Question

a particle is moving in xy plane such that vx=4+4t and vy=4t if the initial position of the particle is 1,2 then the equation of trajectory will be

Answer 1

Answer:

$x^2+y^2-2xy+2x-10y+17=0$

Explanation:

We know that rate of change of displacement is veloctiy in that direction

Therefore in x direction

$\frac{dx}{dt} =v_x$

And in y direction

$\frac{dy}{dt} =v_y$

Given

The x component of the velocity

$v_x=4+4t$

$\implies \frac{dx}{dt} =4+4t$

$\implies dx =(4+4t)dt$

$\implies \int dx =\int (4+4t)dt$

$\implies x=4t+2t^2+c$     where c is constant

Given at t=0, x=1

∴ $1=0+c$

$\implies c=1$

∴ $x=4t+2t^2+1$            ........... (1)

Similarly

$v_y=4t$

$\implies \frac{dy}{dt} =4t$

$\implies \int dy =\int (4t)dt$

$\implies y=2t^2+c'$

Given at t=0, y=2

∴ $2=0+c'$

$\implies c'=2$

$\implies y=2t^2+2$            ........... (2)

Subtracting eq (2) from eq (1)

$x-y=4t-1$

$\implies t=(x-y+1)/4$

Putting the value of t in eq (2)

$y-2=2\frac{(x-y+1)^2}{16}$

$\implies 8y-16={(x-y+1)^2}$

$\implies 8y-16=x^2+y^2+1-2xy+2x-2y\\\implies x^2+y^2-2xy+2x-10y+17=0$

Thus is the required equation of the trajectory