Question

Initially, a particle is moving at 4.60 m/s at an angle of 33.0 ∘ above the horizontal. Two seconds later, its velocity is 6.35 m/s at an angle of 58.0 ∘ below the horizontal. What was the particle's average acceleration during these 2.00 seconds? Enter the x and y components of the average acceleration separated by comma.

Answer 1

v1 = 4.60[cos33 i + sin33 j]

    = 3.8578 i + 2.505 j m/s

v2 = 6.35[cos58 i - sin58 j]

     = 3.364 i - 5.385 j   m/s

time t = 2 s

Average acceleration a = (v2 - v1)/t

                                     = [(3.364 i - 5.385 j)-(3.8578 i + 2.505 j)]/2

                                     = -0.493 i - 7.89 j    m/s^2

X component ax = -0.493 m/s^2

Y component ay = -7.89 m/s^2

I hope help you......