JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Laws of Motion

A force

F = (40 \hat{i} + 10 \hat{j}) N

acts on a body of mass 5 kg. If the body starts from rest, its position vector

r

at time

t = 10 s

, will be :

(A) $(100 \hat{i} + 100 \hat{j}) m$
(B) $(400 \hat{i} + 100 \hat{j}) m$ ✓ Correct
(C) $(400 \hat{i} + 400 \hat{j}) m$
(D) $(100 \hat{i} + 400 \hat{j}) m$

Answer: (B)

Solution

The acceleration is

a = \frac{F}{m} = \frac{40 i ^ + 10 j ^}{5} = (8 \hat{i} + 2 \hat{j})

m/s². Starting from rest, the position vector is

r = \frac{1}{2} a t^{2} = \frac{1}{2} (8 \hat{i} + 2 \hat{j}) (10)^{2} = (400 \hat{i} + 100 \hat{j})

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Single correct option

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