JEE PYQ: Motion in a Plane - Question ID 3aed4f8d2a22 (JEE Main 2020)

In plane motion, the position vector $\overrightarrow{r}(t)$ describes the trajectory of a particle. The velocity $\overrightarrow{v}(t)$ and acceleration $\overrightarrow{a}(t)$ are obtained by differentiating $\overrightarrow{r}(t)$ with respect to time:

• $\overrightarrow{v}(t) = \frac{d\overrightarrow{r}}{dt}$
• $\overrightarrow{a}(t) = \frac{d\overrightarrow{v}}{dt} = \frac{d^2\overrightarrow{r}}{dt^2}$

Two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ are perpendicular if their dot product vanishes: $\overrightarrow{A} \cdot \overrightarrow{B} = 0$. A vector $\overrightarrow{A}$ is directed toward the origin if it is antiparallel to the position vector $\overrightarrow{r}$, i.e. $\overrightarrow{A} = -k\,\overrightarrow{r}$ for some positive scalar $k$.

Step 1: Compute the velocity vector

Given $\displaystyle \overrightarrow{r}(t) = \cos(\omega t)\,\widehat{i} + \sin(\omega t)\,\widehat{j}$, we differentiate with respect to $t$:

Step 2: Check orthogonality of $\overrightarrow{v}$ and $\overrightarrow{r}$

Form the dot product:

Since the dot product is zero, $\overrightarrow{v}$ is perpendicular to $\overrightarrow{r}$.

Step 3: Compute the acceleration vector

Differentiate $\overrightarrow{v}(t)$:

Step 4: Interpret the direction of $\overrightarrow{a}$

The acceleration is a negative scalar multiple of $\overrightarrow{r}$: $\overrightarrow{a} = -\omega^2\,\overrightarrow{r}$. This means $\overrightarrow{a}$ points exactly opposite to $\overrightarrow{r}$, i.e. toward the origin.

Step 5: Match with the given options

• Option A claims both $\overrightarrow{v}$ and $\overrightarrow{a}$ are perpendicular to $\overrightarrow{r}$. We found $\overrightarrow{a}$ is antiparallel, not perpendicular.
• Option B claims both are parallel to $\overrightarrow{r}$. That is false for $\overrightarrow{v}$.
• Option C states $\overrightarrow{v}$ is perpendicular to $\overrightarrow{r}$ and $\overrightarrow{a}$ is directed toward the origin. This matches our results.
• Option D says $\overrightarrow{a}$ is directed away from the origin, which is incorrect.

Therefore the correct choice is C.

1. Sign error in differentiation: Students often forget the negative sign when differentiating $\cos(\omega t)$ or $\sin(\omega t)$, leading to wrong velocity or acceleration expressions. 2. Misinterpreting antiparallel as perpendicular: The acceleration $\overrightarrow{a} = -\omega^2\overrightarrow{r}$ is collinear with $\overrightarrow{r}$ but points inward. Some confuse this with perpendicularity. 3. Skipping the dot-product check: Without computing $\overrightarrow{v}\cdot\overrightarrow{r}$, one might guess that both vectors are either parallel or perpendicular. Always compute the dot product explicitly.

Exam Tip: Whenever you see a position vector of the form $\cos(\omega t)\,\widehat{i} + \sin(\omega t)\,\widehat{j}$, recognize it as uniform circular motion. In such motion the velocity is always tangent (perpendicular to radius) and the acceleration is centripetal (directed toward the center).