Q 11-03-025NEETJEE MainMedium
A particle starts from the origin at $t = 0$ with a velocity of $4\hat{j}$ m/s and moves in the $xy$-plane with a constant acceleration of $(6\hat{i} + 2\hat{j})\ \text{m/s}^2$. Its $y$-coordinate at the instant its $x$-coordinate is $12$ m is
Answer: (B) $12$ m
Concept. Treat $x$ and $y$ separately with constant acceleration.
Step 1. Time. $x = 3t^2 = 12 \Rightarrow t = 2$ s.
Step 2. $y$. $y = 4(2) + \dfrac12(2)(4) = 12$ m.
Answer. $12$ m (Option B).
Solution by Sreeraj P, M.Sc Physics