Q 11-03-059JEE MainJEE Main 2024 (4 Apr, Shift 1)Easy
The coordinates of a particle moving in the $x$-$y$ plane are given by $x = 2 + 4t$, $y = 3t + 8t^2$. The motion of the particle is
Answer: (A) uniformly accelerated having motion along a parabolic path
$v_x = 4$ (constant), $v_y = 3 + 16t$, so $\vec a = 16\hat j$, which is constant: the motion is uniformly accelerated.
The initial velocity $(4\hat i + 3\hat j)$ is not along the acceleration, so the path is curved. Eliminating $t = \dfrac{x - 2}{4}$ gives $y$ as a quadratic in $x$: a parabola.
Solution by Sreeraj P, M.Sc Physics