Question

The coordinates of a particle moving in x-y plane at any instant of time \(t\) is \(x = 4t^2\); \(y = 3t^2\). The speed of the particle at that instant is:

• A. \(10 t\)
• B. \(5 t\)
• C. \(3 t\)
• D. \(2 t\)

Hint:

Notice that \(x \propto t^2\) and \(y \propto t^2\). This means the particle is undergoing constant acceleration in both the x and y directions. Also, by dividing the equations we get \(\frac{y}{x} = \frac{3}{4}\), indicating the particle travels in a straight line trajectory.

Answer 1

The Correct Option is A

Concept: The velocity vector of a particle in a 2D plane is the time derivative of its position vector. The components of velocity are \(v_x = \frac{dx}{dt}\) and \(v_y = \frac{dy}{dt}\). The speed of the particle is the magnitude of the resultant velocity vector, given by \(v = \sqrt{v_x^2 + v_y^2}\).

Step 1: Finding the x and y components of velocity.
Given the coordinates: \[ x = 4t^2 \] \[ y = 3t^2 \] Differentiate with respect to time \(t\) to find velocity components: \[ v_x = \frac{dx}{dt} = \frac{d}{dt}(4t^2) = 8t \] \[ v_y = \frac{dy}{dt} = \frac{d}{dt}(3t^2) = 6t \]

Step 2: Calculating the resultant speed.
The magnitude of velocity (speed) is: \[ v = \sqrt{v_x^2 + v_y^2} \] Substitute the values of \(v_x\) and \(v_y\): \[ v = \sqrt{(8t)^2 + (6t)^2} \] \[ v = \sqrt{64t^2 + 36t^2} \] \[ v = \sqrt{100t^2} \] \[ v = 10t \]

Step 3: Selecting the correct option.
From the calculation above, the speed is: \[ \boxed{10 t} \] Therefore, the correct option is: \[ \boxed{\text{Option (A)}} \]