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Find the magnitude of a particle's acceleration vector at t = 0 for a particle moving with position vector
r(t)-(cos(t),t^3)

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Thank you

User Mattpr
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1 Answer

5 votes

Answer:

1

Explanation:

r(t) = (cos t, t³)

Take first derivative to find velocity:

v(t) = dr/dt

v(t) = (-sin t, 3t²)

Take derivative again to find acceleration:

a(t) = dv/dt

a(t) = (-cos t, 6t)

At t=0, the acceleration's magnitude is:

a(0) = (-1, 0)

|a(0)| = √((-1)² + 0²)

|a(0)| = 1

User Martin Tilsted
by
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