JEE PYQ: Motion in a Straight Line - Question ID 779846aea1be (JEE Main 2024)

Core Formula & Concept:

In kinematics, when the position $x$ is given as a function of time $t$, we can find velocity $v$ and acceleration $a$ by successive differentiation:

• Velocity is the first derivative of position with respect to time: $v = \frac{dx}{dt}$.
• Acceleration is the first derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$.

However, in this problem, time $t$ is expressed as a function of position $x$, i.e., $t = \alpha x^2 + \beta x$. To find velocity and acceleration, we must use implicit differentiation with respect to time.

Key formulas used:

• $v = \frac{dx}{dt}$
• $a = \frac{dv}{dt} = v \frac{dv}{dx}$ (chain rule)

Step-by-Step Derivation:

Step 1: Express $dt/dx$

Given: $t = \alpha x^2 + \beta x$ Differentiate both sides with respect to $x$: $\frac{dt}{dx} = 2\alpha x + \beta$

Step 2: Find velocity $v = dx/dt$

Since $v = \frac{dx}{dt}$, we have: $v = \frac{1}{\frac{dt}{dx}} = \frac{1}{2\alpha x + \beta}$

Step 3: Express $x$ in terms of $v$

From above: $v = \frac{1}{2\alpha x + \beta} \implies 2\alpha x + \beta = \frac{1}{v}$ Solve for $x$: $2\alpha x = \frac{1}{v} - \beta \implies x = \frac{1 - \beta v}{2\alpha v}$

Step 4: Find acceleration $a = dv/dt$ using chain rule

We use: $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$ So, we need $\frac{dv}{dx}$.

From Step 2: $v = (2\alpha x + \beta)^{-1}$ Differentiate with respect to $x$: $\frac{dv}{dx} = -1 \cdot (2\alpha x + \beta)^{-2} \cdot 2\alpha = -\frac{2\alpha}{(2\alpha x + \beta)^2}$ But $2\alpha x + \beta = \frac{1}{v}$, so: $\frac{dv}{dx} = -\frac{2\alpha}{\left(\frac{1}{v}\right)^2} = -2\alpha v^2$

Step 5: Compute acceleration $a$

Now: $a = v \frac{dv}{dx} = v \cdot (-2\alpha v^2) = -2\alpha v^3$

Step 6: Match with given options

We have derived: $a = -2\alpha v^3$ This matches Option C.

Common Traps & Exam Tip:

Common Mistake 1: Students often try to differentiate $t$ with respect to $t$ directly, leading to confusion. Remember: $t$ is given as a function of $x$, so $dt/dx$ is the correct starting point.

Common Mistake 2: Forgetting to use the chain rule $a = v \frac{dv}{dx}$ and instead trying to differentiate $v$ with respect to $t$ directly, which is not straightforward here.

Exam Tip: When time is given as a function of position, always invert the relationship to find $dx/dt$, then proceed carefully using implicit differentiation. Keep track of powers of $v$ — the final expression must be in terms of $v$ and constants only.