JEE PYQ: Motion in a Straight Line - Question ID 23afef9ba3cb (JEE Main 2005)

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 the relation is given as $t = ax^2 + bx$, i.e., time is expressed as a function of position. To find acceleration, we must first express $x$ as a function of $t$, or use implicit differentiation to find $\frac{dx}{dt}$ and $\frac{d^2x}{dt^2}$.

Step 1: Differentiate the given relation implicitly with respect to time $t$.

Given: $t = a x^2 + b x$ Differentiate both sides with respect to $t$: $\frac{dt}{dt} = \frac{d}{dt}(a x^2 + b x)$ $1 = 2a x \frac{dx}{dt} + b \frac{dx}{dt}$ Factor out $\frac{dx}{dt}$: $1 = (2a x + b) \frac{dx}{dt}$

Step 2: Solve for velocity $v = \frac{dx}{dt}$.

$\frac{dx}{dt} = \frac{1}{2a x + b}$ So, $v = \frac{1}{2a x + b}$

Step 3: Differentiate velocity $v$ with respect to time $t$ to find acceleration $a$.

Acceleration is: $a = \frac{dv}{dt} = \frac{d}{dt}\left( \frac{1}{2a x + b} \right)$ Use the chain rule: $\frac{d}{dt}\left( \frac{1}{u} \right) = -\frac{1}{u^2} \frac{du}{dt}, \quad \text{where } u = 2a x + b$ So, $a = -\frac{1}{(2a x + b)^2} \cdot \frac{d}{dt}(2a x + b)$ $= -\frac{1}{(2a x + b)^2} \cdot 2a \frac{dx}{dt}$ But $\frac{dx}{dt} = v$, so: $a = -\frac{2a v}{(2a x + b)^2}$

Step 4: Express $2a x + b$ in terms of $v$.

From Step 2: $v = \frac{1}{2a x + b} \Rightarrow 2a x + b = \frac{1}{v}$ Substitute into the expression for acceleration: $a = -\frac{2a v}{\left( \frac{1}{v} \right)^2} = -\frac{2a v}{v^{-2}} = -2a v \cdot v^2 = -2a v^3$

Step 5: Final expression for acceleration.

Thus, the acceleration is: $a = -2a v^3$ This matches option D.

Students often make the following mistakes:

• Incorrect differentiation: Forgetting to use implicit differentiation when $t$ is given as a function of $x$. Instead, they try to solve for $x$ explicitly, which is messy and error-prone.
• Sign errors: Misplacing negative signs during differentiation, especially when using the chain rule on $\frac{1}{u}$.
• Confusing $a$ (acceleration) with $a$ (constant): The constant $a$ in the equation $t = a x^2 + b x$ is not the same as acceleration. Students sometimes substitute incorrectly or confuse the two.
• Not expressing acceleration in terms of $v$: The question asks for acceleration in terms of velocity $v$. Students may stop at an expression involving $x$, missing the final substitution.

Exam Tip: Always label variables clearly. Use $a_{\text{accel}}$ or $\alpha$ for acceleration if the problem uses $a$ as a constant. Also, practice implicit differentiation—it’s a powerful tool in kinematics when position and time relations are non-standard.