Question

if the speed position relation of a particle moving in a straight line is given as V^2=a-bx^2 (where a and b are constants) then the Acceleration of the particle is proportional to​

Answer 1

Answer ⇒ This means, acceleration is proportional to negative of distance.

Explanation ⇒ Given,

v² = a - bx²

differentiating both sides with respect to t.

2v × dv/dt =   - 2bx

∴ a = -bx/v

∴ a = -bx/√(a - bx²)

This means, acceleration is proportional to negative of distance.

Hop it helps.

Answer 2

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