Position vector is given by r(t)=3sin(4t)i+3cos(4t)j+5tk Determine the velocity and acceleration of the particle at any time t>0.

Question

I understand that the velocity and acceleration are found by the first and second derivatives of the position vector respectively. Also that the magnitude of the velocity is speed, given by ||v(t)||.

What I am getting confused with is how exactly to clarify the velocity and acceleration for any time t>0.

If asked to show the magnitudes of the velocity and acceleration also, how would I go about showing a magnitude for t>0 and not at a point such as t=3?

Answer 1

As you say, the velocity is the derivative with respect to time of the position vector, which gives you a vector in terms of $t$. The speed at any time $t$ is the magnitude of this vector, again as a function of $t$. Differentiating velocity gives you acceleration as a function of $t$, again as a vector.

Answer 2

the velocity vector $$v = 12\cos(4t)i-12\sin (4t)j + 5k$$ and the acceleration vector $$a = -48\sin(4t)i - 48\cos (4t) + 0k$$ this represents a motion around the cylinder of radius $3$ with angular velocity $3$ and a uniform vertical velocity of $5.$