Question

A particle starts from the origin at t=0 and moves in the xy -plane with constant acceleration a0 j_cap .Its equation of motion is y=3x^(2) .The component of velocity in x direction is​

Answer 1

The equation of motion of the particle is,

$\sf{\longrightarrow y=3x^2}$

Differentiating with respect to time,

$\sf{\longrightarrow \dfrac{dy}{dt}=\dfrac{d}{dt}(3x^2)}$

$\sf{\longrightarrow \dfrac{dy}{dt}=3\cdot2x\cdot\dfrac{dx}{dt}}$

Here $\sf{\dfrac{dy}{dt}=v_y}$ is vertical velocity and $\sf{\dfrac{dx}{dt}=v_x}$ is horizontal velocity of the particle. We have to find $\sf{v_x.}$

Then,

$\sf{\longrightarrow v_y=6x\cdot v_x}$

$\sf{\longrightarrow v_x=\dfrac{v_y}{6x}\quad\quad\dots(1)}$

Here the particle has constant acceleration $\sf{a_0\ \hat j,}$ whose direction shows that the acceleration is acting to the particle along y direction.

So it is the vertical acceleration of the particle.

Here the particles starts from rest from origin. So its initial velocity is zero.

By third equation of motion, the vertical velocity after a vertical displacement $\sf{y}$ is given by,

$\sf{\longrightarrow (v_y)^2=0^2+2a_0y}$

$\sf{\longrightarrow v_y=\sqrt{2a_0y}}$

Therefore (1) becomes,

$\sf{\longrightarrow\underline{\underline{v_x=\dfrac{\sqrt{2a_0y}}{6x}}}}$