Question

The equations of motion of a point particle in space are given by: (i) x(t) = 2t2 − 3 ; y(t) = −3t2 ; z(t) = 4t2 − 5 (ii) x(t) = asin(ωt); y(t) = acos(ωt); z(t) = bt In each case, calculate the velocity vector and the acceleration vector for an arbitrary value of t (considering a and ω as fixed parameters). Calculate the absolute values of those vectors.

Answer 1

Motion -> Velocity (Take one derivative of each equation)
Then to get to acceleration, take another derivative.
(i):
\[dx/dt = v(t) = 4t\]
\[dy/dt = v(t) = -6t\]
\[dz/dt = v(t) = 8t\]

\[d^{2}x/dt^{2} = dv/dt = 4\]
\[d^{2}y/dt^{2} = dv/dt = -6\]
\[d^{2}z/dt^{2} = dv/dt = 8\]
(ii):
\[dx/dt = \omega*a*\cos(\omega*t)\]
\[dy/dt = -\omega*a*sin(\omega*t)\]
\[dz/dt = b\]

\[d^{2}x/dt^{2} = -\omega^{2}*a*\sin(\omega*t)\]
\[d^{2}y/dt^{2} = -\omega^{2}*a*\cos(\omega*t)\]
\[d^{2}z/dt^{2} = 0\]

For absolute values of these, just drop the (-) sign

Answer 2

Thank you!