Step 1: Understanding the Concept:
When the position of a particle in a 2D plane is given by two separate functions for \( x \) and \( y \) coordinates in terms of time \( t \), we use vector kinematics to solve the problem.
The instantaneous velocity along any axis is the first derivative of the position coordinate along that axis with respect to time.
Velocity is a vector quantity having two components in this case: \( v_x \) and \( v_y \).
Speed is the magnitude of the velocity vector at that specific instant.
Unlike velocity, speed is a scalar quantity and is always non-negative.
Step 2: Key Formula or Approach:
Horizontal component of velocity: \( v_x = \frac{dx}{dt} \).
Vertical component of velocity: \( v_y = \frac{dy}{dt} \).
Resultant speed: \( v = \sqrt{v_x^2 + v_y^2} \).
Step 3: Detailed Explanation:
Given the equations for coordinates:
\( x = 4t^2 \)
\( y = 3t^2 \)
First, find the \( x \)-component of velocity by differentiating \( x \) with respect to \( t \):
\[ v_x = \frac{d}{dt}(4t^2) = 4(2t) = 8t \]
Next, find the \( y \)-component of velocity by differentiating \( y \) with respect to \( t \):
\[ v_y = \frac{d}{dt}(3t^2) = 3(2t) = 6t \]
Now, calculate the instantaneous speed \( v \) using the Pythagorean theorem for the components:
\[ v = \sqrt{(8t)^2 + (6t)^2} \]
Expand the squares:
\[ v = \sqrt{64t^2 + 36t^2} \]
Combine the like terms inside the square root:
\[ v = \sqrt{100t^2} \]
Take the square root of both terms:
\[ v = 10t \]
The speed depends linearly on time \( t \), implying the particle is undergoing constant acceleration.
The path of the particle is a straight line passing through the origin because \( y/x = 3/4 \), which is a constant.
Step 4: Final Answer:
The speed of the particle at any instant \( t \) is given by \( 10t \).
The correct option is (A).