initially, a particle is moving at 5.33 m/s at an angle of 37.9° above the horizontal. Two seconds later, its velocity is 6.11 m

Question

3 years ago

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initially, a particle is moving at 5.33 m/s at an angle of 37.9° above the horizontal. Two seconds later, its velocity is 6.11 m

/s at an angle of 54.2° below the horizontal. What was the particle's average acceleration during these 2.00 seconds in the x-direction (enter first) and the y-direction?

Physics

1 answer:

Ainat [17] · Answer 1 · 2022-04-30T06:34:30+03:00

Explanation:

Average acceleration is the change in velocity over the change in time:

a = (v − v₀) / t

In the x direction:

aₓ = (6.11 cos (-54.2°) − 5.33 cos (37.9°)) / 2.00

aₓ = -0.316 m/s²

In the y direction:

aᵧ = (6.11 sin (-54.2°) − 5.33 sin (37.9°)) / 2.00

aᵧ = -4.11 m/s²