Physics - Kinematics Question with Solution | TestHub

$\begin{array}{rl}& x=2+4t\\ & \frac{dx}{dt}=v_x=4\\ & \frac{d{v}_x}{dt}=a_x=0\\ & y=3t+8t^2\\ & \frac{dy}{dt}=v_y=3+16t\\ & \frac{d{v}_y}{dt}=a_y=16\end{array}$ the motion will be uniformly accelerated motion. For path $\begin{array}{rl}& \mathrm{x}=2+4\mathrm{t}\\ & \frac{(\mathrm{x}-2)}{4}=\mathrm{t}\end{array}$
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.