A particle moves in the plane xy with constant acceleration a directed along the negative y-axis. The equaiton of motion of the particle has the form y = px - qx^(2) where p and q are positive constants. Find the velocity of the particle at the origin of corrdinates.

A particle moves in the plane xy with constant acceleration a directed

1.	A particle moves in the plane xy with constant acceleration a directed along the negative y-axis. The equaiton of motion of the particle has the form y = px - qx^(2) where p and q are positive constants. Find the velocity of the particle at the origin of corrdinates.
Answer» SOLUTION :GIVEN that `y = px - qx^(2)` `:. (dy)/(DX) = p (dx)/(DT)-q.2x(dx)/(dt)` and `(d^(2)y)/(dt^(2)) =p (d^(2)x)/(dt^(2)) -2qx(d^(2)x)/(dt^(2))-2q ((dx)/(dt))^(2)` or `-a =- 2q ((dx)/(dt))^(2) =-2qv_(x)^(2)` `:. (d^(2)x)/(dt^(2)) =0` and `(d^(2)y)/(dt^(2)) =-a` `:. v_(x)^(2) = (a)/(2q)` or `v_(x) = sqrt((a)/(2q))` Further, `((dy)/(dt))_(x=0) = p(dx)/(dt)` or `v_(p) =pv_(x) :. v_(p) = psqrt(((a)/(2q)))` Now `v - sqrt((v_(x)^(2)+v_(y)^(2))) =sqrt(((a)/(2q)+(ap^(2))/(2q)))rArr` `v=sqrt([(a(p^(2)+1))/(2q)])`