A particle is projected in XY-plane with a velocity $\mathbf{u}=(2\hat{\mathrm{i}}+3\hat{\mathrm{j}}){\mathrm{ms}}^{-1}$ at t = 0 from origin. Its acceleration is $\mathbf{a}=\hat{\mathrm{i}}+4\hat{\mathrm{j}}$. Then,

(1) Here, acceleration is constant but not directed in the direction of initial velocity.

So, path is parabolic.

(3) $\mathrm{r}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$

or $\mathrm{r}=\left(2\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}\right)\times 1+\frac{1}{2}\left(\widehat{\mathrm{i}}+4\widehat{\mathrm{j}}\right)\times 1^2$

$=\left(2.5\widehat{\mathrm{i}}+5\widehat{\mathrm{j}}\right)\mathrm{m}$

(4) $\because \mathrm{v}=\mathrm{u}+\mathrm{at}$

$=2\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}+\widehat{\mathrm{i}}+4\widehat{\mathrm{j}}$

$=3\widehat{\mathrm{i}}+7\widehat{\mathrm{j}}$

$\therefore \mathrm{Speed}=\left|\mathrm{v}\right|=\sqrt{58}{\mathrm{ms}}^{-1}$