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) 1 0 2 3 Thank you

Question

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)`

1
0
2
3

Thank you

Answer 1

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