Q56 / 75 · JEE Main 2026 (21 January Shift 1) · Physics
A 4 kg mass moves under the influence of a force $\vec{F}=\left(4 t^{3} \hat{i}-3 t \hat{j}\right) \mathrm{N}$ where $t$ is the time in second. If mass starts from origin at $t=0$, the velocity and position after $t=2 \mathrm{~s}$ will be:
- A$\vec{v}=4 \hat{i}-\frac{3}{2} \hat{j} \quad \vec{r}=\frac{6}{5} \hat{i}-\hat{j}$
- B$\vec{v}=3 \hat{i}+\frac{3}{2} \hat{j} \quad \vec{r}=\frac{6}{5} \hat{i}+\hat{j}$
- Correct answerC$\vec{v}=4 \hat{i}-\frac{3}{2} \hat{j} \quad \vec{r}=\frac{8}{5} \hat{i}-\hat{j}$
- D$\vec{v}=4 \hat{i}+\frac{5}{2} \hat{j} \quad \vec{r}=\frac{8}{5} \hat{i}+2 \hat{j}$
Solution
Given $\vec{F} = (4t^3\hat{i} - 3t\hat{j})$ N, m = 4 kg
$\vec{a} = \frac{\vec{F}}{m} = t^3\hat{i} - \frac{3t}{4}\hat{j}$
Integrating with $\vec{v}(0) = 0$:
$\vec{v} = \frac{t^4}{4}\hat{i} - \frac{3t^2}{8}\hat{j}$
At $t = 2$: $\vec{v} = \frac{16}{4}\hat{i} - \frac{12}{8}\hat{j} = 4\hat{i} - \frac{3}{2}\hat{j}$
Integrating with $\vec{r}(0) = 0$:
$\vec{r} = \frac{t^5}{20}\hat{i} - \frac{t^3}{8}\hat{j}$
At $t = 2$: $\vec{r} = \frac{32}{20}\hat{i} - \frac{8}{8}\hat{j} = \frac{8}{5}\hat{i} - \hat{j}$
$\vec{a} = \frac{\vec{F}}{m} = t^3\hat{i} - \frac{3t}{4}\hat{j}$
Integrating with $\vec{v}(0) = 0$:
$\vec{v} = \frac{t^4}{4}\hat{i} - \frac{3t^2}{8}\hat{j}$
At $t = 2$: $\vec{v} = \frac{16}{4}\hat{i} - \frac{12}{8}\hat{j} = 4\hat{i} - \frac{3}{2}\hat{j}$
Integrating with $\vec{r}(0) = 0$:
$\vec{r} = \frac{t^5}{20}\hat{i} - \frac{t^3}{8}\hat{j}$
At $t = 2$: $\vec{r} = \frac{32}{20}\hat{i} - \frac{8}{8}\hat{j} = \frac{8}{5}\hat{i} - \hat{j}$
Hence, the correct answer is option: 'C'.