Question

A particle starts its journey from rest have uniform
acceleration a = 3i+4j Then the equation of the
trajectory of the particle is

Answer 1

First consider the horizontal motion of the particle.

• Initial velocity, $\sf{u_x=0.}$

• Acceleration, $\sf{a_x=3}$

By second equation of motion, the displacement,

$\sf{\longrightarrow s_x=u_xt+\dfrac{1}{2}\,a_xt^2}$

$\sf{\longrightarrow x=\dfrac{3}{2}t^2}$

$\sf{\longrightarrow t^2=\dfrac{2}{3}x\quad\quad\dots(1)}$

Now consider the vertical motion of the particle.

• Initial velocity, $\sf{u_y=0.}$

• Acceleration, $\sf{a_y=4}$

By second equation of motion, the displacement,

$\sf{\longrightarrow s_y=u_yt+\dfrac{1}{2}\,a_yt^2}$

$\sf{\longrightarrow y=2t^2}$

From (1),

$\sf{\longrightarrow\underline{\underline{y=\dfrac{4}{3}x}}}$

This is the equation of trajectory of the particle. It is a straight line passing through origin and having slope $\sf{\dfrac{4}{3}.}$