The position of a particle is given by

$r =3.0 t \hat{ i }-2.0 t^{2} \hat{ j }+4.0 \hat{ k } \;m$

where $t$ is in seconds and the coefficients have the proper units for $r$ to be in metres.

$(a)$ Find the $v$ and a of the particle?

$(b)$ What is the magnitude and direction of velocity of the particle at $t=2.0 \;s ?$

$(a)$ The position of the particle is given by:

$\vec{r}=3.0 t \hat{ i }-2.0 t^{2} \hat{ j }+4.0 \hat{ k }$

Velocity $\vec{v},$ of the particle is given as:

$\vec{v}=\frac{d \vec{r}}{d t}=\frac{d}{d t}\left(3.0 t \hat{ i }-2.0 t^{2} \hat{ j }+4.0 \hat{ k }\right)$

$\therefore \vec{v}=3.0 \hat{ i }-4.0 t \hat{ j }$

Acceleration $\vec{a}$, of the particle is given as:

$\vec{a}=\frac{d \vec{v}}{d t}=\frac{d}{d t}(3.0 \hat{ i }-4.0 t \hat{ j })$

$\therefore \vec{a}=-4.0 \hat{ j }$

$8.54 \,m / s , 69.45^{\circ}\, below$ the $x$ -axis

$(b)$ We have velocity vector, $\vec{v}=3.0 \hat{ i }-4.0 t \hat{ j }$

At $t=2.0 \,s$

$\vec{v}=3.0 \hat{ i }-8.0 \hat{ j }$

The magnitude of velocity is given by:

$|\vec{v}|=\sqrt{3^{2}+(-8)^{2}}=\sqrt{73}=8.54 \,m / s$

Direction, $\theta=\tan ^{-1}\left(\frac{v_{y}}{v_{x}}\right)$

$=\tan ^{-1}\left(\frac{-8}{3}\right)=-\tan ^{-1}(2.667)$

$=-69.45^{\circ}$

The negative sign indicates that the direction of velocity is below the $x$ -axis.