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

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Kinematics

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 2s?

(A)15 m
(B) $202 m$ ✓ Correct
(C)5 m
(D) $102 m$

Answer: (B)

Solution

The displacement over

2

s is

S = v_{0} t + \frac{1}{2} a t^{2} = (5 \hat{i} + 4 \hat{j}) (2) + \frac{1}{2} (4 \hat{i} + 4 \hat{j}) (4) = (10 \hat{i} + 8 \hat{j}) + (8 \hat{i} + 8 \hat{j}) = 18 \hat{i} + 16 \hat{j}

. Adding the starting position,

r = (2 \hat{i} + 4 \hat{j}) + (18 \hat{i} + 16 \hat{j}) = 20 \hat{i} + 20 \hat{j}

, whose distance from the origin is

2 0^{2} + 2 0^{2} = 202

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option

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