JEE PYQ: Motion in a Straight Line - Question ID f03f124a3067 (JEE Main 2022)

Core Formula & Concept:

In kinematics, the velocity of an object moving along a straight line is the first derivative of its position function with respect to time. If the position of a particle is given by $x(t)$, then its instantaneous velocity $v(t)$ is: $v(t) = \frac{dx}{dt}$

In this problem, we are given the position functions of two buses, $X_P(t)$ and $X_Q(t)$. To find the time $t$ at which both buses have the same velocity, we must:

1. Differentiate each position function to obtain the velocity functions.
2. Set the two velocity expressions equal to each other.
3. Solve the resulting equation for $t$.

Step-by-Step Derivation:

Step 1: Write the given position functions

$X_P(t) = \alpha t + \beta t^2$ $X_Q(t) = f t - t^2$

Step 2: Compute the velocity of each bus

Velocity is the time derivative of position: $v_P(t) = \frac{dX_P}{dt} = \frac{d}{dt}(\alpha t + \beta t^2) = \alpha + 2\beta t$ $v_Q(t) = \frac{dX_Q}{dt} = \frac{d}{dt}(f t - t^2) = f - 2t$

Step 3: Set the velocities equal and solve for $t$

We want the time $t$ when $v_P(t) = v_Q(t)$: $\alpha + 2\beta t = f - 2t$

Step 4: Rearrange the equation

Bring all terms involving $t$ to one side and constants to the other: $2\beta t + 2t = f - \alpha$ Factor out $t$ on the left: $t(2\beta + 2) = f - \alpha$ Simplify the coefficient of $t$: $t \cdot 2(\beta + 1) = f - \alpha$

Step 5: Solve for $t$

Divide both sides by $2(\beta + 1)$: $t = \frac{f - \alpha}{2(1 + \beta)}$

Step 6: Match with the given options

The derived expression matches option D: $\boxed{D}$ Common Traps & Exam Tip:

Trap 1: Misidentifying the derivative
Students often forget that velocity is the first derivative of position, not the second. Taking the second derivative would give acceleration, which is irrelevant here.

Trap 2: Sign errors in differentiation
In $X_Q(t) = f t - t^2$, the derivative of $-t^2$ is $-2t$. A common mistake is to write $+2t$, leading to an incorrect velocity expression.

Trap 3: Algebraic rearrangement errors
When rearranging $\alpha + 2\beta t = f - 2t$, students sometimes incorrectly group terms, such as $2\beta t - 2t = f + \alpha$, which flips signs and leads to the wrong answer.

Exam Tip:
Always double-check the differentiation step and ensure that signs are preserved. After solving, plug the derived $t$ back into both velocity expressions to verify equality.