Question

At any time \( t \), the co-ordinates of moving particle are \( x = at^2 \) and \( y = bt^2 \). The speed of the particle is

• A. \( 2t\sqrt{a^2 + b^2} \)
• B. \( 2t\sqrt{a^2 - b^2} \)
• C. \( 2t(a + b) \)
• D. \( \frac{2t}{\sqrt{a^2+b^2}} \)

Hint:

Speed is the magnitude of the velocity vector: $|v| = \sqrt{\dot{x}^2 + \dot{y}^2}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The position of a particle in a two-dimensional plane is given by its \( x \) and \( y \) coordinates as functions of time \( t \).
The velocity vector \( \vec{v} \) has components \( v_x \) and \( v_y \), which are the time derivatives of the respective coordinates.
The speed of the particle is the magnitude of the velocity vector.
Step 2: Key Formula or Approach:
The velocity components are calculated as: \[ v_x = \frac{dx}{dt} \] \[ v_y = \frac{dy}{dt} \] The speed \( v \) is given by: \[ v = |\vec{v}| = \sqrt{v_x^2 + v_y^2} \] Step 3: Detailed Explanation:
Given the equations of motion: \[ x = at^2 \] \[ y = bt^2 \] First, we find the x-component of the velocity by differentiating \( x \) with respect to \( t \): \[ v_x = \frac{d}{dt}(at^2) = 2at \] Next, we find the y-component of the velocity by differentiating \( y \) with respect to \( t \): \[ v_y = \frac{d}{dt}(bt^2) = 2bt \] Now, we calculate the magnitude of the velocity (speed) using the components: \[ v = \sqrt{(2at)^2 + (2bt)^2} \] \[ v = \sqrt{4a^2t^2 + 4b^2t^2} \] We can factor out \( 4t^2 \) from the terms under the square root: \[ v = \sqrt{4t^2(a^2 + b^2)} \] Taking the square root of \( 4t^2 \), we get: \[ v = 2t\sqrt{a^2 + b^2} \] This matches option (A).
Step 4: Final Answer:
The speed of the particle is \( 2t\sqrt{a^2 + b^2} \).