Question

A particle undergoes one-dimensional motion such that its velocity varies according to (x) = BX-2n, where B and n are positive constants and x is the position of the particle. The ratio of velocity to acceleration (for n = 1) varies with position xas​

Answer 1

Answer:

c

Explanation:

Given,

v(x)=ßx^-2n

a(x)=dv/dt =v.dv/dx

so, dv/dx=-2nßx^-2n-1

v/a = v/v.2nßx^-2n-1

= x^2n+1 (n=1)

=x^3