JEE PYQ: Motion In Two Dimensions Question 24
Question 24 - 2019 (11 Jan Shift 2)
A particle moves from the point $(2.0\hat{i} + 4.0\hat{j})$ m, at $t = 0$, with an initial velocity $(5.0\hat{i} + 4.0\hat{j})$ ms$^{-1}$. It is acted upon by a constant force which produces a constant acceleration $(4.0\hat{i} + 4.0\hat{j})$ ms$^{-2}$. What is the distance of the particle from the origin at time 2 s?
(a) 15 m (b) $20\sqrt{2}$ m (c) 5 m (d) $10\sqrt{2}$ m
Show Answer
Answer: (b)
Solution
$\vec{S} = (5\hat{i} + 4\hat{j})(2) + \frac{1}{2}(4\hat{i} + 4\hat{j})(4) = 18\hat{i} + 16\hat{j}$. Position: $(20\hat{i} + 20\hat{j})$. Distance $= 20\sqrt{2}$ m.
BETA
