→
→
51. We first note that a1 (the acceleration at t1 = 2.00 s) is perpendicular to a2 (the acceleration at t2=5.00 s), by taking their scalar (dot) product.:
& & ˆ ˆ ⋅ [(4.00 m/s 2 )i+( ˆ − 6.00 m/s 2 )j]=0. ˆ a1 ⋅ a2 = [(6.00 m/s 2 )i+(4.00 m/s 2 )j] Since the acceleration vectors are in the (negative) radial directions, then the two positions (at t1 and t2) are a quarter-circle apart (or three-quarters of a circle, depending on whether one measures clockwise or counterclockwise). A quick sketch leads to the conclusion that if the particle is moving counterclockwise (as the problem states) then it travels three-quarters of a circumference in moving from the position at time t1 to the position at time t2 . Letting T stand for the period, then t2 – t1 = 3.00 s = 3T/4. This gives T = 4.00 s. The magnitude of the acceleration is a = ax2 + a y2 = (6.00) 2 + (4.00) 2 = 7.21 m/s 2 . Using Eq. 4-34 and 4-35, we have a = 4π 2 r / T 2 , which yields aT 2 (7.21 m/s 2 )(4.00 s) 2 r= = = 2.92 m. 4π 2 4π 2