JEE PYQ: Motion in a Straight Line - Question ID 2b08731dabf7 (JEE Main 2003)

To determine the speed of a particle moving in a plane, we use the following fundamental concepts from kinematics:

• Position Vector: The coordinates \( x(t) \) and \( y(t) \) describe the particle’s location at time \( t \).
• Velocity Components: The velocity of the particle is the time derivative of its position. The \( x \)-component of velocity is \( v_x = \frac{dx}{dt} \), and the \( y \)-component is \( v_y = \frac{dy}{dt} \).
• Speed as Magnitude of Velocity: The speed \( v \) is the magnitude of the velocity vector, given by: $v = \sqrt{v_x^2 + v_y^2}$

In this problem, the position coordinates are given as: $x = \alpha t^3, \quad y = \beta t^3$ We will compute the velocity components and then the speed.

Step 1: Compute the \( x \)-component of velocity (\( v_x \))

Given \( x = \alpha t^3 \), the \( x \)-component of velocity is: $v_x = \frac{dx}{dt} = \frac{d}{dt} (\alpha t^3) = 3 \alpha t^2$

Step 2: Compute the \( y \)-component of velocity (\( v_y \))

Given \( y = \beta t^3 \), the \( y \)-component of velocity is: $v_y = \frac{dy}{dt} = \frac{d}{dt} (\beta t^3) = 3 \beta t^2$

Step 3: Compute the speed \( v \) as the magnitude of the velocity vector

The speed is: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{(3 \alpha t^2)^2 + (3 \beta t^2)^2}$ Simplify the expression inside the square root: $v = \sqrt{9 \alpha^2 t^4 + 9 \beta^2 t^4} = \sqrt{9 t^4 (\alpha^2 + \beta^2)}$ Factor out the common term: $v = \sqrt{9 t^4} \sqrt{\alpha^2 + \beta^2} = 3 t^2 \sqrt{\alpha^2 + \beta^2}$

Step 4: Match with the given options

The derived expression for speed is: $v = 3 t^2 \sqrt{\alpha^2 + \beta^2}$ This matches Option B.

Students often make the following mistakes in this question:

• Incorrect Differentiation: Some students forget to apply the power rule correctly and compute \( \frac{dx}{dt} \) as \( \alpha t^2 \) instead of \( 3 \alpha t^2 \). This leads to an incorrect speed expression.
• Confusing Speed with Velocity: Speed is a scalar (magnitude of velocity), while velocity is a vector. Students sometimes stop at computing \( v_x \) and \( v_y \) and forget to take the magnitude.
• Algebraic Errors in Simplification: When simplifying \( \sqrt{(3 \alpha t^2)^2 + (3 \beta t^2)^2} \), students may incorrectly factor out terms, leading to expressions like \( 3 t \sqrt{\alpha^2 + \beta^2} \) (Option A) instead of \( 3 t^2 \sqrt{\alpha^2 + \beta^2} \).
• Ignoring the Time Dependence: Some students overlook the \( t^2 \) term and select Option D, which is independent of time. This is incorrect because the speed clearly depends on \( t \).

Exam Tip: Always double-check your differentiation and simplification steps. For problems involving parametric equations, ensure you compute both components of velocity before taking the magnitude. Also, verify the units of your final answer to catch dimensional inconsistencies.