Question

The equation r(t)= (3t+9)i+(sqrt(2)t)j+(t^2)k is the position of a particle in space at time t. Find the angle between the velocity and acceleration vectors at time t=0
What is the angle? ______radians

Answer 1

Answer:

$\theta= \frac{\pi}{2} +\pi \cdot i$, for all $i = \mathbb{Z} \cup\{0\}$

Step-by-step explanation:

The velocity vector is found by deriving the position vector depending on the time:

$\dot r(t)= v (t) = 3 \cdot i +\sqrt{2} \cdot j + 2\cdot t \cdot k$

In turn, acceleration vector is found by deriving the velocity vector depending on time:

$\ddot r(t) = \dot v(t) = a(t) = 2 \cdot k$

Velocity and acceleration vectors at $t = 0$ are:

$v(0) = 3\cdot i + \sqrt{2} \cdot j\\a(0) = 2 \cdot k$

Norms of both vectors are, respectively:

$||v(0)||\approx 3.317\\||a(0)|| \approx 2$

The angle between both vectors is determined by using the following characteristic of a Dot Product:

$\theta = \cos^{-1}(\frac{v(0) \bullet a(0)}{||v(0)||\cdot ||a(0)||})$

Given that cosine has a periodicity of $\pi$. There is a family of solutions with the form:

$\theta= \frac{\pi}{2} +\pi \cdot i$, for all $i = \mathbb{Z} \cup\{0\}$