JEE PYQ: Motion in a Straight Line - Question ID fbe579312d0c (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 velocity of the particle at any instant is defined as the first derivative of the position with respect to time:

$v(t) = \frac{dx}{dt}$

This formula gives the instantaneous velocity at time $t$. The magnitude of velocity is simply the absolute value of $v(t)$, i.e., $|v(t)|$, since velocity is a vector quantity and its magnitude represents speed.

In this problem, the position function is: $x(t) = 5t^2 - 4t + 5 \quad \text{(in meters)}$ We are to find the magnitude of velocity at $t = 2$ seconds.

--- Step-by-Step Derivation:

Step 1: Differentiate the position function to find velocity

Given: $x(t) = 5t^2 - 4t + 5$ We compute the derivative of $x(t)$ with respect to $t$: $v(t) = \frac{dx}{dt} = \frac{d}{dt}(5t^2) - \frac{d}{dt}(4t) + \frac{d}{dt}(5)$ Using basic differentiation rules: - $\frac{d}{dt}(t^n) = n t^{n-1}$ - Derivative of a constant is zero

So, $v(t) = 5 \cdot 2t^{2-1} - 4 \cdot 1 + 0 = 10t - 4$

Step 2: Evaluate velocity at $t = 2$ seconds

Substitute $t = 2$ into $v(t)$: $v(2) = 10(2) - 4 = 20 - 4 = 16 \text{ m/s}$

Step 3: Determine the magnitude of velocity

Since velocity is a scalar in one-dimensional motion (along a straight line), its magnitude is the absolute value of $v(t)$. Here, $v(2) = 16$ m/s, which is positive, so: $|v(2)| = 16 \text{ m/s}$

Step 4: Match with given options

The magnitude of velocity at $t = 2$ s is $16 \text{ m/s}$, which corresponds to Option B. --- Common Traps & Exam Tip:

Trap 1: Forgetting to differentiate – Some students mistakenly plug $t = 2$ directly into the position function and think that $x(2)$ gives velocity. But velocity is not position; it's the rate of change of position. Always differentiate first.

Trap 2: Misapplying differentiation rules – Errors like $\frac{d}{dt}(5t^2) = 5t$ (forgetting to multiply by 2) or $\frac{d}{dt}(-4t) = -4t$ (forgetting the derivative of $t$ is 1) are common. Double-check each term.

Trap 3: Ignoring the sign of velocity – The question asks for the magnitude of velocity, not the velocity itself. Even if $v(t)$ were negative, the magnitude would be positive. However, in this case, $v(2) = 16$ m/s is already positive, so no issue arises.

Exam Tip: Always write down the velocity function explicitly before substituting values. This helps avoid calculation errors and makes your reasoning clear to examiners.

Final Answer: B: $16 \text{ m/s}$