Given mass \( m = 4 \) kg.
\[ \vec{F} = (4t^3 \hat{i} - 3t \hat{j}) \]
Using Newton’s second law:
\[ \vec{a} = \frac{\vec{F}}{m} \]
\[ \vec{a} = \left(t^3 \hat{i} - \frac{3}{4}t \hat{j}\right) \]
Velocity is integral of acceleration:
\[ \vec{v} = \int \vec{a} \, dt \]
\[ v_x = \int t^3 dt = \frac{t^4}{4} \]
\[ v_y = \int -\frac{3}{4}t dt = -\frac{3}{8}t^2 \]
At \( t = 2 \):
\[ v_x = \frac{16}{4} = 4 \]
\[ v_y = -\frac{3}{8} \times 4 = -\frac{3}{2} \]
\[ \vec{v} = 4\hat{i} - \frac{3}{2}\hat{j} \]
Position is integral of velocity:
\[ x = \int \frac{t^4}{4} dt = \frac{t^5}{20} \]
\[ y = \int -\frac{3}{8}t^2 dt = -\frac{t^3}{8} \]
At \( t = 2 \):
\[ x = \frac{32}{20} = \frac{8}{5} \]
\[ y = -\frac{8}{8} = -1 \]
\[ \vec{r} = \frac{8}{5}\hat{i} - \hat{j} \]
Therefore correct answer is (C).
When force depends on time, acceleration also becomes time dependent. Newton’s second law in vector form is:
\[ \vec{F} = m\vec{a} \]
If force varies with time, we must integrate acceleration to obtain velocity and integrate velocity to obtain position. Initial conditions are extremely important.
For vector motion:
\[ \vec{a} = a_x \hat{i} + a_y \hat{j} \]
Each component is treated independently.
Velocity:
\[ \vec{v} = \int \vec{a} dt \]
Position:
\[ \vec{r} = \int \vec{v} dt \]
This topic connects calculus with mechanics and is highly important for JEE Main and Advanced. Many students forget to divide force by mass before integrating.
Common mistakes:
• Not dividing by mass
• Missing integration constants
• Arithmetic errors in powers
• Confusing vector components
This concept is foundation for advanced mechanics problems including variable force fields and motion in electric fields.