A particle moves along a parabolic path y = 9x^2 in such a way that the x component of velocity remains constant. If, the acceleration of the particle is 2j ms⁻² , find the x component of velocity.

Step-by-step solution

The equation of the path is given by y = 9x^2 . Differentiating with respect to time t , we get the y-component of velocity: v_y = dy dt = 18x dx dt = 18x v_x Differentiating again with respect to time t , we get the y-component of acceleration: a_y = dv_y dt = 18 ( dx dt v_x + x dv_x dt ) Since the x-component of velocity v_x is constant, its derivative dv_x dt = 0 . Substituting dx dt = v_x and dv_x dt = 0 , we get: a_y = 18 v_x^2 It is given that the acceleration of the particle is 2 j ms⁻² , which means a_y = 2