Question

For a particle moving in x-y plane, if at any instant of time ‘t’ (in second) its displacements (in metre) are \(x=2t^{2}-t\) and \(y=4t^{2}-4t\), then the velocity of the particle at a time \(t=1\,s\) is:

• A. \(3\,m/s\)
• B. \(7\,m/s\)
• C. \(1\,m/s\)
• D. \(5\,m/s\)

Hint:

Velocity magnitude = \(\sqrt{v_x^2 + v_y^2}\) from component velocities.

Answer 1

The Correct Option is D

Velocity components: \[ v_x = \frac{dx}{dt} = \frac{d}{dt}(2t^2 - t) = 4t - 1 \] \[ v_y = \frac{dy}{dt} = \frac{d}{dt}(4t^2 - 4t) = 8t - 4 \] At \(t=1\): \[ v_x = 4(1) - 1 = 3\, m/s, v_y = 8(1) - 4 = 4\, m/s \] Magnitude of velocity: \[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5\, m/s \]