JEE PYQ: Motion in a Straight Line - Question ID edaed8a68625 (JEE Main 2021)

In kinematics, when the position \( x \) is given as a function of time \( t \), we can find velocity \( v \) and acceleration \( a \) (or retardation, which is negative acceleration) by successive differentiation.

• Velocity: \( v = \frac{dx}{dt} \). Since \( t \) is expressed in terms of \( x \), we use the chain rule: \( v = \frac{dx}{dt} = \frac{1}{\frac{dt}{dx}} \).
• Acceleration: \( a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx} \). Retardation is simply the magnitude of negative acceleration.

Given the relation \( t = m x^2 + n x \), we will differentiate it with respect to \( x \) to find \( \frac{dt}{dx} \), then proceed to find velocity and acceleration.

Step 1: Differentiate \( t \) with respect to \( x \)

Given: $t = m x^2 + n x$ Differentiate both sides with respect to \( x \): $\frac{dt}{dx} = 2 m x + n$

Step 2: Find velocity \( v \)

Velocity is the reciprocal of \( \frac{dt}{dx} \): $v = \frac{dx}{dt} = \frac{1}{\frac{dt}{dx}} = \frac{1}{2 m x + n}$

Step 3: Express \( x \) in terms of \( v \)

From the velocity expression: $v = \frac{1}{2 m x + n} \implies 2 m x + n = \frac{1}{v}$ Solve for \( x \): $2 m x = \frac{1}{v} - n \implies x = \frac{\frac{1}{v} - n}{2 m}$

Step 4: Find acceleration \( a \)

Acceleration is given by: $a = \frac{dv}{dt} = v \frac{dv}{dx}$ We need \( \frac{dv}{dx} \). Differentiate \( v \) with respect to \( x \): $v = \frac{1}{2 m x + n} = (2 m x + n)^{-1}$ $\frac{dv}{dx} = -1 \cdot (2 m x + n)^{-2} \cdot 2 m = -\frac{2 m}{(2 m x + n)^2}$ But \( 2 m x + n = \frac{1}{v} \), so: $\frac{dv}{dx} = -\frac{2 m}{\left(\frac{1}{v}\right)^2} = -2 m v^2$ Now, acceleration: $a = v \frac{dv}{dx} = v \cdot (-2 m v^2) = -2 m v^3$

Step 5: Interpret retardation

Retardation is the magnitude of negative acceleration: $\text{Retardation} = |a| = 2 m v^3$ This matches option A.

Students often make these mistakes:

• Incorrect differentiation: Forgetting to apply the chain rule when differentiating \( v \) with respect to \( x \), leading to wrong expressions for \( \frac{dv}{dx} \).
• Sign confusion: Misinterpreting retardation as acceleration instead of its magnitude. Retardation is always positive, so the negative sign in acceleration must be dropped.
• Algebraic errors: Mishandling the expression \( 2 m x + n = \frac{1}{v} \), especially when solving for \( x \) or substituting back.
• Option confusion: Overlooking the constants \( m \) and \( n \) in the options. The correct answer must include \( m \), not just \( n \).

Exam Tip: Always verify units. Retardation has units of acceleration (\( \text{m/s}^2 \)). The expression \( 2 m v^3 \) must yield \( \text{m/s}^2 \). Since \( v \) is in \( \text{m/s} \), \( v^3 \) is \( \text{m}^3/\text{s}^3 \). For the units to match, \( m \) must have units of \( \text{s}^2/\text{m}^2 \), which is consistent with the given relation \( t = m x^2 + n x \) (where \( m x^2 \) must have units of time).