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Physics · MOTION IN A PLANE
हिन्दीThe position coordinates of a particle moving in the \( x\text{-}y \) plane are \( x(t) = 2t^3 - 3t^2 \) and \( y(t) = t^3 - 12t \) (where \( x, y \) are in metres and \( t \) in seconds). At what non-zero time \( t \) is the \( y \)-component of velocity equal to zero, and what is the speed of the particle at that instant?
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Correct answer · Option C
\( t = 2\text{ s} \), speed \( = 12\text{ m/s} \)
Explanation
\( v_y = \frac{dy}{dt} = 3t^2 - 12 = 0 \implies t = 2\text{ s} \). At \( t = 2\text{ s} \), \( v_x = \frac{dx}{dt} = 6t^2 - 6t = 6(4) - 12 = 12\text{ m/s} \). Since \( v_y = 0 \), speed \( v = \sqrt{v_x^2 + v_y^2} = 12\text{ m/s} \).
