A moving particle starts at an initial position r(0) = ‹1, 0, 0› with initial velocity v(0) = i - j + k. Its acceleration a(t) = 8ti + 8tj + k. Find its velocity and position at time t. SOLUTION Since a(t) = v'(t), we have v(t) = ∫a(t)dt = ∫(8ti + 8tj + k)dt = i + j + tk + C

v(t) = (4t² + 1)î + (4t² - 1)j + (t + 1)k

r(t) = [(4t³/3) + t + 1]î + [(4t³/3) - t]j + [(t²/2) + t]

Explanation:

r(0) = ‹1, 0, 0› = î

v(0) = î - j + k

a(t) = 8ti + 8tj + k

To find the velocity vector at any time t

Since a(t) = v'(t),

a(t) = (dv/dt)

a(t) = (dv/dt) = 8ti + 8tj + k

(dv/dt) = 8ti + 8tj + k

dv = (8ti + 8tj + k)dt

∫ dv = ∫ (8ti + 8tj + k)dt

Integrating the left hand side from from v(0) to v(t) and the right hand side from 0 to t

v - v(0) = (4t²î + 4t²j + tk)

v(t) = v(0) + (4t²î + 4t²j + tk) = (î - j + k) + ((4t²î + 4t²j + tk))

v(t) = (4t² + 1)î + (4t² - 1)j + (t + 1)k

For its position vector at any time

v(t) = r'(t)

v(t) = (dr/dt) = (4t² + 1)î + (4t² - 1)j + (t + 1)k

(dr/dt) = (4t² + 1)î + (4t² - 1)j + (t + 1)k

dr = [(4t² + 1)î + (4t² - 1)j + (t + 1)k] dt

∫ dr = ∫ [(4t² + 1)î + (4t² - 1)j + (t + 1)k] dt

Integrating the left hand side r(0) to r(t) and the right hand side from 0 to t

r - r(0) = [(4t³/3) + t]î + [(4t³/3) - t]j + [(t²/2) + t]

r(t) = r(0) + [(4t³/3) + t]î + [(4t³/3) - t]j + [(t²/2) + t]

r(t) = î + [(4t³/3) + t]î + [(4t³/3) - t]j + [(t²/2) + t]

r(t) = [(4t³/3) + t + 1]î + [(4t³/3) - t]j + [(t²/2) + t]

f=ma newton's second law 7x8=56n

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