JEE Main 2023 — Laws Of Motion Question with Solution

Given the velocity vector of a particle $v=(2t\hat{i}+3t^2\hat{j}){\text{ms}}^{-1}$, the acceleration $a$ is the derivative of the velocity vector with respect to time. So, we have:

$a=\frac{dv}{dt}=(2\hat{i}+6t\hat{j}){\text{ms}}^{-2}$.

At $t=1\text{s}$, the acceleration $a$ is $(2\hat{i}+6\hat{j}){\text{ms}}^{-2}$.

According to Newton'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, $0.5\text{kg}$.

Therefore, the force $F$ on the particle at $t=1\text{s}$ is:

$F=m\cdot a=0.5\cdot (2\hat{i}+6\hat{j})=(1\hat{i}+3\hat{j})\text{N}$.

So, the force acting on the particle at $t=1\text{s}$ is $(\hat{i}+x\hat{j})\text{N}$, where $x=3$.

Therefore, the answer is $x=3$.