A point moves along the curve y2= 16 x (x and y in meters) in such a manner that the y coordinate of its position at any time is y=t2+4t, where t is in seconds, determine the velocity of the point when t= 5 sec.

Your answer is partially correct. The velocity of a particle moving in the x-y plane is given by (7.70i + 4.91j) m/s at time t = 7.38 s. Its average acceleration during the next 0.029 s is (3.3i + 2.6j) m/s?. Determine the velocity v of the particle at t = 7.409 s and the angle 8 between the average- acceleration vector and the velocity vector at t = 7.409 s. Answers: v = ( 7.7957 i+ 4.9754 j) m/s 5.73

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At time t = 12 s, the velocity of a particle moving in the x-y plane is v=0.08i + 2.28j m/s. By time t = 12.12 s, its velocity has become -0.08i + 2.14j m/s. Determine the magnitude day of its average acceleration during this interval and the angle 0 made by the average acceleration with the positive x-axis. The angle is measured counterclockwise from the positive x-axis. Answers: dav 0= Mi i m/s² O