A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t1 = 3.30 s, it is at point (4.40 m, 5.90 m) with velocity (2.70 m/s)ĵ and acceleration in the positive x direction. At time t2 = 12.0 s, it has velocity (–2.70 m/s)î and acceleration in the positive y direction. What are the (a) x and (b) y coordinates of the center of the circular path? Assume at both times that the particle is on the same orbit.

Name: A particle moves along a circular path over a horizontal xy coordinate
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Description: A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t1 = 3.30 s, it is at point (4.40 m, 5.90 m) with velocity (2.70 m/s)ĵ and acceleration in the positive x direction. At time t2 = 12.0 s, it ha

(given) Velocity of particle = (2.70 m/s)j^ at time t1= 3.30 s and at point (4.40 m, 5.90 m)…

Answered: A particle moves along a circular path…

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter3: Motion In Two Dimensions

Section: Chapter Questions

Problem 6P: At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of...

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A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t₁ = 3.30 s, it is at point (4.40 m, 5.90 m) with velocity (2.70 m/s)ĵ and acceleration in the positive x direction. At time t₂ = 12.0 s, it has velocity (–2.70 m/s)î and acceleration in the positive y direction. What are the (a) x and (b) y coordinates of the center of the circular path? Assume at both times that the particle is on the same orbit.

Expert Solution

Step 1

(given)

Velocity of particle = (2.70 m/s) $\hat{j}$ at time t₁= 3.30 s and at point (4.40 m, 5.90 m)

Velocity of particle = (-2.70 m/s) $\hat{j}$ at time t₂= 12.0 s

and acceleration in the positive direction.

Particle covers (3/4) whole of the circle.

Therefore,

$L = \frac{3}{4} \times (2 πR) = \frac{3}{2} \times (πR)$

Velocity $= \frac{D i s p l a c e m e n t}{t i m e}$ = $\frac{L}{t}$

L = v $\times t$

where t = t₂-t₁=(12-3.30) = 8.7 s

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