Answer to Question #111645 in Vector Calculus for Felix Coleman

Question

The position vector r of a particle at time t is
r = 2t
2
i + (t
2 − 4t)j + (3t − 5)k.
Find the velocity and the acceleration of the particle at time t. Show that when t =
2
5
the velocity
and the acceleration are perpendicular to each other. The velocity and the acceleration are resolved
into components along and perpendicular to the vector i − 3j + 2k. Find the velocity and acceleration
components parallel to this vector when t =
2
5

Answer 1

Given $\vec{r} = 2 t ^2 \hat{i} +(t^2 - 4t)\hat{j}+(3t-5)\hat{k}.$

So, Velocity $\vec{v} = \frac{d\vec{r}}{dt} = 4t \hat{i} +(2t-4)\hat{j}+3\hat{k}$

and acceleration $\vec{a} = \frac{d\vec{v}}{dt} = 4\hat{i} + 2\hat{j}$.

Now velocity and acceleration are perpendicular when $\vec{v}\cdot\vec{a} = 0$.

So, $(4t)\times 4 +(2t-4)\times 2 = 0, so \hspace{0.1 in} 20t = 8, so \hspace{0.1 in} t =\frac{8}{20}=\frac{2}{5}.$

Now, given another vector is $\vec{b} = \hat{i}-3\hat{j}+2\hat{k}.$

Component of velocity along this

$\vec{v_a} = \frac{(\vec{v}.\vec{b})\vec{b}}{|\vec{b}|^2} = \frac{(4t-6t+12+6)(\hat{i}-3\hat{j}+2\hat{k})}{1^2+3^2+2^2} = (\frac{-t+9}{7})(\hat{i}-3\hat{j}+2\hat{k})$ .

Component of acceleration along given vector

$\vec{a_a} = \frac{(\vec{a}.\vec{b})\vec{b}}{|\vec{b}|^2} = \frac{(-1)}{7} (\hat{i}-3\hat{j}+2\hat{k})$ .

At t = 2/5, $\vec{v}_a = \frac{43}{35} (\hat{i}-3\hat{j}+2\hat{k}), and \hspace{0.1 in} \vec{a}_a=\frac{-1}{7} (\hat{i}-3\hat{j}+2\hat{k})$.

Now, Component of velocity perpendicular to given vector = $\vec{v}-\vec{v}_a$.

And component of acceleration perpendicular to given vector = $\vec{a} - \vec{a}_a$