Question

If velocity v of a particle at a distance x from origin is given as v=√(a^2-x^2)/√(x^2), then it's acceleration is
proportional to
(1)1/x
(2)1/x^2
(3)1/x^3 (4)1/x^4​

Answer 1

Answer:

3

Explanation:

Tell me if i am correct or not.

Answer 2

Answer:

Acceleration will be proportional to 1/x^3.

Explanation:

From the question we have the velocity to be v=√(a^2-x^2)/√(x^2) so we can also write it as v = (a^2/x^2 -1)^1/2 again as v = (a^2x^-2 -1)^1/2. So, now we have to find the acceleration.

To get the acceleration we know that differentiation of velocity with respect to time gives us acceleration so on dv/dx we will get that the value of dv/dx to be

(-a^2/vx^3).

So, now to get acceleration we know that dv/dt is acceleration so dv/dx * dx/dt will also give acceleration so on solving we will get the value of acceleration to be (a^2/x^3) which we get that acceleration is proportional to 1/x^3.