Question

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

Answer 1

Final Answer : $50 m/s^2$

Steps:
1) Y - component of velocity is constant.
$a_y = 0$

2) Trajectory of Particle :
$x = ( {y}^{2} + 2y + 2) \\ = > \frac{dx}{dt} = 2y \frac{dy}{dt} + 2 \frac{dy}{dt} + 0 \\ = > v_x = 2yv_y +2v_y\\ => v_x = 2y*5+2*5\\ => v_x = 10y +10\\ => a_x = 10v_y +0 \\ => a_x = 10*5 \\ => a_x =50 m/s^2$

Hence, Net acceleration is
$a_{net} = a_x = 50m/s^2$

Answer 2

Given data states a parabolic path equation as $x = y^2 + 2y + 2$, and the component along y axis is given constant. Thereby the acceleration is given by differentiation of the velocity so, the acceleration in y axis will be zero, as differentiation of constant is zero. Now, for the x axis, we have,

           $\frac { dx }{ dt } =\frac { d(y^{ 2 }+2y+2) }{ dt } \\\Rightarrow v_{ x }=2yv_{ y }+2v_{ y } \\\Rightarrow v_{ x }=2y\times 5+2\times 5 \\\Rightarrow v_{ x }=10y+10.$

And,

            $a_{ x }=\frac { dv_{ x } }{ dt } \\\Rightarrow a_{ x }=10v_{ y }+0\\\Rightarrow a_{ x }=10\times 5\\\Rightarrow a_{ x }=50\frac { m }{ { s }^{ 2 } }$