Question

If $ x = 5t + 3t^{2} $ and $ y = 4 t $ 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 $
$v_{y}=\frac{dy}{dt}=\frac{d}{dt}\left(4t\right)=4$
$a_{x}=\frac{dv_{x}}{dt}=\frac{d}{dt}\left(5+6t\right)=6$
and $a_{y}=\frac{dv_{y}}{dt}=\frac{d}{dt}\left(4\right)=0$
$\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.