Question

If velocity of a moving point is given by the law v^(2)=ax-bx^(3) where v is its velocity and x is its position,prove that 27bv^(4)=4(a-2f)(a+f)^(2) , where a and b are constants and f is the the acceleration​

Answer 1

Explanation:

we know that

rate of change of velocity is acceleration

differentiate the given equation with respect to time

v=(dx/dt); a=(dv/dt)

hence the given equation becomes

2v(dv/dt) =a(dx/dt) - 3bx^2(dx/dt)

2vf=av-3bx^2(v)

v gets cancelled out and equation becomes

2f=a-3bx^2

from this x^2 can be written as

x^2=(a-2f) /3b

from given equation

v^2=ax-bx^3 can also be written as

v^2=x{a-bx^2}

substituting x^2 in above equation

we will get

v^2=2x(a+f) /3

from this x can be written as

x=3v^2/2(a+f)

squaring on both sides

9v^4/4((a+f)^2)=x^2

by substituting x^2 we get

27bv^4=4(a-2f)(a+f) ^2