Q 11-04-173JEE MainJEE Main 2021 (25 Jul, Shift 2)Easy
A force $\vec{F} = (40\hat{i} + 10\hat{j})$ N acts on a body of mass $5$ kg. If the body starts from rest, its position vector $\vec{r}$ at time $t = 10$ s will be
Answer: (C) $(400\hat{i} + 100\hat{j})$ m
$\vec{a} = \dfrac{\vec{F}}{m} = (8\hat{i} + 2\hat{j})$ m s$^{-2}$.
Starting from rest (at the origin): $\vec{r} = \frac{1}{2}\vec{a}t^2 = 50(8\hat{i} + 2\hat{j}) = (400\hat{i} + 100\hat{j})$ m.
Solution by Sreeraj P, M.Sc Physics