A particle moves in the xy-plane with constant acceleration `a` directed along the negative y-axis. The equation of path of the particle has the form `y= bx - cx^2`, where b and c are positive constants. Find the velocity of the particle at the origin of coordinates.

A particle moves in the xy-plane with constant acceleration `a` direct

1.	A particle moves in the xy-plane with constant acceleration `a` directed along the negative y-axis. The equation of path of the particle has the form `y= bx - cx^2`, where b and c are positive constants. Find the velocity of the particle at the origin of coordinates.
Answer» Correct Answer - `v_0=sqrt((1+a^2)w//2b)`. According to the problem `vecw=w(-vecj)` So, `w_x=(dv_x)/(dt)=0` and `w_y=(dv_y)/(dt)=-w` (1) Differentiating Eq. of trajectory, `y=ax-bx^2`, with respect to time `(dy)/(dt)=(adx)/(dt)-2bx(dx)/(dt)` (2) `(dy)/(dt):\|:_(x=0)=a(dx)/(dt):\|_(x=0)` Again differentiating with respect to time `(d^2y)/(dt^2)=(ad^2x)/(dt^2)-2b((dx)/(dt))^2-2bx(d^2x)/(dt^2)` or, `-w=a(0)-2b((dx)/(dt))^2-2bx(0)` (using 1) or, `(dx)/(dt)=sqrt((w)/(2b))` (using 1) Using (3) in (2) `(dy)/(dt):\|_(x=0)=asqrt((w)/(2b))` Hence, the velocity of the particle at the origin `v=sqrt(((dx)/(dt))_(x=0)^(2)+((dy)/(dt))_(x=0)^2)=sqrt((w)/(2b)+a^2(w)/(2b))` (using Eqns (3) and (4)) Hence, `v=sqrt((w)/(2b)(1+a^2))`

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A particle moves in the xy-plane with constant acceleration `a` directed along the negative y-axis. The equation of path of the particle has the form `y= bx - cx^2`, where b and c are positive constants. Find the velocity of the particle at the origin of coordinates.

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