MHTCET 22 Apr 2025 Shift 2 Paper

Concept:Speed is the magnitude of the velocity vector; it is calculated from the time derivatives of the given coordinate functions.Explanation:At any instant, the velocity of a particle is the rate of change of its position with time.Since the position coordinates are given as functions of time, the velocity components are found by differentiating each coordinate with respect to $t$.The $x$-component of velocity is:$$v_x = \frac{dx}{dt} = \frac{d}{dt}(a t^2) = 2 a t$$This represents the horizontal component of the particle's velocity.The $y$-component of velocity is:$$v_y = \frac{dy}{dt} = \frac{d}{dt}(b t^2) = 2 b t$$This represents the vertical component of the particle's velocity, perpendicular to the horizontal one.The speed is the magnitude of the total velocity vector:$$v = \sqrt{v_x^2 + v_y^2}$$Substituting the two components into this expression:$$v = \sqrt{(2at)^2 + (2bt)^2} = \sqrt{4a^2 t^2 + 4b^2 t^2}$$Taking $4t^2$ out of the square root:$$v = 2t \sqrt{a^2 + b^2}$$Thus, the speed of the particle is $2t \sqrt{a^2 + b^2}$.This matches option A.Answer:$2 t \sqrt{a^{2}+b^{2}}$ (Option A)