JEE Main 2024 — 4 April, Shift 1 — Question 47

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

$\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}_{\mathrm{x}}=4$ $\frac{\mathrm{dv}_{\mathrm{x}}}{\mathrm{dt}}=\mathrm{a}_{\mathrm{x}}=0$

$y=3 t+8 t^{2}$

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

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 $y$

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

this is a quadratic equation so path will be parabola.