Study Area Document Sharing User Settings A particle starting from rest revolves with uniformly increasing speed in a clockwise circle in the xy plane The center of the circle is at the origin of an zy coordinate system. Att=0 the particle is at x=0.0,y=2.9mAtt=1.0 s, it has made one- quarter of a revolution and is at x=y_(0),y=0.0 Part A Part B - Part C Determine the average ac acceleration vector during this interval Express your answer using two significant figures. Enter the and y components of the acceleration comma. (-7 ? a_(av)=(4.6,0) Submit Previous Answers RequestAnswer & Incorrect; Try Again;3 attempts remaining Provide Feedback

To determine the average acceleration vector during the given time interval, we need to find the change in velocity and divide it by the time taken.<br /><br />Given:<br />Initial position at t=0: (x0, y0) = (0, 2.9)<br />Final position at t=1.0 s: (x1, y1) = (2.9, 0)<br /><br />Step 1: Calculate the initial and final velocities.<br />Initial velocity (v0) can be calculated using the formula:<br />v0 = (x1 - x0) / (t1 - t0), (y1 - y0) / (t1 - t0)<br /><br />v0 = (2.9 - 0) / (1.0 - 0), (0 - 2.9) / (1.0 - 0)<br />v0 = (2.9, -2.9) m/s<br /><br />Final velocity (v1) can be calculated using the formula:<br />v1 = (x2 - x1) / (t2 - t1), (y2 - y1) / (t2 - t1)<br /><br />v1 = (0 - 2.9) / (1.0 - 1.0), (2.9 - 0) / (1.0 - 1.0)<br />v1 = (-2.9, 2.9) m/s<br /><br />Step 2: Calculate the change in velocity.<br />Δv = v1 - v0<br />Δv = (-2.9 - 2.9, 2.9 - (-2.9))<br />Δv = (-5.8, 5.8) m/s<br /><br />Step 3: Calculate the time interval.<br />Δt = t2 - t1<br />Δt = 1.0 - 0<br />Δt = 1.0 s<br /><br />Step 4: Calculate the average acceleration.<br />a_avg = Δv / Δt<br />a_avg = (-5.8 / 1.0, 5.8 / 1.0)<br />a_avg = (-5.8, 5.8) m/s²<br /><br />Therefore, the average acceleration vector during this interval is (-5.8, 5.8) m/s².