Question

The acceleration 'a' of a particle moving along
X-axis is given as a = 2x. Assume all Sl units. The
velocity-position (v-x) graph is best represented by​

Answer 1

Answer:

Given:

Acceleration vs Position equation is given as :

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \huge{ \sf{ \green{a = 2x}}}}$

To find:

Velocity vs Position graph

Concept:

Whenever we have an acceleration-position equation we can convert it into velocity -position terms in the following manner :

$\therefore \: a = 2x$

$= > \dfrac{dv}{dt} = 2x$

Applying chain rule :

$= > \dfrac{dv}{dx} \times \dfrac{dx}{dt} = 2x$

Now we know that dx/dt is velocity :

$= > \dfrac{dv}{dx} \times v = 2x$

Shifting differential terms to respective sides :

$= > v \: dv = 2x \: dx$

Integrating to the limits :

$= > \int v \: dv = 2 \int x \: dx$

$= > \dfrac{ {v}^{2} }{2} = 2 \times \dfrac{ {x}^{2} }{2}$

$= > v = \sqrt{2} \: x$

So final relationship is :

$\boxed{ \huge{ \sf{ \green{v = \sqrt{2} \: x }}}}$

Now look at the graph to understand better.