Physics · 2019
JEE Main · 11 January 2019, Shift 2 · Q2
A particle moves from the point (2.0 i +4.0 j ) m, at t =0, with an initial velocity (5.0 i +4.0 j ) ms^-1. It is acted upon by a constant force…
A particle moves from the point $\displaystyle (2.0 \hat{i}+4.0 \hat{j}) \mathrm{m}$, at $\displaystyle \mathrm{t}=0$, with an initial velocity $\displaystyle (5.0 \hat{i}+4.0 \hat{j}) \mathrm{ms}^{-1}$. It is acted upon by a constant force which produces a constant acceleration $\displaystyle (4.0 \hat{i}+4.0 \hat{j}) \mathrm{ms}^{-2}$. What is the distance of the particle from the origin at time $\displaystyle 2$ s ?
Official answer
(1)
$\displaystyle 20 \sqrt{2} \mathrm{~m}$
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JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.