Question

Initially, a particle is moving at 4.20 m/s at an angle of 30.5 ∘ above the horizontal. Two seconds later, its velocity is 6.35 m/s at an angle of 51.0 ∘ below the horizontal. What was the particle's average acceleration during these 2.00 seconds?

Answer 1

Solution:-

v1 = 4.2[cos30.5 i + sin30.5 j]

    = 3.6188 i + 2.1316 j m/s

v2 = 6.35[cos51 i - sin51 j]

     = 3.996 i - 4.9348j m/s

time t = 2 s

Average acceleration (a) = (v2 - v1)/t

= [(3.6188 i - 2.1316 j)-(3.996 i + 4.934 j)]/2

                                     = -0.1886 i - 3.5328 j    m/s^2

X component ax = -0.1886 m/s^2

Y component ay = -3.5328 m/s^2