JEE Physics 2026 Question with Solution

Given: $m = 4 \, \text{kg}$, $\vec{F}(t) = (4t^3\hat{i} - 3t^2\hat{j}) \, \text{N}$, initial conditions $\vec{v}_0 = 0$ and $\vec{r}_0 = 0$.

Find: The velocity vector and position vector at $t = 2 \, \text{s}$.

For a time-dependent force, use the chain:

$$\vec{F} \rightarrow \vec{a} \rightarrow \vec{v} \rightarrow \vec{r}$$

First divide force by mass to get acceleration, then integrate twice using the initial conditions.

From Newton's second law,

$$\vec{a}(t) = \frac{\vec{F}}{m} = \frac{1}{4}(4t^3\hat{i} - 3t^2\hat{j}) = \left(t^3\hat{i} - \frac{3}{4}t^2\hat{j}\right) \, \text{m/s}^2$$

Now integrate acceleration to get velocity:

$$\vec{v}(t) = \int_0^t \left(t'^3\hat{i} - \frac{3}{4}t'^2\hat{j}\right) dt'$$ $$\vec{v}(t) = \left[\frac{t'^4}{4}\hat{i} - \frac{3}{4}\frac{t'^3}{3}\hat{j}\right]_0^t = \left(\frac{t^4}{4}\hat{i} - \frac{t^3}{4}\hat{j}\right) \, \text{m/s}$$

At $t = 2 \, \text{s}$,

$$\vec{v}(2) = \frac{2^4}{4}\hat{i} - \frac{2^3}{4}\hat{j} = \frac{16}{4}\hat{i} - \frac{8}{4}\hat{j} = (4\hat{i} - 2\hat{j}) \, \text{m/s}$$

Next integrate velocity to get position:

$$\vec{r}(t) = \int_0^t \left(\frac{t'^4}{4}\hat{i} - \frac{t'^3}{4}\hat{j}\right) dt'$$ $$\vec{r}(t) = \left[\frac{1}{4}\frac{t'^5}{5}\hat{i} - \frac{1}{4}\frac{t'^4}{4}\hat{j}\right]_0^t = \left(\frac{t^5}{20}\hat{i} - \frac{t^4}{16}\hat{j}\right) \, \text{m}$$

At $t = 2 \, \text{s}$,

$$\vec{r}(2) = \frac{2^5}{20}\hat{i} - \frac{2^4}{16}\hat{j} = \frac{32}{20}\hat{i} - \frac{16}{16}\hat{j} = \left(\frac{8}{5}\hat{i} - \hat{j}\right) \, \text{m}$$

Therefore, the velocity is $(4\hat{i} - 2\hat{j}) \, \text{m/s}$ and the position is $\left(\frac{8}{5}\hat{i} - \hat{j}\right) \, \text{m}$. The correct option is B.

Given: A particle of mass $4 \, \text{kg}$ is acted upon by $\vec{F} = (4t^3\hat{i} - 3t^2\hat{j})$ and starts from rest at the origin.

Find: $\vec{v}(2)$ and $\vec{r}(2)$.

Resolve the force into components and divide each by mass:

$$a_x = \frac{4t^3}{4} = t^3, \qquad a_y = \frac{-3t^2}{4} = -\frac{3}{4}t^2$$

Integrating componentwise,

$$v_x = \int_0^t t'^3 \, dt' = \frac{t^4}{4}, \qquad v_y = \int_0^t -\frac{3}{4}t'^2 \, dt' = -\frac{t^3}{4}$$

So,

$$\vec{v}(t) = \frac{t^4}{4}\hat{i} - \frac{t^3}{4}\hat{j}$$

At $t=2$,

$$\vec{v}(2) = 4\hat{i} - 2\hat{j}$$

Now integrate again for position:

$$x(t) = \int_0^t \frac{t'^4}{4} \, dt' = \frac{t^5}{20}, \qquad y(t) = \int_0^t -\frac{t'^3}{4} \, dt' = -\frac{t^4}{16}$$

Hence,

$$\vec{r}(t) = \frac{t^5}{20}\hat{i} - \frac{t^4}{16}\hat{j}$$

At $t=2$,

$$\vec{r}(2) = \frac{8}{5}\hat{i} - \hat{j}$$

Therefore, the required pair matches option B.