A particle moves such that its position vector $\overrightarrow{\mathrm{r}}(\mathrm{t})=\cos \omega \mathrm{t} \hat{\mathrm{i}}+\sin \omega \mathrm{t} \hat{\mathrm{j}}$ where $\omega$ is a constant and $t$ is time. Then which of the following statements is true for the velocity $\overrightarrow{\mathrm{v}}(\mathrm{t})$ and acceleration $\overrightarrow{\mathrm{a}}(\mathrm{t})$ of the particle
$\overrightarrow{\mathrm{v}}$ is perpendicular to $\overrightarrow{\mathrm{r}}$ and $\overrightarrow{\mathrm{a}}$ is directed towards the origin
$\overrightarrow{\mathrm{v}}$ and $\overrightarrow{\mathrm{a}}$ both are parallel to $\overrightarrow{\mathrm{r}}$
$\overrightarrow{\mathrm{v}}$ and $\overrightarrow{\mathrm{a}}$ both are perpendicular to $\overrightarrow{\mathrm{r}}$
$\overrightarrow{\mathrm{v}}$ is perpendicular to $\overrightarrow{\mathrm{r}}$ and $\overrightarrow{\mathrm{a}}$ is directed away from the origin
Which of the following is not a projectile?
A river is flowing due east with a speed $3\, ms^{-1}$. A swimmer can swim in still water at a speed of $4\, ms^{-1}$ (figure).
$(a)$ If swimmer starts swimming due north, what will be his resultant velocity (magnitude and direction) ?
$(b)$ If he wants to start from point A on south bank and reach opposite point $B$ on north bank,
$(i)$ Which direction should he swim ?
$(ii)$ What will be his resultant speed ?
$(c)$ From two different cases as mentioned in $(a)$ and $(b)$ above, in which case will he reach opposite bank in shorter time ?
Which physical quantity can be found by first differntiation and second differentiation of position vector ?