The position of a particle is given by

solution:

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

The position of the particle is given by:

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

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

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

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

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

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

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

(b) $8.54 \mathrm{~m} / \mathrm{s}, 69.45^{\circ}$ below the $x$-axis

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

At $t=2.0 \mathrm{~s}:$

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

The magnitude of velocity is given by:

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

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.