JEE PYQ: Motion in a Straight Line - Question ID f669169d7cee (JEE Main 2019)

Let's solve the problem systematically, ensuring clarity at each step.

In kinematics, the position vector $\overrightarrow{r}(t)$ of a particle describes its location in space as a function of time. The velocity $\overrightarrow{v}(t)$ is the first time derivative of the position vector, and the acceleration $\overrightarrow{a}(t)$ is the first time derivative of the velocity vector (or the second time derivative of the position vector). Mathematically:

The magnitude of the acceleration vector is given by:

where $a_x$ and $a_y$ are the components of the acceleration vector along the $\widehat{i}$ and $\widehat{j}$ directions, respectively.

Step 1: Write the given position vector

The position vector of the particle is: $\overrightarrow{r}(t) = 15t^2 \widehat{i} + (4 - 20t^2) \widehat{j}$

Step 2: Compute the velocity vector

Differentiate $\overrightarrow{r}(t)$ with respect to time $t$ to obtain the velocity vector $\overrightarrow{v}(t)$:

Differentiating each component:

Thus, the velocity vector is: $\overrightarrow{v}(t) = 30t \widehat{i} - 40t \widehat{j}$

Step 3: Compute the acceleration vector

Differentiate $\overrightarrow{v}(t)$ with respect to time $t$ to obtain the acceleration vector $\overrightarrow{a}(t)$:

Differentiating each component:

Thus, the acceleration vector is: $\overrightarrow{a}(t) = 30 \widehat{i} - 40 \widehat{j}$

Step 4: Compute the magnitude of the acceleration at $t = 1$

The acceleration vector is constant (does not depend on $t$), so its magnitude at any time $t$ (including $t = 1$) is:

Step 5: Match with the given options

The magnitude of the acceleration at $t = 1$ is $50$, which corresponds to option A.

1. Mistaking velocity for acceleration: Some students stop at finding the velocity vector and mistakenly compute its magnitude instead of proceeding to find acceleration. Always ensure you differentiate twice to obtain acceleration.

2. Incorrect differentiation: Forgetting to apply the chain rule or misapplying differentiation rules (e.g., treating $t^2$ as a constant) can lead to wrong components. Double-check each differentiation step.

3. Sign errors in components: The $\widehat{j}$ component of the position vector is $(4 - 20t^2)$. Students sometimes overlook the negative sign when differentiating, leading to $v_y = -40t$ instead of $+40t$. This propagates to the acceleration calculation.

4. Assuming acceleration depends on time: In this problem, the acceleration is constant because the position vector is quadratic in $t$. However, students might unnecessarily substitute $t = 1$ into the acceleration components before computing the magnitude, which is redundant here but could be critical in other problems.

Exam Tip: Always verify whether the acceleration is time-dependent or constant. If it's constant, compute its magnitude once; if not, substitute the given time before computing the magnitude.