Question

A particle moves in the plane x y with constant acceleration a directed along the negative y-axis. The equation of motion of the particle has the form y = k_(1)x-k_(2)x^(2), where k_(1) and k_(2) are positive constants. Find the velocity of the particle at the origin of coordinates.

Answer 1

Correct Question

A particle moves in the plane x y with constant acceleration a directed along the negative y-axis. The equation of motion of the particle has the form y = mx-nx², where m and n are positive constants. Find the velocity of the particle at the origin of coordinates.

Solution

Velocity of the particle at origin is $\sf v_x \sqrt{1 + m}$

Given

The position vector of the particle along y axis is given as

$\sf \: y = mx - {nx}^{2}$

Differentiating y w.r.t to y,we get velocity of the particle along y axis :

$\sf \: v_y = \dfrac{dy}{dt} \\ \\ \longrightarrow \: \sf \: v_y = \dfrac{d(mx - {nx}^{2} )}{dt} \\ \\ \longrightarrow \: \sf v_y = \dfrac{d(mx)}{dt} - \dfrac{d( {nx}^{2}) }{dt} \\ \\ \longrightarrow \: \sf \: v_y = m \times \frac{dx}{dt} - 2nx \dfrac{dx}{dt} \\ \\ \longrightarrow \: \sf \: v_y = mv_x - 2nxv_x$

If the particle is at origin,the position coordinates would be (0,0)

Thus,

$\longrightarrow\: \boxed{ \boxed{ \sf v_y = mv_x}}$

Now,

Velocity at the origin would be given by :

$\sf \: v = \sqrt{ {v_x}^{2} + {v_y}^{2} } \\ \\ \dashrightarrow \: \sf \: v = \sqrt{ {v_x}^{2} + {mv_x}^{2} } \\ \\ \dashrightarrow \: \sf \: v = \sqrt{ {v_x}^{2}(1 + m) } \\ \\ \dashrightarrow \: \boxed{ \boxed{ \sf v = v_x \sqrt{1 + m} }}$