Question

A particle traverses the path shown from point A to point B. At point A, the tangent to the path is eA 65° from the horizontal and the particle has speed 12 m/s. The particle increases speed and 3 seconds later reaches 15 m/s at point B, where the path's tangent is ee 25 magnitude of the average acceleration experienced by the particle between points A and B. from the horizontal. Determine the A

Answer 1

OA= 65° Given : VA = 12 m/s VB = 15 mis Find the horizontal and vertical components of velocity at A and B !. (VA) si = VA COS65 = 1200865 = 5.071 mis (VA)y = VA sin 65 = 12 sin 65 = 10.875 ms 5 cc) 23 = 13:594 m/s (Ve)y = 15 sin 25º = . 6.339 mis Hence acceleration of particle b/w A & B in horizontal dirt Ax = (Ve)x - (VA)X = 13.594-5-071 3

ax = 2.841 m/s2 In vertical direction! ON ay = (VB)y - (VA)y 6-333 - 10.875 ay = 1.5 12 m/s2 acceleration experienced by the Hence, magnitude of average barticle 6W A and B. az vax? + Qy2 - 1 (2-841)2 + (-1-512)2 a = 3-218 m/s2