The $x$ and $y$ coordinates of a particle at any time are $x=3 t^{2}$ and $y=4 t^{2}$ respectively,

Question

The $x$ and $y$ coordinates of a particle at any time are $x=3 t^{2}$ and $y=4 t^{2}$ respectively, where $x$ and $y$ are in metres and $t$ in seconds. The magnitude of the acceleration of the particle is

Professor PC Thomas Classes · Answer

(a) : $x=3t^{2} \implies v_x = \frac{dx}{dt} = 6t \implies a_x = \frac{dv_x}{dt} = 6 \mathrm{~m} \mathrm{~s}^{-2}$. \ $y=4t^{2} \implies v_y = \frac{dy}{dt} = 8t \implies a_y = \frac{dv_y}{dt} = 8 \mathrm{~m} \mathrm{~s}^{-2}$. \ Magnitude of acceleration $|\vec{a}| = \sqrt{a_x^2 + a_y^2} = \sqrt{6^2 + 8^2} = 10 \mathrm{~m} \mathrm{~s}^{-2}$.