JEE PYQ: Motion in a Straight Line - Question ID 6253f2c609ed (JEE Main 2019)

Core Formula & Concept:

The problem involves a particle moving along the positive x-axis with a speed that depends on its position: $v = b \sqrt{x}$. We are asked to find the speed of the particle at time $t = \tau$, given that it starts from the origin ($x = 0$) at $t = 0$.

Key concepts and formulas used:

• Relation between velocity, position, and time: Since velocity is the time derivative of position, we have $v = \frac{dx}{dt}$ Given $v = b \sqrt{x}$, this becomes a differential equation: $\frac{dx}{dt} = b \sqrt{x}$
• Separation of variables: This is a first-order ordinary differential equation (ODE) that can be solved by separating variables: $\frac{dx}{\sqrt{x}} = b \, dt$
• Integration: Integrate both sides to find $x(t)$, then differentiate to find $v(t)$, or directly find $v(t)$ using the relation between $v$ and $x$.

Step-by-Step Derivation:

Step 1: Write the given velocity-position relation

We are given: $v = b \sqrt{x}$ But $v = \frac{dx}{dt}$, so: $\frac{dx}{dt} = b \sqrt{x}$

Step 2: Separate variables and integrate

Rewrite the equation: $\frac{dx}{\sqrt{x}} = b \, dt$ Integrate both sides. The left side integrates with respect to $x$, the right with respect to $t$: $\int \frac{dx}{\sqrt{x}} = \int b \, dt$ We know: $\int x^{-1/2} dx = 2 x^{1/2} + C_1$ $\int b \, dt = b t + C_2$ So: $2 \sqrt{x} = b t + C$

Step 3: Apply initial condition to find constant

At $t = 0$, $x = 0$: $2 \sqrt{0} = b \cdot 0 + C \Rightarrow C = 0$ Thus: $2 \sqrt{x} = b t$ $\Rightarrow \sqrt{x} = \frac{b t}{2}$ $\Rightarrow x = \left( \frac{b t}{2} \right)^2 = \frac{b^2 t^2}{4}$

Step 4: Find velocity as a function of time

Recall $v = b \sqrt{x}$. Substitute $\sqrt{x} = \frac{b t}{2}$: $v = b \cdot \frac{b t}{2} = \frac{b^2 t}{2}$

Step 5: Evaluate velocity at $t = \tau$

$v(\tau) = \frac{b^2 \tau}{2}$

This matches option C.

Common Traps & Exam Tip:

Trap 1: Forgetting to apply initial conditions. Many students integrate but forget to use $x = 0$ at $t = 0$, leading to an incorrect constant of integration and wrong final expression.

Trap 2: Misinterpreting the velocity expression. Some students confuse $v = b \sqrt{x}$ as a function of time directly and try to integrate $v$ over time without first relating $x$ and $t$. This leads to incorrect results.

Trap 3: Incorrect integration of $1/\sqrt{x}$. A common mistake is to integrate $1/\sqrt{x}$ as $\ln|\sqrt{x}|$ instead of $2\sqrt{x}$. Remember: $\int x^n dx = \frac{x^{n+1}}{n+1} \quad \text{for } n \neq -1$ Here $n = -1/2$, so: $\int x^{-1/2} dx = \frac{x^{1/2}}{1/2} = 2 \sqrt{x}$

Exam Tip: Always write down the relation $v = dx/dt$ and substitute the given expression. Then separate variables and integrate carefully. Double-check integration limits and initial conditions. In such problems, the final answer is often a simple function of time, so if your expression looks overly complex, revisit your steps.

Final Answer: Option C: $\frac{b^2 \tau}{2}$