JEE PYQ: Motion in a Straight Line - Question ID de5a28d2d9f5 (JEE Main 2006)

In kinematics, when a particle moves along a straight line, its velocity $v$ is defined as the time derivative of its displacement $x$: $v = \frac{dx}{dt}.$ The problem gives the velocity as a function of position: $v = \alpha \sqrt{x},$ where $\alpha$ is a positive constant. Our goal is to find how the displacement $x$ varies with time $t$, starting from $x = 0$ at $t = 0$.

Step 1: Express the differential equation
Given $v = \frac{dx}{dt} = \alpha \sqrt{x}$, we rewrite this as: $\frac{dx}{dt} = \alpha x^{1/2}.$ This is a separable first-order ordinary differential equation.

Step 2: Separate variables
Rearrange the equation to isolate $x$ and $t$: $\frac{dx}{x^{1/2}} = \alpha \, dt.$

Step 3: Integrate both sides
Integrate the left side with respect to $x$ and the right side with respect to $t$: $\int x^{-1/2} \, dx = \int \alpha \, dt.$ The integrals evaluate to: $2 x^{1/2} = \alpha t + C,$ where $C$ is the constant of integration.

Step 4: Apply the initial condition
At $t = 0$, $x = 0$. Substituting these values: $2 \cdot 0^{1/2} = \alpha \cdot 0 + C \implies C = 0.$ Thus, the equation simplifies to: $2 x^{1/2} = \alpha t.$

Step 5: Solve for $x$
Square both sides to eliminate the square root: $(2 x^{1/2})^2 = (\alpha t)^2 \implies 4 x = \alpha^2 t^2.$ Therefore, $x = \frac{\alpha^2}{4} t^2.$ This shows that the displacement $x$ varies as $t^2$.

Students often make the following mistakes:

• Incorrect separation of variables: Forgetting to separate $dx$ and $dt$ properly, leading to incorrect integration.
• Ignoring the initial condition: Failing to apply $x = 0$ at $t = 0$, which results in an incorrect constant of integration.
• Misinterpreting the velocity expression: Assuming $v = \alpha \sqrt{x}$ implies constant acceleration, which is not the case here. The acceleration is not constant.
• Algebraic errors in solving for $x$: Squaring both sides incorrectly or forgetting to divide by the coefficient of $x$.