JEE PYQ: Motion in a Plane - Question ID a93335844523 (JEE Main 2005)

Core Formula & Concept:

In two-dimensional motion, acceleration is defined as the rate of change of velocity with respect to time. Since velocity is a vector quantity, both its magnitude and direction matter. The average acceleration $\vec{a}_{avg}$ over a time interval $\Delta t$ is given by:

$\vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t} = \frac{\vec{v}_f - \vec{v}_i}{\Delta t}$

where:

• $\vec{v}_i$ is the initial velocity vector,
• $\vec{v}_f$ is the final velocity vector,
• $\Delta t$ is the time interval.

In this problem, the particle changes its velocity from eastward to northward. We must treat velocity as a vector and compute the vector difference $\Delta \vec{v} = \vec{v}_f - \vec{v}_i$. Then, divide by the time interval to get the average acceleration vector. The direction of $\vec{a}_{avg}$ is the same as the direction of $\Delta \vec{v}$.

Step-by-Step Derivation:

Let’s define the coordinate system:

• East direction: positive $x$-axis
• North direction: positive $y$-axis

Initial velocity: $\vec{v}_i = 5\, \text{m/s}$ east $\Rightarrow \vec{v}_i = (5, 0)\, \text{m/s}$
Final velocity: $\vec{v}_f = 5\, \text{m/s}$ north $\Rightarrow \vec{v}_f = (0, 5)\, \text{m/s}$
Time interval: $\Delta t = 10\, \text{s}$

Compute the change in velocity: $\Delta \vec{v} = \vec{v}_f - \vec{v}_i = (0, 5) - (5, 0) = (-5, 5)\, \text{m/s}$

Now, compute the average acceleration: $\vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t} = \frac{(-5, 5)}{10} = \left(-\frac{1}{2}, \frac{1}{2}\right)\, \text{m/s}^2$

Find the magnitude of $\vec{a}_{avg}$: $|\vec{a}_{avg}| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}\, \text{m/s}^2$

Determine the direction of $\vec{a}_{avg}$: The vector $\vec{a}_{avg} = \left(-\frac{1}{2}, \frac{1}{2}\right)$ points in the second quadrant, i.e., towards the north-west direction (since $x$ is negative and $y$ is positive).

Thus, the average acceleration is $\frac{1}{\sqrt{2}}\, \text{m/s}^2$ towards north-west.

Common Traps & Exam Tip:

Students often make the following mistakes:

• Treating velocity as a scalar: They compute the difference in magnitudes only, ignoring direction. This leads to incorrect acceleration values.
• Incorrect vector subtraction: Subtracting $\vec{v}_i$ from $\vec{v}_f$ incorrectly, e.g., $(5,0) - (0,5) = (5,-5)$, which reverses the direction of $\Delta \vec{v}$.
• Direction confusion: Misidentifying the resultant direction of $\Delta \vec{v}$. The vector $(-5,5)$ points north-west, not north-east.
• Magnitude error: Forgetting to take the square root when computing the magnitude of $\vec{a}_{avg}$.

Exam Tip: Always draw a vector diagram. Represent $\vec{v}_i$ and $\vec{v}_f$ as arrows. The vector $\Delta \vec{v} = \vec{v}_f - \vec{v}_i$ is the arrow from the tip of $\vec{v}_i$ to the tip of $\vec{v}_f$. This visual helps avoid direction errors.

Hence, the correct answer is Option C.