Question

If $x=5t+3t^2$ and $y=4t$ are the $x$ and $y$ co-ordinates of a particle at any time $t$ second where $x$ and $y$ are in metre, then the acceleration of the particle

• A. is zero throughout its motion
• B. is a constant throughout its motion
• C. depends only on its $y$ component
• D. varies along both $x$ and $y$ direction

Answer 1

Answer: (B)

As $x=5t+3t^2$ and $y=4t$
$\therefore v_x=\frac{dx}{dt}=\frac{d}{dt}\left(5t+3t^2\right)=5+6t$
$v_y=\frac{dy}{dt}=\frac{d}{dt}\left(4t\right)=4$
$a_x=\frac{d{v}_x}{dt}=\frac{d}{dt}\left(5+6t\right)=6$
and $a_y=\frac{d{v}_y}{dt}=\frac{d}{dt}\left(4\right)=0$
$\therefore$ The acceleration of the particle is
$a=\sqrt{a_x^2+a_y^2}=\sqrt{{\left(6\right)}^2+{\left(0\right)}^2}=6$
Thus, it is constant through its motion.