Question

The velocity-displacement (v-x) graph for a particle
moving along a straight line is shown in figure. The
relation between acceleration(a) and
displacement(x) is given by​

Answer 1

Answer:

solved

Explanation:

slope of the line = -Vo/Xo

equation of the line is

v = -Vo/Xo(x) + Vo

v =  -Vo/Xo(x) + Vo

a = dv/dt =  -Vo/Xo(dx/dt) + dVo/dt =  -Vo/Xo(dx/dt) + 0

a =  -Vo/Xo(dx/dt) =  -Vo/Xo(v) =  -Vo/Xo( -Vo/Xo(x) + Vo  )

a = (Vo/Xo)² x - Vo²/Xo

Answer 2

The relation between acceleration a and displacement x

$\boxed{a=\frac{v_0^2}{x_0^2}(x-x_0)}$

Explanation:

We know that rate of change of displacement is velocity and rate of change of velocity is acceleration

Thus,

$v=\frac{ds}{dt}$

And, $a=\frac{dv}{dt}$

or, a can be written as

$a=v\frac{dv}{dx}$

Where dv/dx is slope of the line $=-\frac{v_0}{x_0}$

Now for the graph shown in the figure, the equation of the straight line will be

$\frac{x}{x_0}+\frac{v}{v_0}=1$

Therefore, at any instant x, the velocity will be

$v=v_0(1-\frac{x}{x_0})$

Thus, at that instant the acceleration is

$a=v\frac{dv}{dx}$

$\implies a=v_0(1-\frac{x}{x_0})\times (-\frac{v_0}{x_0})$

$\implies a=-\frac{v_0^2(x_0-x)}{x_0^2}$

$\implies a=\frac{v_0^2}{x_0^2}(x-x_0)$

This is the relation between acceleration a and displacement x

Hope this answer is helpful.

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