JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Laws of Motion

A 4 kg mass moves under the influence of a force

F = (4 t^{3} \hat{i} - 3 t \hat{j}) N

where

t

is the time in second.
If mass starts from origin at

t = 0

, the velocity and position after

t = 2 s

will be :

(A) $v = 3 \hat{i} + \frac{3}{2} \hat{j} r = \frac{6}{5} \hat{i} + \hat{j}$
(B) $v = 4 \hat{i} - \frac{3}{2} \hat{j} r = \frac{8}{5} \hat{i} - \hat{j}$ ✓ Correct
(C) $v = 4 \hat{i} + \frac{5}{2} \hat{j} r = \frac{8}{5} \hat{i} + 2 \hat{j}$
(D) $v = 4 \hat{i} - \frac{3}{2} \hat{j} r = \frac{6}{5} \hat{i} - \hat{j}$

Answer: (B)

Solution

Dividing the force by the mass,

a = t^{3} \hat{i} - \frac{3}{4} t \hat{j}

. Integrating from

0

2 s

from rest,

v_{x} = \int_{0}^{2} t^{3} d t = 4

and

v_{y} = - \frac{3}{4} \int_{0}^{2} t d t = - \frac{3}{2}

, so

v = 4 \hat{i} - \frac{3}{2} \hat{j}

. Integrating once more from the origin,

x = \int_{0}^{2} \frac{t ^{4}}{4} d t = \frac{8}{5}

and

y = - \frac{3}{8} \int_{0}^{2} t^{2} d t = - 1

, so

r = \frac{8}{5} \hat{i} - \hat{j}

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Single correct option