A particle, originally at rest and located at point (3 ft, 2 ft, 5 ft), is subjected to an acceleration of a = 56t i + 12 t 2k6 ft>s2. Determine the particle’s position (x, y, z) at t = 1 s.

What I understand that acceleration a={56ti+12t²k}ft/s²

The particle position is (12.333 ft, 2 ft, 6 ft )

Explanation:

Given data

Particle located at point (3 ft, 2 ft,5 ft)

Acceleration a={56ti+12t²k}ft/s²

To find

Particle position

Solution

As given acceleration is

r₀ is given as=(3i+2j+5k)

So

So the particle position is (12.333 ft, 2 ft, 6 ft )

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