A particle is moving on a linear pathway whose acceleration varies as $a = 2 x$ $a=2x$ (where, x is its distance covered) If it had started from rest(zero velocity when x=0),find an equation to suggest its velocity-displacement relationship? How to solve? Pls help

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Answer 1 · 2018-03-05T15:07:08

As the particle is moving in a linear pathway,we can say its distance covered will be equals to displacement,also because it only accelerates.

We know,accleration = rate of change in velocity = $\frac{d v}{d t}$ $(dv)/(dt)$

So,we can write, $\frac{d v}{d t} = 2 x$ $(dv)/(dt) =2x$

or, $\frac{d v}{d x} \frac{d x}{d t} = v \frac{d v}{d x} = 2 x$ $(dv)/(dx) (dx)/(dt) =v(dv)/(dx)=2x$ (as, velocity = rate of change in velocity)

So,we can write,

$v d v = 2 x d x$ $vdv =2xdx$

or, $\int_{0}^{V} v d v = 2 \int_{0}^{x} x d x$ $int_0^V vdv =2 int_0^x xdx$ (as, at $x = 0, v = 0$ $x=0,v=0$ )

so, $\frac{V^{2}}{2} = 2 (\frac{x^{2}}{2}) = x^{2}$ $V^2/2 = 2 (x^2/2)=x^2$

or, $V^{2} = 2 x^{2}$ $V^2 =2x^2$

or, $V = \sqrt{2} x$ $V=sqrt(2) x$

this is the required velocity displacement relationship.

A particle is moving on a linear pathway whose acceleration varies as a=2xa=2x(where, x is its distance covered) If it had started from rest(zero velocity when x=0),find an equation to suggest its velocity-displacement relationship? How to solve? Pls help