Question

A particle moves along the parabolic path $x=y^2+2y+2$ in such a way that y-component of velocity vector remains $5$ $m{s}^{-1}$ during the motion. The magnitude of the acceleration of the particle is

• A. 50 $m{s}^{-2}$
• B. 100 $m{s}^{-2}$
• C. $10\sqrt{2}m{s}^{-2}$
• D. 0.1 $m{s}^{-2}$

Answer 1

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

Since velocity in y is constant, thus acceleration in y=0. Thus magnitude of acceleration is given by the x-component of acceleration.
Now, $x=y^2+2y+2$.
Differentiate with time to get,
$v_x=2y{v}_y+2v_y=10y+10$
Again differentiate with time to get,
$a_x=10v_y$
Since $v_y=5$, thus $a_x=50m/s^2$