At any instant the velocity of a particle of mass $500 \mathrm{~g}$ is $\left(2 t \hat{i}+3 t^{2} \hat{j}\right) \mathrm{ms}^{-1}$. If the force acting on the particle at $t=1 \mathrm{~s}$ is $(\hat{i}+x \hat{j}) \mathrm{N}$. Then the value of $x$ will be:

<p>Given the velocity vector of a particle $v = (2t \hat{i}+3 t^{2} \hat{j}) \, \text{ms}^{-1}$, the acceleration $a$ is the derivative of the velocity vector with respect to time. So, we have:</p> <p>$a = \frac{dv}{dt} = (2 \hat{i} + 6t \hat{j}) \, \text{ms}^{-2}$.</p> <p>At $t=1 \, \text{s}$, the acceleration $a$ is $(2 \hat{i} + 6 \hat{j}) \, \text{ms}^{-2}$.</p> <p>According to Newton&#39;s second law, the force $F$ is equal to the mass $m$ times acceleration $a$. The mass $m$ is given as $500 \, \text{g}$, or equivalently, $0.5 \, \text{kg}$.</p> <p>Therefore, the force $F$ on the particle at $t=1 \, \text{s}$ is:</p> <p>$F = m \cdot a = 0.5 \cdot (2 \hat{i} + 6 \hat{j}) = (1 \hat{i} + 3 \hat{j}) \, \text{N}$.</p> <p>So, the force acting on the particle at $t=1 \, \text{s}$ is $(\hat{i} + x \hat{j}) \, \text{N}$, where $x=3$.</p> <p>Therefore, the answer is $x=3$.</p>