θ = \frac{3 π}{2}, Δ t = 5 s

=

\frac{3 π}{2 \times 5} = \frac{3 π}{10} S^{- 1}

u s i n g v = ω r

r = \frac{v}{ω}

| \to v | = 3 π m / s g i v e n

∴ r = \frac{3 π}{\frac{3 π}{10}} = \frac{30 π}{3 π} = 10 m

∴ i f r = 10 m

, coordinates of the centre (15 m,6m)

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