$x=2+4 t$

$\frac{d x}{d t}=v_x=4$

$\frac{d v_x}{d t}=a_x=0$

$y=3 t+8 t^2$

$\frac{d y}{d t}=v_y=3+16 t$

$\frac{d v_y}{d t}=a_y=16$

the motion will be uniformly accelerated motion.

For path

$\mathrm{x}=2+4 \mathrm{t}$

$\frac{(\mathrm{x}-2)}{4}=\mathrm{t}$

Put this value of $t$ is equation of $y$

$y=3\left(\frac{x-2}{4}\right)+8\left(\frac{x-2}{4}\right)^2$

this is a quadratic equation so path will be parabola.

Correct answer $(4)$