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 constant ) then the acceleration of the particle is proportional to

Answer 1

answer : acceleration of the particle is proportional to x.

explanation : the speed - position relation of a particle moving in a straight line is given as v² = a - bx²

differentiating with respect to x,

d(v²)/dx = d(a - bx²)/dx

⇒ 2v × dv/dx = 0 - b × d(x²)/dx

⇒ 2v dv/dx = -2bx

⇒ v dv/dx = -bx. ..........(1)

we know, acceleration is the rate of change of velocity with respect to time.

so, a = dv/dt

= dv/dx × dx/dt

= (dx/dt) (dv/dx)

= v dv/dx .....(2)

now from equations (1) and (2),

v dv/dx = a = -bx

⇒ $a\propto x$

hence, magnitude of acceleration is directly proportional to x.