JEE Main 2026 — Jan 21, Shift 1 — Physics solution

JEE Main2026Jan 21, Shift 1Physics

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 = 2s will be :

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

Solution

Acceleration: $a = F / m = t^{3} \hat{i} - \frac{3 t}{4} \hat{j}$ .

Integrate to get velocity (with $v (0) = 0$ ): $v (t) = \frac{t ^{4}}{4} \hat{i} - \frac{3 t ^{2}}{8} \hat{j}$ .

At $t = 2$ : $v = 4 \hat{i} - \frac{3}{2} \hat{j}$ .

Integrate again (with $r (0) = 0$ ): $r (t) = \frac{t ^{5}}{20} \hat{i} - \frac{t ^{3}}{8} \hat{j}$ . At $t = 2$ : $r = \frac{8}{5} \hat{i} - \hat{j}$ .

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