Question

If x = 5t + 3$t^2$ 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

The correct option is B is a constant throughout its motion

$D=5t+3t^2$ , $y=4t$

$D,y$ are co-ordinates that double divides gives acceleration so,

$\frac{dD}{dt}=5+6t\frac{d^{2}D}{{dt}^2}=6\frac{dy}{dt}=4\frac{d^{2}y}{{dt}^2}=0$

So acceleration comes to be constant that is $6m/s^2$ (throughout motion)