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
$x = \dfrac{1}{2}(6)t^2 = 3t^2 = 12 \;\Rightarrow\; t = 2$ s.
$$y = 4t + \frac{1}{2}(2)t^2 = 8 + 4 = 12\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics