Q 11-03-149JEE MainJEE Main 2020 (4 Sep, Shift 1)Easy
Starting from the origin at time $t = 0$, with initial velocity $5\hat j\ \text{m s}^{-1}$, a particle moves in the $x$-$y$ plane with a constant acceleration of $(10\hat i + 4\hat j)\ \text{m s}^{-2}$. At time $t$, its coordinates are $(20\ \text{m}, y_0\ \text{m})$. The values of $t$ and $y_0$ are, respectively:
Answer: (A) $2\ \text{s}$ and $18\ \text{m}$
$x = \tfrac12(10)t^{2} = 20 \Rightarrow t = 2$ s.
$y_0 = 5t + \tfrac12(4)t^{2} = 10 + 8 = 18$ m.
Solution by Sreeraj P, M.Sc Physics