A particle moves in the xy plane in a circle centered on the origin.At a certain instant the velocity and acceleration of the particle are 6.0 î m/s and (3.0 î + 4.0 ĵ1 )m/s².What are the x and y coordinates of the particle at this moment? A)x = 0,y = -9.0 m B)x = 0,y = +7.2 m C)x = 0,y = +9.0 m D)x = 0,y = -7.2 m E)x = 6.0 m,y = -9.0 m

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The particle is moving in a circle, so its acceleration is always perpendicular to its velocity. The acceleration has a positive y-component, so the particle must be moving in the positive y-direction. The velocity has a positive x-component, so the particle must be moving in the positive x-direction. Therefore, the particle is in the first quadrant.The magnitude of the acceleration is:$$a = \sqrt{a_x^2 + a_y^2} = \sqrt{(3.0 	ext{ m/s}^2)^2 + (4.0 	ext{ m/s}^2)^2} = 5.0 	ext{ m/s}^2$$The magnitude of the velocity is:$$v = \sqrt{v_x^2 + v_y^2} = \sqrt{(6.0 	ext{ m/s})^2 + (0 	ext{ m/s})^2} = 6.0 	ext{ m/s}$$The angle between the velocity and the acceleration is:$$	heta = 	an^{-1}\left(\frac{a_y}{a_x}ight) = 	an^{-1}\left(\frac{4.0 	ext{ m/s}^2}{3.0 	ext{ m/s}^2}ight) = 53.1^\circ$$The x-coordinate of the particle is:$$x = v_x t = (6.0 	ext{ m/s})t$$The y-coordinate of the particle is:$$y = v_y t + \frac{1}{2}a_y t^2 = (0 	ext{ m/s})t + \frac{1}{2}(4.0 	ext{ m/s}^2)t^2 = (2.0 	ext{ m/s}^2)t^2$$At the instant when the velocity and acceleration are given, we have:$$v_x = 6.0 	ext{ m/s}$$$$a_x = 3.0 	ext{ m/s}^2$$$$a_y = 4.0 	ext{ m/s}^2$$Substituting these values into the equations for x and y, we get:$$x = (6.0 	ext{ m/s})t = 6.0t$$$$y = (2.0 	ext{ m/s}^2)t^2 = 2.0t^2$$To find the x and y coordinates of the particle at this moment, we need to find the value of t. We can do this by using the fact that the particle is moving in a circle. The centripetal acceleration of a particle moving in a circle is given by:$$a_c = \frac{v^2}{r}$$where v is the speed of the particle and r is the radius of the circle. In this case, the speed of the particle is 6.0 m/s and the radius of the circle is unknown. However, we can use the fact that the acceleration of the particle is perpendicular to its velocity to find the radius of the circle. The acceleration of the particle is:$$a = \sqrt{a_x^2 + a_y^2} = 5.0 	ext{ m/s}^2$$The centripetal acceleration of the particle is:$$a_c = \frac{v^2}{r}$$Setting these two equations equal to each other, we get:$$5.0 	ext{ m/s}^2 = \frac{(6.0 	ext{ m/s})^2}{r}$$Solving for r, we get:$$r = \frac{(6.0 	ext{ m/s})^2}{5.0 	ext{ m/s}^2} = 7.2 	ext{ m}$$Now that we know the radius of the circle, we can find the value of t by using the equation for the x-coordinate of the particle:$$x = (6.0 	ext{ m/s})t = 6.0t$$Substituting r = 7.2 m into this equation, we get:$$7.2 	ext{ m} = (6.0 	ext{ m/s})t$$Solving for t, we get:$$t = 1.2 	ext{ s}$$Substituting this value of t into the equation for the y-coordinate of the particle, we get:$$y = (2.0 	ext{ m/s}^2)(1.2 	ext{ s})^2 = 2.88 	ext{ m}$$Therefore, the x and y coordinates of the particle at this moment are x = 7.2 m and y = 2.88 m.

A Particle Moves in the Xy Plane in a Circle