Q. The co-ordinates of a moving particle at time $t$ are given by $x = ct^{2}$ and $y = bt^{2}.$ The instantaneous speed of the particle is
A) $2t(b + c)$
B) $2t(b + c)^{1/2}$
C) $2t(c^{2} - b^{2})$
D) $2t\left( c^{2} + b^{2} \right)^{1/2}$
Explanation
(d) $v_{x} = \frac{dx}{dt} = 2ct$ and $v_{y} = \frac{dy}{dt} = 2bt$
$\therefore v = \sqrt{v_{x}^{2} + v_{y}^{2}} = 2t\left( c^{2} + b^{2} \right)^{1/2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.