Q 11-03-164JEE MainJEE Main 2019 (11 Jan, Shift 2)Easy
A particle moves from the point $(2.0\hat{i} + 4.0\hat{j})$ m, at $t = 0$, with an initial velocity $(5.0\hat{i} + 4.0\hat{j})\ \text{m s}^{-1}$. It is acted upon by a constant force which produces a constant acceleration $(4.0\hat{i} + 4.0\hat{j})\ \text{m s}^{-2}$. What is the distance of the particle from the origin at time $2$ s?
Answer: (B) $20\sqrt2$ m
$$\vec{r} = \vec{r}_0 + \vec{u}t + \frac12\vec{a}t^2 = (2\hat{i} + 4\hat{j}) + (10\hat{i} + 8\hat{j}) + (8\hat{i} + 8\hat{j}) = 20\hat{i} + 20\hat{j}$$
$$|\vec{r}| = 20\sqrt2\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics