Question

If the speed-position (v-x) 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
1)x^4
2)x
3)x^3
4)x^2
Please explain with a pic, if you can.​

Answer 1

Answer:

2) x

Explanation:

1) We know,

acceleration, A

$A=\frac{dv}{dt}=\frac{dv*dx}{dt*dx}\\ \\ =\frac{dv}{dx}*\frac{dx}{dt}\\ \\ =\frac{dv}{dx}*v=\frac{vdv}{dx}$

2) Now,

Speed -position relation of of a particle in s straight line is given by ,

$v^2=a-bx^2$

Using Implicit differentiation,

$d(v^2)=d(a-bx^2)\\ \\=>2vdv=0-2bxdx\\ \\ =>vdv=-bxdx\\ \\=>\frac{vdv}{dx}=-bx\\ \\=>A=-bx\\ \\=>A \propto x$

Hence, acceleration of a particle is proportional to 'x'.

Option 2) x is correct answer.

Answer 2

Explanation:

refer to image...........