A particle rotates in a circle of radius 430 m. At a particular instant its acceleration is 1.05m/s^2 in a direction that makes an angle of 38.0^circ to its direction of motion. Part A Part B Determine its speed 2.30 s later assuming ng constant tangential acceleration. v= square x Incorrect; Try Again;5 attempts remaining

To determine the speed of the particle 2.30 seconds later, we need to consider both the tangential and centripetal accelerations.<br /><br />Given:<br />- Radius of the circle, \( r = 430 \) m<br />- Initial acceleration, \( a = 1.05 \) m/s²<br />- Angle between acceleration and direction of motion, \( \theta = 38.0^\circ \)<br />- Time interval, \( t = 2.30 \) s<br /><br />First, let's find the tangential and centripetal components of the acceleration.<br /><br />### Tangential Acceleration<br />The tangential component of the acceleration is given by:<br />\[ a_t = a \cos(\theta) \]<br /><br />Substituting the given values:<br />\[ a_t = 1.05 \cos(38.0^\circ) \]<br />\[ a_t \approx 1.05 \times 0.7880 \]<br />\[ a_t \approx 0.826 \text{ m/s}^2 \]<br /><br />### Centripetal Acceleration<br />The centripetal acceleration is given by:<br />\[ a_c = \frac{v^2}{r} \]<br /><br />However, we need to find the velocity first. Since the tangential acceleration is constant, we can use it to find the change in velocity over time.<br /><br />The tangential acceleration causes a change in velocity:<br />\[ \Delta v = a_t \times t \]<br />\[ \Delta v = 0.826 \times 2.30 \]<br />\[ \Delta v \approx 1.9008 \text{ m/s} \]<br /><br />### Velocity After 2.30 Seconds<br />If the particle starts from rest, the velocity after 2.30 seconds will be:<br />\[ v = \Delta v \]<br />\[ v \approx 1.9008 \text{ m/s} \]<br /><br />Therefore, the speed of the particle 2.30 seconds later is approximately:<br />\[ v \approx 1.90 \text{ m/s} \]