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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=ax 2 _bx 2 ,where a and b are constants. Find the velocity of the particle when it passess through the origin.

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.

Grade:12th pass

1 Answers

Vikas TU
14149 Points
7 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 
(Minus sign indicates that it is along the -ve y axis) 

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} 

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