JEE Main 2024 — Motion In Two Dimensions Question with Solution

$\begin{aligned} & x=2+4 t \\ & \frac{d x}{d t}=v_x=4 \\ & \frac{d v_x}{d t}=a_x=0 \\ & y=3 t+8 t^2 \\ & \frac{d y}{d t}=v_y=3+16 t \\ & \frac{d v_y}{d t}=a_y=16 \end{aligned}$ the motion will be uniformly accelerated motion. For path $\begin{aligned} & \mathrm{x}=2+4 \mathrm{t} \\ & \frac{(\mathrm{x}-2)}{4}=\mathrm{t} \end{aligned}$<br/>Putthisvalueof$t$isequationof$y$ $y=3\left(\frac{x-2}{4}\right)+8{\left(\frac{x-2}{4}\right)}^2$ this is a quadratic equation so path will be parabola.