ng A particle starting from rest revolves with uniformly increasing speed in a clockwise circle in the zy plane The center of the circle is at the origin of an ay coordinate system. At t=0 the particle is at x=0.0,y=2.9mcdot Att=1.0s , it has made one- quarter of a revolution and is at x=y_(0),y=0.0 Part A Determine its speed at 1.0 s. Express your answer to two significant figures and include the appropriate units. ? m v=4.56 square Submit Previous Answers Request Answer & Incorrect; Try Again;One attempt remaining

To determine the speed of the particle at \( t = 1.0 \, \text{s} \), we need to use the information given about its motion and the geometry of the circle.<br /><br />1. **Initial Position and Speed:**<br /> - At \( t = 0 \), the particle is at \( (x_0, y_0) = (0.0, 2.9 \, \text{m}) \).<br /> - The particle starts from rest, so its initial velocity \( v_0 = 0 \).<br /><br />2Quarter Revolution:**<br /> - At \( t = 1.0 \, \text{s} \), the particle has made one-quarter of a revolution.<br /> - One-quarter of a revolution corresponds to an angle of \( \theta = \frac{\pi}{2} \) radians.<br /><br />3. **Circle Geometry:**<br /> - The center of the circle is at the origin.<br /> - The radius \( r \) of the circle can be found using the initial position: \( r = \sqrt{x_0^2 + y_0^2} = \sqrt{0^2 + 2.9^2} = 2.9 \, \text{m} \).<br /><br />4. **Velocity after One-Quarter Revolution:**<br /> - After one-quarter of a revolution, the particle will be at \( (y_0, -x_0) = (2.9 \, \text{m}, -0.0) \).<br /> - The velocity component in the x-direction is \( v_x = -\frac{y_0}{r} \omega \).<br /> - The velocity component in the y-direction is \( v_y = \frac{x_0}{r} \omega \).<br /><br />5. **Angular Velocity \( \omega \):**<br /> - The angular velocity \( \omega \) can using the fact that the particle makes one-quarter of a revolution in 1 second: \( \omega = \frac{2\pi}{T} = \frac{2\pi}{1 \, \text{s}} = 2\pi \, \text{rad/s} \).<br /><br />6. **Components of Velocity:**<br /> - \( v_x = -\frac{2.9}{2.9} \cdot 2\pi = -2\pi \, \text{m/s} \).<br /> - \( v_y = \frac{0}{2.9} \cdot 2\pi = 0 \, \text{m/s} \).<br /><br />7. **Speed Calculation:**<br /> - The speed \( v \) is the magnitude of the velocity vector: \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{(-2\pi)^2 + 0^2} = 2\pi \, \text{m/s} \).<br /><br />Therefore, the speed of the particle at \( t = 1.0 \, \text{s} \) is \( 2\pi \, \text{m/s} \).<br /><br />Expressing this to two significant figures:<br />\[ v \approx 6.3 \, \text{m \]<br /><br />So, the correct answer is:<br />\[ v = 6.3 \, \text{m/s} \]