A particle rotates in a circle of radius 430 m. Ata 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 Determine its speed at this moment. v=1.67m/s Previous Answers correct Part B Determine its speed 230 s later, assuming constant tangential acceleration. ? v=1.90 Submit Previous Answers Request Answer & Incorrect; Try Again;4 attempts remaining Nex

To determine the speed of the particle 230 seconds later, assuming constant tangential acceleration, we can use the following formula:<br /><br />$v = v_0 + at$<br /><br />where:<br />- $v$ is the final velocity<br />- $v_0$ is the initial velocity<br />- $a$ is the tangential acceleration<br />- $t$ is the time<br /><br />Given that the particle's acceleration is $1.05 m/s^2$ in a direction that makes an angle of $38.0^\circ$ to its direction of motion, we can calculate the tangential acceleration as:<br /><br />$a_t = a \cdot \sin(38.0^\circ)$<br /><br />Assuming the particle starts from rest, the initial velocity $v_0$ is $0 m/s$. The time $t$ is given as $230 s$.<br /><br />Substituting the values into the formula, we get:<br /><br />$v = 0 + (1.05 m/s^2 \cdot \sin(38.0^\circ)) \cdot 230 s$<br /><br />Calculating this expression gives us the speed of the particle 230 seconds later, assuming constant tangential acceleration.