Question

A particle moves in such a way that its position vector at any time $t$ is $\overrightarrow{r}=t\hat{i}+\frac{1}{2}{t}^2\hat{j}+t\hat{k}$. Find as a function of time
(i) the velocity $\left(\begin{array}{c}\frac{d\overrightarrow{r}}{dt}\end{array}\right)$
(ii) the speed $\left(\begin{array}{c}\left|\begin{array}{c}\frac{d\overrightarrow{r}}{dt}\end{array}\right|\end{array}\right)$
(iii) the acceleration $\left(\begin{array}{c}\frac{d\overrightarrow{v}}{dt}\end{array}\right)$
(iv) the magnitude of the acceleration
(v) the magnitude of the component of acceleration along velocity (called tangential acceleration)
(vi) the magnitude of the component of acceleration perpendicular to velocity (called normal acceleration).

Answer 1

$\overrightarrow{r}=t\hat{i}+\frac{1}{2}{t}^2\hat{j}+t\hat{k}\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=\hat{i}+t\hat{j}+\hat{k}\left|\overrightarrow{v}\right|=\left|\frac{d\overrightarrow{r}}{dt}\right|=\sqrt{1+t^2+1}=\sqrt{2+t^2}$

Acceleration, $\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=\hat{j}$

$\left|\overrightarrow{a}\right|=\left|\frac{d\overrightarrow{v}}{dt}\right|=\sqrt{1}=1$

First find the angle between acceleration and velocity:

$\overrightarrow{a}.\overrightarrow{v}=\left|a\right|\left|v\right|cos\theta$

$\overrightarrow{a}.\overrightarrow{v}=t$

$t=\sqrt{2+t^2}cos\theta$

$\frac{t}{\sqrt{2+t^2}}=cos\theta$

So, the component of acceleration in direction of velocity is $\frac{t}{\sqrt{2+t^2}}$ and the acceleration component perpendicular to velocity is $\overrightarrow{a}sin\theta$.

$\overrightarrow{a}sin\theta$ $=\frac{\sqrt{2}}{\sqrt{2+t^2}}$ normal acceleration