Q 11-03-143JEE MainJEE Main 2020 (9 Jan, Shift 2)Easy
A particle starts from the origin at $t = 0$ with an initial velocity of $3.0\hat{i}$ m/s and moves in the $x$-$y$ plane with a constant acceleration $\left(6.0\hat{i} + 4.0\hat{j}\right)$ m/s$^2$. The $x$-coordinate of the particle at the instant when its $y$-coordinate is $32$ m is $D$ meters. The value of $D$ is:
Answer: (C) $60$
Along $y$ (starts from rest): $32 = \tfrac{1}{2}(4)t^2 \Rightarrow t = 4$ s.
Along $x$: $D = 3(4) + \tfrac{1}{2}(6)(4)^2 = 12 + 48 = 60$ m.
Solution by Sreeraj P, M.Sc Physics