JEE PYQ: Motion in a Straight Line - Question ID 56c5f124e7b3 (JEE Main 2019)

This problem involves kinematics in two dimensions under constant acceleration. The key concepts and formulas are:

• Position vector as a function of time: When a particle starts at position $\vec{r}_0$ with initial velocity $\vec{v}_0$ and undergoes constant acceleration $\vec{a}$, its position at time $t$ is given by: $\vec{r}(t) = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2} \vec{a} t^2$
• Distance from the origin: The distance $d$ of the particle from the origin at time $t$ is the magnitude of the position vector: $d = \left| \vec{r}(t) \right|$

Since the motion is in two dimensions (along $\widehat{i}$ and $\widehat{j}$), we treat each component independently and then combine them using the Pythagorean theorem.

Given data:

• Initial position: $\vec{r}_0 = 2.0 \widehat{i} + 4.0 \widehat{j}$ m
• Initial velocity: $\vec{v}_0 = 5.0 \widehat{i} + 4.0 \widehat{j}$ m/s
• Constant acceleration: $\vec{a} = 4.0 \widehat{i} + 4.0 \widehat{j}$ m/s²
• Time: $t = 2$ s

Step 1: Write the position vector at time $t = 2$ s

Step 2: Compute each term

• $\vec{v}_0 t = (5.0 \widehat{i} + 4.0 \widehat{j}) \cdot 2 = 10.0 \widehat{i} + 8.0 \widehat{j}$
• $\frac{1}{2} \vec{a} t^2 = \frac{1}{2} (4.0 \widehat{i} + 4.0 \widehat{j}) \cdot 4 = (2.0 \widehat{i} + 2.0 \widehat{j}) \cdot 4 = 8.0 \widehat{i} + 8.0 \widehat{j}$

Step 3: Sum all components

Step 4: Compute the distance from the origin

Students often make the following mistakes:

• Ignoring vector components: Treating the motion as one-dimensional and adding magnitudes directly, leading to incorrect results.
• Misapplying the kinematic equation: Forgetting to multiply acceleration by $\frac{1}{2} t^2$ or misplacing terms in the equation.
• Arithmetic errors: Especially in squaring and square root calculations. Always double-check calculations involving $\sqrt{2}$.
• Unit confusion: Ensure all units are consistent (meters and seconds here).

Exam Tip: Always write down the vector components separately and verify each arithmetic step. This question tests both conceptual understanding and numerical accuracy.

The correct answer is B: $20\sqrt{2}$ m.