Question

if the speed-position relation of a particle moving in a straight line is given asV square is equal to a minus b x square where A and B are constant then the acceleration of the particle is proportional to ​

Answer 1

Answer:

The acceleration of the particle is proportional to ​x

Explanation:

The speed-position relation of a particle moving in a straight line is given asV square is equal to a minus b x square where A and B are constant.

The equation is given as:

V^2= a- bx^2 [ where A and B are constant]............ (i)

The acceleration of the particle is a

a = Vdv/dx

Derivation both side of (i) we get

2V dv/dx =-2 bx

V dv/dx = -bx

putting the value of a

a = -bx

Therefore a ∝ x

Answer 2

Answer:

Acceleration, $a=-bx$

Explanation:

The relation between the speed and the position is given by :

$v^2=a-bx^2$..................(1)

Where

a and b are constant

We know that the acceleration of an object is given by :

$a=\dfrac{dv}{dt}$

And velocity of an object is given by :

$v=\dfrac{dx}{dt}$

Differentiating equation (1) wrt t we get :

$2v\ \dfrac{dv}{dt}=-bx.\dfrac{dx}{dt}$

$2v\ a=-2bx\ v$

$a=-bx$

It is clear that, acceleration is directly proportional to the position of an object i.e. x, $a\propto -x$