The velocity of a particle moving in the x-y plane is given by (4.63i + 4.41j) m/s at time t = 6.04 s. Its average acceleration during the next 0.027 s is (3.1i +7.9j) m/s. Determine the velocity v of the particle at t = 6.067 s and the angle between the average- acceleration vector and the velocity vector at t = 6.067 s. Answers: v = (i 0= Mi Save for Later i+ i j) m/s Attempts: 0 of 1 used Submit Answer