Question

A particle moves along a parabolic path $x=y^2+y+1$ in such a way that the y-component of the velocity vector remain $4m{s}^{-1}$ during the motion. The magnitude of the acceleration of the prtile is

• A. $4m{s}^{-2}$
• B. $m{s}^{-2}$
• C. $32m{s}^{-2}$
• D. $50m{s}^2$

Answer 1

The correct option is D $50m{s}^2$

Let in $x$ and $y$ the equations be,

$x=t$ and $y^2+2y+2=t$

Now $y$ component of velocity vector is :

$2ydy/dt+2dy/dt+0=1$

$dy/dt=1/(2y+2)=5$

On solving we get,

$10y+10=1$

$y=-9/10=0.9$

acceleration $=-0.5/{(y+1)}^2$

on putting $y=0.9$, we get

acceleration $=50$ $m/s^2$