As S=ut+21at2 S=(5i^+4j^)2+21(4i^+4j^)4 =10i^+8j^+8i^+8j^ rf−ri=18i^+16j^ [ass= change in position =rf−ri] rr=20i^+20j^ rr=202
JEE Main 2019 — Physics Mechanics
A particle moves from the point (2.0i^+4.0j^)m, at t=0, with an initial velocity (5.0i^+4.0j^)ms−1. It is acted upon by a constant force which produces a constant acceleration (4.0i^+4.0j^)ms−2. What is the distance of the particle from the origin at time 2 s?
Held on 11 Jan 2019 · Verified 1 Jul 2026.
15 m
202 m
5 m
102 m
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