Answer:
The particle's position at t = 2 is
r = (11ft, 2ft, 21ft) = (11, 2, 21) ft
Explanation:
r₀ = (3, 2, 5) ft
a = (6t, 0, 12t²) ft/s²
a = dv/dt
dv/dt = (6tî + 0j + 12t²ķ)
dv = (6tî + 0j + 12t²ķ) st
Integrating the left hand side from 0 to v (the particle was originally at rest) and the right hand side from 0 to t,
We obtain,
v = (3t²î + 0j + 4t³ķ) ft/s
v = dr/dt
dr/dt = (3t²î + 0j + 4t³ķ)
dr = (3t²î + 0j + 4t³ķ) st
Integrating the left hand side from r₀ (the original position) to r and the right hand side from 0 to t,
r - r₀ = (t³î + 0j + t⁴ķ) ft
r = (t³î + 0j + t⁴ķ) + r₀
r = (t³î + 0j + t⁴ķ) + (3î + 2j + 5ķ)
At t=2s, t³ = 8 and t⁴ = 16
r = (8î + 0j + 16ķ) + (3î + 2j + 5ķ)
r = (11î + 2j + 21ķ) ft
r = (11ft, 2ft, 21ft)
Hope this helps!
