JEE PYQ: Motion in a Straight Line - Question ID 8ce9a4360053 (JEE Main 2023)

Core Formula & Concept:

In kinematics, the position of a particle moving along a straight line is often given as a function of time, $x(t)$. The instantaneous velocity of the particle at any time $t$ is defined as the derivative of the position function with respect to time. Mathematically, $v(t) = \frac{dx}{dt}$ This formula arises from the fundamental definition of velocity as the rate of change of displacement.

In this problem, the position function is given as $x = 4t^2$. To find the velocity at a specific time (here, $t = 5$ s), we must differentiate $x(t)$ with respect to $t$ and then substitute the value of $t$.

Step-by-Step Derivation:

Step 1: Write the given position function.
The distance travelled by the particle is given by: $x = 4t^2$ where $x$ is in meters and $t$ is in seconds.

Step 2: Differentiate the position function to find velocity.
Velocity is the time derivative of position: $v(t) = \frac{dx}{dt} = \frac{d}{dt}(4t^2)$ Using the power rule of differentiation, $\frac{d}{dt}(t^n) = nt^{n-1}$, we get: $v(t) = 4 \cdot 2t^{2-1} = 8t$ So, the velocity as a function of time is: $v(t) = 8t \quad \text{ms}^{-1}$

Step 3: Substitute $t = 5$ s to find the velocity at that instant.
$v(5) = 8 \times 5 = 40 \quad \text{ms}^{-1}$

Step 4: Match with the given options.
The calculated velocity at $t = 5$ s is $40 \text{ ms}^{-1}$, which corresponds to option D.

Common Traps & Exam Tip:

Trap 1: Confusing distance with displacement.
Some students mistakenly think that $x = 4t^2$ represents displacement, but in this context, it's given as the distance travelled. However, since the motion is along a straight line and the function is increasing, distance and displacement are numerically equal. Still, it's crucial to recognize that velocity is the derivative of position (or displacement), not necessarily distance.

Trap 2: Forgetting to differentiate.
A common error is to directly substitute $t = 5$ into $x = 4t^2$ and assume that $x(5) = 100$ m is the velocity. This is incorrect because velocity is the rate of change of position, not the position itself.

Trap 3: Misapplying differentiation rules.
Students sometimes differentiate incorrectly, such as writing $\frac{d}{dt}(4t^2) = 4t$ or $8t^2$. Always apply the power rule carefully: $\frac{d}{dt}(t^n) = nt^{n-1}$.

Exam Tip:
Whenever you see a position-time function, immediately think: "Velocity is the derivative of position with respect to time." This is a fundamental concept in kinematics and appears frequently in JEE problems. Practice differentiating polynomial functions quickly and accurately.