JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Laws of Motion

A body of mass 2 kg moves under a force of

(2 \hat{i} + 3 \hat{j} + 5 \hat{k}) N

. It starts from rest and was at the origin initially. After 4 s , its new coordinates are (

8, b, 20

). The value of

b

is ____ (Round off the Nearest Integer)

Answer: 12

Solution

The force is constant, so the acceleration is constant as well:

a = \frac{F}{m} = \frac{2 i ^ + 3 j ^ + 5 k ^}{2} = \hat{i} + 1.5 \hat{j} + 2.5 \hat{k} m s^{- 2}

Starting from rest at the origin, the displacement after time

t

r = \frac{1}{2} a t^{2}

, and at

t = 4 s

r = \frac{1}{2} (\hat{i} + 1.5 \hat{j} + 2.5 \hat{k}) \times 16 = 8 \hat{i} + 12 \hat{j} + 20 \hat{k}

The coordinates after

4 s

are

(8, 12, 20)

, so

b = 12

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Integer answer

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