a particle is projected with velocity Vo along X-axis The deceleration on the particle is proportional to the square of the distance from the origin i.,e a=axsquare. the distance at which the particle stops is
a = dv/dt
a = (dv/dx) x (dx/dt)
Now, we know that dx/dt = v, so,
a = (dv/dx) . v
a.dx = v.dv
Here, ax^2 dx = v dv
On integrating both sides, we have,
(ax^3)/3 = (v^2)/2
x = ((3v^2)/2a))^1/3.
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