JEE Main 2019 · 11 Jan · Shift II

JEE Main 2019 Physics question on Kinematics

A particle moves from the point (2.0i^+4.0j^)(2.0\hat{i} + 4.0\hat{j}) m at t=0t = 0 with an initial velocity (5.0i^+4.0j^)(5.0\hat{i} + 4.0\hat{j}) ms−1^{-1}. It is acted upon by a constant force which produces a constant acceleration (4.0i^+4.0j^)(4.0\hat{i} + 4.0\hat{j}) ms−2^{-2}. What is the distance of the particle from the origin at time 2 s?
  1. A.
    5 m
  2. B.
    20220\sqrt{2} m
  3. C.
    10210\sqrt{2} m
  4. D.
    15 m
Show answer

Correct option: B

[JEE Main 2019 · 11 Jan · Shift II]

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