### Finding the Velocity, Acceleration, and Speed of a Vector Particle in 3D (Example 2)

Find the velocity, acceleration, and speed of a particle. The particle's position function is given by the vector function \(r(t) = {t^2}i + 2tj + \ln (t)k\).

Here i,j,k are just the components, we can also just write it in another vector notation as follows:

\[r(t) = \left\langle {{t^2},2t,\ln (t)} \right\rangle \]

The velocity is just the differentiation of the position function, and the acceleration is the second derivative of the position function. The speed is just the velocity function's norm (you remove the directional components from the velocity equation by finding the magnitude/norm).

First we can find the velocity function v(t). To differentiate a vector function just differentiate each component.

\[r'(t) = v(t) = \left\langle {2t,2,\frac{1}{t}} \right\rangle \]

And the speed function is just the norm of this.

\[\left| {v(t)} \right| = \sqrt {4{t^2} + 4 + \frac{1}{{{t^2}}}} \]

We can simplify this by completing the square and applying the square root.

\[\left| {v(t)} \right| = \sqrt {{{\left( {2t + \frac{1}{t}} \right)}^2}} \]

\[\left| {v(t)} \right| = 2t + \frac{1}{t}\]

And acceleration is just the differentiation of the velocity function (second derivative of the position function):

\[r''(t) = a(t) = \left\langle {2,0, - \frac{1}{{{t^2}}}} \right\rangle \]

Done! Another variation of this question is to give the acceleration function and ask you to find the position function, so in that case you'd just work backwards by integrating and using given initial conditions.

Please see our other worked-out examples on finding displacement, velocity, speed, and acceleration of a moving particle:

Velocity Acceleration Speed of Vector Particle 3D (Example 1)

Velocity Acceleration Speed of Vector Particle 3D (Example 3)

Alternatively return to the particle in motion hub page:

Speed Velocity Acceleration of Vector Particle