A particle moves from the point (2.0i + 4.0j) m, at t = 0, with an… | JEE Main 2019 (11 Jan, Shift 2)

Q30JEE Main 2019 · 11 JAN · SHIFT 2Medium

Equations of Kinematics

A particle moves from the point $\left(2.0\hat{i} + 4.0\hat{j}\right)$ m, at t = 0, with an initial velocity $\left(5.0\hat{i} + 4.0\hat{j}\right)$ ms ${}^{-1}$ . It is acted upon by a constant force which produces a constant acceleration $\left(4.0\hat{i} + 4.0\hat{j}\right)$ ms ${}^{-2}$ . What is the distance of the particle from the origin at time 2 s?

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Answer: (b)

$\vec{r} = \vec{r}_0 + \vec{u}t + \frac{1}{2}\vec{a}t^2$ with $t = 2\text{ s}$ .
x: $2 + 5(2) + \frac{1}{2}(4)(4) = 20\text{ m}$ ; y: $4 + 4(2) + \frac{1}{2}(4)(4) = 20\text{ m}$
Distance from origin $= \sqrt{20^2 + 20^2} = 20\sqrt{2}\text{ m}$
So the answer is (b) $20\sqrt{2}$ m.

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Equations of Kinematics question from JEE Main 2019 (11 Jan, Shift 2)

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