Question

please solve it Physical science experts​

Answer 1

Correct Question:-

The coordinates of a moving particle at any time $\sf{t}$ are given by $\sf{x=\alpha\,t^3}$ and $\sf{y=\beta\,t^3.}$ The speed of the particle at time $\sf{t}$ is given by,

$\sf{1)\quad 3t\sqrt{\alpha^2+\beta^2}\quad\quad\quad\quad2)\quad3t^2\sqrt{\alpha^2+\beta^2}}$

$\sf{3)\quad t^2\sqrt{\alpha^2+\beta^2}\quad\quad\quad\quad4)\quad\sqrt{\alpha^2+\beta^2}}$

Solution:-

The velocity of the particle along x direction is,

$\longrightarrow\sf{v_x=\dfrac{dx}{dt}}$

$\longrightarrow\sf{v_x=\dfrac{d}{dt}\,\left[\alpha\,t^3\right]}$

$\longrightarrow\sf{v_x=3\alpha\,t^2}$

The velocity of the particle along y direction is,

$\longrightarrow\sf{v_y=\dfrac{dy}{dt}}$

$\longrightarrow\sf{v_y=\dfrac{d}{dt}\,\left[\beta\,t^3\right]}$

$\longrightarrow\sf{v_x=3\beta\,t^2}$

Hence speed of the particle is,

$\longrightarrow\sf{v=\sqrt{(v_x)^2+(v_y)^2}}$

$\longrightarrow\sf{v=\sqrt{(3\alpha\,t^2)^2+(3\beta\,t^2)^2}}$

$\longrightarrow\sf{v=\sqrt{9\alpha^2\,t^4+9\beta^2\,t^4}}$

$\longrightarrow\sf{v=\sqrt{9t^4\left(\alpha^2+\beta^2\right)}}$

$\longrightarrow\underline{\underline{\sf{v=3t^2\sqrt{\alpha^2+\beta^2}}}}$

Hence (2) is the answer.