# A particle moves in the xy plane with constant acceleration ∂ directed along the negative direction of y axis .The equation of trajectory of the particle is y=ax2 _bx2,where a and b are constants. Find the velocity of the particle when it passess through the origin.

Vikas TU
14149 Points
8 years ago
and i think the eqn. is y = ax – bx^2

Let v(x) & v(y) be the velocities of the particle in the x and y directions respectively at any given time.

So
v(x) = dx / dt
v(y) = dy / dt

Let a(x) & a(y) be the acceleration of the particle in the x and y directions respectively at any given time.

So
a(x) = d v(x) / dt
a(y) = d v(y) / dt

But it is given that the acceleration a is directed along the -ve y axis.
So the acceleration along the x axis = 0
So
a(x) = 0
a(y) = -a

Consider y = ax - bx^2
Differentiate w.r.t. t
v(y) = a v(x) - 2bx v(x) ...............(1)

Again differentiate w.r.t. t
a(y) = a a(x) - 2b { [v(x)]^2 + x a(x) } .............(2)

Put a(x) = 0 and a(y) = -a in (2)
v(x) = sqrt. (a / 2b)

Put v(x) = a/2b in (1)
v(y) = a * sqrt.(a / 2b) - ax

These are the general expressions for v(x) and v(y).
Note that as x increases, v(x) remains constant as a(x) = 0 and v(y) decreases as a(y)

To find the components of velocity of the particle at the origin, put x = 0 and y = 0 in the expressions of v(x) and v(y).

So
v(x) (at x = 0) = sqrt.(a / 2b)
v(y) (at x = 0) = a * sqrt.(a / 2b)

Velocity of the particle at origin = sqrt.{v(x)^2 + v(y)^2}
= sqrt.{a(1 + a^2)/2b}