$\begin{aligned}
& x=2+4 t \
& \frac{d x}{d t}=v_x=4 \
& \frac{d v_x}{d t}=a_x=0 \
& y=3 t+8 t^2 \
& \frac{d y}{d t}=v_y=3+16 t \
& \frac{d v_y}{d t}=a_y=16
\end{aligned}themotionwillbeuniformlyacceleratedmotion.Forpath\begin{aligned}
& \mathrm{x}=2+4 \mathrm{t} \
& \frac{(\mathrm{x}-2)}{4}=\mathrm{t}
\end{aligned}$
Put this value of t is equation of y y=3(4x−2)+8(4x−2)2 this is a quadratic equation so path will be parabola.