Kinematics- Velocity at Origin

A particle moves in a the place x x - y y with constant acceleration a directed along the negative y y -axis. The equation of motion of the particle has the form y = p x q x 2 y=px-q{ x }^{ 2 } where p p and q q are positive constants. Then the velocity of the particle at the origin is?

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a ( p 2 + 1 ) 2 q \sqrt { \cfrac { a({ p }^{ 2 }+1) }{ 2q } } q ( p 2 + 1 ) 2 a \sqrt { \cfrac { q({ p }^{ 2 }+1) }{ 2a } } a ( p 2 1 ) 2 q \sqrt { \cfrac { a({ p }^{ 2 }-1) }{ 2q } } a ( q 2 + 1 ) 2 p \sqrt { \cfrac { a({ q }^{ 2 }+1) }{ 2p } }

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1 solution

Arjen Vreugdenhil
Sep 21, 2015

The general equations for motion with constant acceleration are { x = x 0 + v x , 0 t + 1 2 a x t 2 y = y 0 + v y , 0 t + 1 2 a y t 2 \left\{\begin{array}{l} x = x_0 + v_{x,0}t + \tfrac12a_xt^2 \\ y = y_0 + v_{y,0}t + \tfrac12a_yt^2 \end{array}\right. In this case we have x 0 = y 0 = 0 x_0 = y_0 = 0 , a x = 0 a_x = 0 , a y = a a_y = -a , so that { x = v x , 0 t y = v y , 0 t 1 2 a t 2 \left\{\begin{array}{l} x = v_{x,0}t \\ y = v_{y,0}t - \tfrac12at^2 \end{array}\right. Write t = x / v x , 0 t = x/v_{x,0} to eliminate t t in the second equation: y = v y , 0 v x , 0 x a 2 v x , 0 2 x 2 . y = \frac{v_{y,0}}{v_{x,0}}x - \frac{a}{2v_{x,0}^2}x^2. Compare with the given form y = p x q x 2 y = px - qx^2 to conclude p = v y , 0 v x , 0 , q = a 2 v x , 0 2 . p = \frac{v_{y,0}}{v_{x,0}}, q = \frac{a}{2v_{x,0}^2}. Solve for the initial velocity components: { v x , 0 = a 2 q ; v y , 0 = p v x , 0 . \left\{\begin{array}{l} v_{x,0} = \sqrt{\frac{a}{2q}}; \\ v_{y,0} = pv_{x,0}.\end{array}\right. The magnitude of the initial velocity is then v 0 = v x , 0 2 + v y , 0 2 = ( 1 + p 2 ) v x , 0 2 = ( 1 + p 2 ) a 2 q = a ( 1 + p 2 2 ) 2 q . v_0 = \sqrt{v_{x,0}^2 + v_{y,0}^2} = \sqrt{(1+p^2)v_{x,0}^2} = \sqrt{(1+p^2)\frac{a}{2q}} = \sqrt{\frac{a(1+p2^2)}{2q}}.

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